sample problem solving on force

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High School Physics : Calculating Force

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : net force.

Plug these into the equation to solve for acceleration.

Example Question #2 : Calculating Force

Plug in the values given to us and solve for the force.

Plug in the given values to solve for the mass.

Example Question #1 : Calculating Force

(Assume the only two forces acting on the object are friction and Derek).

Plug in the information we've been given so far to find the force of friction.

Friction will be negative because it acts in the direction opposite to the force of Derek.

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

Example Question #6 : Calculating Force

We can find the net force by adding the individual force together.

Example Question #7 : Calculating Force

If the object has a constant velocity, that means that the net acceleration must be zero.

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

We are given the mass, but we will need to calculate the acceleration to use in the formula.

Plug in our given values and solve for acceleration.

Now we know both the acceleration and the mass, allowing us to solve for the force.

Example Question #9 : Calculating Force

We can calculate the gravitational force using the mass.

Example Question #10 : Calculating Force

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Physics Problems with Solutions

Forces in physics, tutorials and problems with solutions.

Free tutorials on forces with questions and problems with detailed solutions and examples. The concepts of forces, friction forces, action and reaction forces, free body diagrams, tension of string, inclined planes, etc. are discussed and through examples, questions with solutions and clear and self explanatory diagrams. Questions to practice for the SAT Physics test on forces are also included with their detailed solutions. The discussions of applications of forces engineering system are also included.

Forces: Tutorials with Examples and Detailed Solutions

• Forces in Physics .
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• Newton's Laws in Physics .
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Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide

This step-by-step guide is meant to show you how to approach problems where you have to deal with moving objects subject to friction and other forces, and you need to apply Newton's Laws. We will go through many problems, so you can have a clear idea of the process involved in solving them.

The problems we will examine include objects that

• are pushed/pulled horizontally with an angle
• move up or down an incline
• hang from ropes attached to the ceiling
• hang from ropes that run over pulleys
• move connected by a string
• are pushed in contact with each other (Coming soon!)
• Box pulled at an angle over a horizontal surface
• Block pushed over the floor with a downward and forward force
• Object moving at constant velocity over a horizontal surface
• Block pushed up a frictionless ramp
• Mass pulled up an incline with friction
• A mass hanging from two ropes
• Two hanging objects connected by a rope
• Two masses on a pulley
• Two blocks connected by a string are pulled horizontally

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6.2: Solving Problems with Newton's Laws (Part 1)

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Learning Objectives

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion. Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy: Applying Newton’s Laws of Motion

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure \(\PageIndex{1a}\). Then, as in Figure \(\PageIndex{1b}\), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

Figure \(\PageIndex{1}\): (a) A grand piano is being lifted to a second-story apartment. (b) Arrows are used to represent all forces: \(\vec{T}\) is the tension in the rope above the piano, \(\vec{F}_{T}\) is the force that the piano exerts on the rope, and \(\vec{w}\) is the weight of the piano. All other forces, such as the nudge of a breeze, are assumed to be negligible. (c) Suppose we are given the piano’s mass and asked to find the tension in the rope. We then define the system of interest as shown and draw a free-body diagram. Now \(\vec{F}_{T}\) is no longer shown, because it is not a force acting on the system of interest; rather, \(\vec{F}_{T}\) acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that if the piano is stationary, \(\vec{T}\) = \(- \vec{w}\).

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure \(\PageIndex{1c}\).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure \(\PageIndex{1c}\) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure \(\PageIndex{1d}\) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

\[\sum F_{x} = m a_{x}, \quad \sum F_{y} = m a_{y}\ldotp\]

(If, for example, the system is accelerating horizontally, then you can then set ay = 0.) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium.

Example 6.1: Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure \(\PageIndex{2}\). Find the tension in each wire, neglecting the masses of the wires.

Figure \(\PageIndex{2}\): A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light.

The system of interest is the traffic light, and its free-body diagram is shown in Figure \(\PageIndex{2c}\). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in Figure \(\PageIndex{2d}\). There are two unknowns in this problem (T 1 and T 2 ), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

First consider the horizontal or x-axis:

\[F_{net x} = T_{2x} - T_{1x} = 0 \ldotp\]

Thus, as you might expect,

\[T_{1x} = T_{2x} \ldotp\]

This give us the following relationship:

\[T_{1} \cos 30^{o} = T_{2} \cos 45^{o} \ldotp\]

\[T_{2} = 1.225 T_{1} \ldotp\]

Note that T 1 and T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 ends up being greater than T 1 because it is exerted more vertically than T 1 .

Now consider the force components along the vertical or y-axis:

\[F_{net y} = T_{1y} + T_{1x} - w = 0 \ldotp\]

This implies

\[T_{1y} + T_{2y} = w \ldotp\]

Substituting the expressions for the vertical components gives

\[T_{1} \sin 30^{o} + T_{2} \sin 45^{o} = w \ldotp\]

There are two unknowns in this equation, but substituting the expression for T 2 in terms of T 1 reduces this to one equation with one unknown:

\[T_{1} (0.500) + (1.225 T_{1})(0.707) = w = mg,\]

which yields

\[1.366 T_{1} = (15.0\; kg)(9.80\; m/s^{2}) \ldotp\]

Solving this last equation gives the magnitude of T 1 to be

\[T_{1} = 108\; N \ldotp\]

Finally, we find the magnitude of T 2 by using the relationship between them, T 2 = 1.225 T 1 , found above. Thus we obtain

\[T_{2} = 132\; N \ldotp\]

Significance

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Example 6.2: Drag Force on a Barge

Two tugboats push on a barge at different angles (Figure \(\PageIndex{3}\)). The first tugboat exerts a force of 2.7 x 10 5 N in the x-direction, and the second tugboat exerts a force of 3.6 x 10 5 N in the y-direction. The mass of the barge is 5.0 × 106 kg and its acceleration is observed to be 7.5 x 10 −2 m/s 2 in the direction shown. What is the drag force of the water on the barge resisting the motion? ( Note : Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.)

Figure \(\PageIndex{3}\): (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces—the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Note that \(\vec{F}_{app}\) is the total applied force of the tugboats.

The directions and magnitudes of acceleration and the applied forces are given in Figure \(\PageIndex{3a}\). We define the total force of the tugboats on the barge as \(\vec{F}_{app}\) so that

\[\vec{F}_{app} = \vec{F}_{1} + \vec{F}_{2} \ldotp\]

The drag of the water \(\vec{F}_{D}\) is in the direction opposite to the direction of motion of the boat; this force thus works against \(\vec{F}_{app}\), as shown in the free-body diagram in Figure \(\PageIndex{3b}\). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x- and y-axes are in the same direction as \(\vec{F}_{1}\) and \(\vec{F}_{2}\). The problem quickly becomes a one-dimensional problem along the direction of \(\vec{F}_{app}\), since friction is in the direction opposite to \(\vec{F}_{app}\). Our strategy is to find the magnitude and direction of the net applied force \(\vec{F}_{app}\) and then apply Newton’s second law to solve for the drag force \(\vec{F}_{D}\).

Since F x and F y are perpendicular, we can find the magnitude and direction of \(\vec{F}_{app}\) directly. First, the resultant magnitude is given by the Pythagorean theorem:

\[ \vec{F}_{app} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \sqrt{(2.7 \times 10^{5}\; N)^{2} + (3.6 \times 10^{5}\; N)^{2}} = 4.5 \times 10^{5} \; N \ldotp\]

The angle is given by

\[\theta = \tan^{-1} \left(\dfrac{F_{2}}{F_{1}}\right) = \tan^{-1} \left(\dfrac{3.6 \times 10^{5}\; N}{2.7 \times 10^{5}\; N}\right) = 53.1^{o} \ldotp\]

From Newton’s first law, we know this is the same direction as the acceleration. We also know that \(\vec{F}_{D}\) is in the opposite direction of \(\vec{F}_{app}\), since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as \(\vec{F}_{app}\), but its magnitude is slightly less than \(\vec{F}_{app}\). The problem is now one-dimensional. From the free-body diagram, we can see that

\[F_{net} = F_{app} - F_{D} \ldotp\]

However, Newton's second law states that

\[F_{net} = ma \ldotp\]

\[F_{app} - F_{D} = ma \ldotp\]

This can be solved for the magnitude of the drag force of the water F D in terms of known quantities:

\[F_{D} = F_{app} - ma \ldotp\]

Substituting known values gives

\[F_{D} = (4.5 \times 10^{5}\; N) - (5.0 \times 10^{6}\; kg)(7.5 \times 10^{-2}\; m/s^{2}) = 7.5 \times 10^{4}\; N \ldotp\]

The direction of \(\vec{F}_{D}\) has already been determined to be in the direction opposite to \(\vec{F}_{app}\), or at an angle of 53° south of west.

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where F D is less than 1/600th of the weight of the ship.

In Newton’s Laws of Motion , we discussed the normal force, which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride?

Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3: What does the Bathroom Scale Read in an Elevator?

Figure \(\PageIndex{4}\) shows a 75.0-kg man (weight of about 165 lb.) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of 1.20 m/s 2 , and (b) if the elevator moves upward at a constant speed of 1 m/s.

Figure \(\PageIndex{4}\): (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is accelerating upward—broken arrows represent forces too large to be drawn to scale. \(\vec{T}\) is the tension in the supporting cable, \(\vec{w}\) is the weight of the person, \(\vec{w}_{s}\) is the weight of the scale, \(\vec{w}_{e}\) is the weight of the elevator, \(\vec{F}_{s}\) is the force of the scale on the person, \(\vec{F}_{p}\) is the force of the person on the scale, \(\vec{F}_{t}\) is the force of the scale on the floor of the elevator, and \(\vec{N}\) is the force of the floor upward on the scale. (b) The free-body diagram shows only the external forces acting on the designated system of interest—the person—and is the diagram we use for the solution of the problem.

If the scale at rest is accurate, its reading equals \(\vec{F}_{p}\), the magnitude of the force the person exerts downward on it. Figure \(\PageIndex{4a}\) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn, as in Figure \(\PageIndex{4b}\). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure \(\PageIndex{4a}\) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight \(\vec{w}\) and the upward force of the scale \(\vec{F}_{s}\). According to Newton’s third law, \(\vec{F}_{p}\) and \(\vec{F}_{s}\) are equal in magnitude and opposite in direction, so that we need to find F s in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

\[\vec{F}_{net} = m \vec{a} \ldotp\]

From the free-body diagram, we see that \(\vec{F}_{net} = \vec{F}_{s} - \vec{w}\), so we have

\[F_{s} - w = ma \ldotp\]

Solving for F s gives us an equation with only one unknown:

\[F_{s} = ma + w,\]

or, because w = mg, simply

\[F_{s} = ma + mg \ldotp\]

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note : We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = −ma.)

• We have a = 1.20 m/s 2 , so that $$F_{s} = (75.0\; kg)(9.80\; m/s^{2}) + (75.0\; kg)(1.20\; m/s^{2})$$yielding $$F_{s} = 825\; N \ldotp$$
• Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because \(a = \frac{\Delta v}{\Delta t}\) and \(\Delta v = 0\). Thus, $$F_{s} = ma + mg = 0 + mg$$or $$F_{s} = (75.0\; kg)(9.80\; m/s^{2}),$$which gives $$F_{s} = 735\; N \ldotp$$

The scale reading in Figure \(\PageIndex{4a}\) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

\[F_{net} = ma = 0 = F_{s} − w\]

\[F_{s} = w = mg\]

\[F_{s} = (75.0\; kg)(9.80\; m/s^{2}) = 735\; N \ldotp\]

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward.

Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure \(\PageIndex{4b}\), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Exercise 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g, then the scale reading is zero and the person appears to be weightless.

Example 6.4: Two Attached Blocks

Figure \(\PageIndex{5}\) shows a block of mass m 1 on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass m 2 . Find the acceleration of the blocks and the tension in the string in terms of m 1 , m 2 , and g.

Figure \(\PageIndex{5}\): (a) Block 1 is connected by a light string to block 2. (b) The free-body diagrams of the blocks.

We draw a free-body diagram for each mass separately, as shown in Figure \(\PageIndex{5}\). Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

For block 1: \(\vec{T} + \vec{w}_{1} + \vec{N} = m_{1} \vec{a}_{1}\)

For block 2: \(\vec{T} + \vec{w}_{2} = m_{2} \vec{a}_{2}\).

Notice that \(\vec{T}\) is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects.

The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x-components. There are no horizontal forces on block 2, so only the y-equation is written. We obtain these results:

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1x = −a 2y . Writing the common acceleration of the blocks as a = a 1x = −a 2y , we now have

\[T = m_{1}a\]

\[T − m_{2}g = −m_{2}a \ldotp\]

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g:

\[a = \frac{m_{2}}{m_{1} + m_{2}}g\]

\[T = \frac{m_{1} m_{2}}{m_{1} + m_{2}} g \ldotp\]

Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set T = m 2 g. You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg and m 2 = 3.00 kg.

Example 6.5: Atwood Machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine, which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure \(\PageIndex{6}\), m 1 = 2.00 kg and m 2 = 4.00 kg. Consider the pulley to be frictionless. (a) If m 2 is released, what will its acceleration be? (b) What is the tension in the string?

Figure \(\PageIndex{6}\): An Atwood machine and free-body diagrams for each of the two blocks.

We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration a 2 in the downward direction, block 1 accelerates upward with acceleration a 1 . Thus, a = a 1 = −a 2 .

• We have $$For\; m_{1}, \sum F_{y} = T − m_{1}g = m_{1}a \ldotp \quad For\; m_{2}, \sum F_{y} = T − m_{2}g = −m_{2}a \ldotp$$(The negative sign in front of m 2 a indicates that m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is $$(m_{2} - m_{1})g = (m_{1} + m_{2})a \ldotp$$Solving for a: $$a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}}g = \frac{4\; kg - 2\; kg}{4\; kg + 2\; kg} (9.8\; m/s^{2}) = 3.27\; m/s^{2} \ldotp$$
• Observing the first block, we see that $$T − m_{1}g = m_{1}a$$ $$T = m_{1}(g + a) = (2\; kg)(9.8\; m/s^{2} + 3.27\; m/s^{2}) = 26.1\; N \ldotp$$

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, (m 2 − m 1 )g, to the total mass of the system, m 1 + m 2 . We can also use the Atwood machine to measure local gravitational field strength.

Exercise 6.3

Determine a general formula in terms of m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

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Simplifying a Difficult Problem

Consider the situation below in which a force is directed at an angle to the horizontal. In such a situation, the applied force could be resolved into two components. These two components can be considered to replace the applied force at an angle. By doing so, the situation simplifies into a familiar situation in which all the forces are directed horizontally and vertically.

Once the situation has been simplified, the problem can be solved like any other problem. The task of determining the acceleration involves first determining the net force by adding up all the forces as vectors and then dividing the net force by the mass to determine the acceleration. In the above situation, the vertical forces are balanced (i.e., F grav , F y , and F norm add up to 0 N), and the horizontal forces add up to 29.3 N, right (i.e., 69.3 N, right + 40 N, left = 29.3 N, right). The net force is 29.3 N, right and the mass is 10 kg (m = F grav /g); therefore, the acceleration is 2.93 m/s/s, right.

Your Turn to Practice

To test your understanding, analyze the two situations below to determine the net force and the acceleration. When finished, click the button to view the answers.

The net force is 69.9 N, right and the acceleration is 3.5 m/s/s, right .

Note that the vertical forces balance but the horizontal forces do not. The net force is

F net = 129.9 N, right - 60 N, left = 69.9 N, right

The mass is

m = (F grav / g) = 20 kg

So the acceleration is

a = (69.9 N) / (20 kg) =3.50 m/s/s.

The net force is 30.7 N, right and the acceleration is 1.23 m/s/s, right .

F net = 70.7 N, right - 40 N, left = 30.7 N, right

m = (F grav / g) = 25 kg

a = (30.7 N) / (25 kg) =1.23 m/s/s.

What's Up with the Normal Force?

There is one peculiarity about these types of problems that you need to be aware of. The normal force (F norm ) is not necessarily equal to the gravitational force (F grav ) as it has been in problems that we have previously seen. The principle is that the vertical forces must balance if there is no vertical acceleration. If an object is being dragged across a horizontal surface, then there is no vertical acceleration. For this reason, the normal force (F norm ) plus the vertical component (F y ) of the applied force must balance the gravitational force (F grav ). A quick review of these problems shows that this is the case. If there is an acceleration for an object being pulled across a floor, then it is a horizontal acceleration; and thus the only imbalance of force would be in the horizontal direction .

Now consider the following situation in which a force analysis must be conducted to fill in all the blanks and to determine the net force and acceleration. In a case such as this, a thorough understanding of the relationships between the various quantities must be fully understood. Make an effort to solve this problem. When finished, click the button to view the answers. (When you run into difficulties, consult the help from a previous unit .)

 The F grav is

F grav = m • g = (10 kg) • (9.8 m/s/s) = 98 N

Using the sine function,

F y = (60 N) • sine (30 degrees) = 30 N

Since vertical forces are balanced, F norm = 68 N.

Now F frict can be found

F frict = mu • F norm = ( 0.3) • (68 N) = 20.4 N

Using the cosine function,

F x = (60 N) • cosine (30 degrees) = 52.0 N

Now since all the individual force values are known, the F net can be found:

F net = 52.0 N,right + 20.4 N, left = 31.6 N, right.

The acceleration is

a = 3.16 m/s/ s, right.

In conclusion, a situation involving a force at an angle can be simplified by using trigonometric relations to resolve that force into two components. Such a situation can be analyzed like any other situation involving individual forces. The net force can be determined by adding all the forces as vectors and the acceleration can be determined as the ratio of Fnet/mass.

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Check your understanding.

The following problems provide plenty of practice with F net = m • a problems involving forces at angles. Try each problem and then click the button to view the answers.

Glen Brook and Warren Peace are incorrect. Warren Peace perhaps believes that the Fnorm = Fgrav; but this is only the case when there are only two vertical forces and no vertical acceleration; sorry Warren - there is a second vertical force in this problem (F app ).

Glen Brook perhaps thinks that the F app force is 50 N upwards and thus the F norm must be 50 N upwards to balance the Fgrav. Sorry Glen - the F app is only 25 N upwards (50 N) • sine 30 degrees).

"Datagal Olive!"

2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks.

First use the mass to determine the force of gravity.

F grav = m • g = (20 kg) • (9.8 m/s/s) = 196 N

Now find the vertical component of the applied force using a trigonometric function.

F y = (80 N) • sine (45 degrees) = 56.7 N

Thus, F norm =139.3 N in order for the vertical forces to balance.

The horizontal component of the applied force can be found as

F x = (80 N) • cosine (45 degrees) = 56.7 N

Since the speed is constant, the horizontal forces must also balance; and so F frict = 56.7 N.

The value of "mu" can be found using the equation

3. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Answer: "mu" = 0.25

 The F grav can be calculated from the mass of the object.

The vertical component of the applied force can be calculated using a trigonometric function:

F y = (80 N) • sine (30 degrees) = 40 N

In order for the vertical forces to balance, F norm + F y = F grav . Thus,

F norm = F grav - F y = = 196 N - 40 N = 156 N

The horizontal component of the applied force can be calculated using a trigonometric function:

F x = (80 N) • cosine (30 degrees) = 69.2 N

The net force is the sum of all the forces when added as vectors. Thus,

F net = (69.2 N, right) + (40 N,left) = 29.2 N, right

a = F net / m = (29.2 N, right) / (20 kg) = 1.46 m/s/s, right.

The value of "mu" can be found using the equation "mu" = F frict / F norm .

4. The 5-kg mass below is moving with a constant speed of 4 m/s to the right. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The F grav can be calculated from the mass of the object.

F grav = m • g = (5 kg) • (9.8 m/s/s) = 49 N

The vertical component of the applied force can be calculated using a trigonometric function.

F y = (15 N) • sine (45 degrees) =10.6 N

F norm = F grav - F y = = 49 N - 10.6 N = 38.4 N

F x = (15 N) • cosine (45 degrees) = 10.6 N

Since the speed is constant, the horizontal forces must balance. Therefore, F frict = 10.6 N.

The value of "mu" can be found using the equation "mu"= F frict / F norm :

"mu" = 0.276

5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Since the velocity is constant, the acceleration and the net force are 0 m/s/s and 0 N respectively. The F grav can be calculated from the mass of the object.

The object moves at constant speed; thus, the horizontal forces must balance. For this reason, F x = 10 N.

The applied force can now be found using a trigonometric function and the horizontal component:

cosine (60 degrees) = (10 N) / (F app )

Proper algebra yields

F app = (10 N) / [cosine (60 degrees) ] = 20 N

F y = (20 N) • sine (60 degrees) = 17.3 N

F norm = F grav - F y = 49 N - 17.3 N = 31.7 N

6. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +2 m/s/s.

The net force can be calculated as:

F net = m • a = (10 kg) • (2 m/s/s) = 20 N, right.

F x = (70 N) • cosine (45 degrees) = 49.5 N

The net force is the vector sum of all the forces. Thus, F net = F x + F frict . That is,

20 N, right = 49.5 N, right + F frict

Therefore, F frict must be 29.5 N, left.

F y = (70 N) • sine (45 degrees) = 49.5 N

F norm = F grav - F y = = 98 N - 49.5 N = 48.5 N

The value of "mu" can be found using the equation "mu" = F frict / F norm :

7. Study the diagram below and determine the acceleration of the box and its velocity after being pulled by the applied force for 2.0 seconds.

F y = (50 N) • sine (45 degrees) = 35.4 N

In order for the vertical forces to balance, F norm + F y = F grav . Algebraic rearrangement leads to:

F norm = F grav - F y = 196 N - 35.4 N = 160.6 N

F x = (50 N) • cosine (45 degrees) = 35.4 N

The F net is 35.4 N, right since the only force which is not balanced is F x .

The acceleration is:

a = F net / m = (35.4 N) / (20 kg) = 1.77 m/s/s

The velocity after 2.0 seconds can be calculated using a kinematic equation:

v f = v i + a • t = 0 m/s + (1.77 m/s/s) • (2.0 s) v f = 3.54 m/s

8. A student pulls a 2-kg backpack across the ice (assume friction-free) by pulling at a 30-degree angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force.

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +0.125 m/s/s.

F net = m • a = (2 kg) • (0.125 m/s/s) = 0.250 N, right

Since the acceleration is horizontal, the vertical forces balance each other. The horizontal component of the applied force (F x ) supplies the horizontal force required for the acceleration. Thus, the horizontal component of the applied force is 0.250 N.

Using trigonometry, the applied force (F app ) can be calculated:

cosine (30 degrees) = ( 0.250 N) / (F app )

Algebraic rearrangement of this equation leads to:

F app = 0.289 N

9. The following object is moving to the right and encountering the following forces. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F grav = m • g = (10 kg) • (10 m/s/s) = 100 N

F norm = F grav - F y = 100 N - 35.4 N = 64.6 N

This x-component of the applied force (F x ) is directed leftward. This horizontal component of force is not counteracted by a rightward force. For this reason, the net force is 35.4 N.  Knowing F net , allows us to determine the acceleration of the object:

a = F net / m = (35.4 N) / (10 kg) = ~3.5 m/s/s

The acceleration of an object is the velocity change per time. For an acceleration of 3.5 m/s/s, the velocity change should be 3.5 m/s for each second of time change. In the velocity-time table, the velocity is decreasing by 3.5 m/s each second. Thus, the values should read 17.5 m/s, 14.0 m/s, 10.5 m/s, 7.0 m/s, 3.5 m/s, 0 m/s.

10. The 10-kg object is being pulled to the left at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = F grav - F norm = 98 N - 80 N = 18 N

The applied force can be determined using a trigonometric function:

sine 30 (degrees) = (18 N) / F app

Algebraic rearrangement leads to:

F app = (18 N) / [ sine (30 degrees) ] = 36 N

Similar trigonometry allows one to determine the x-component of the applied force:

F x = (36 N) • cosine(30 degrees) = 31.2 N

Since the speed is constant, the horizontal forces must also balance. Thus the force of friction is equal to the F x value. F frict = 31.2 N

The value of "mu" can be found using the equation "mu" = F frict / F norm

11. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = (100 N) • sine (45 degrees) = 70.7 N

F norm = F grav - F y = 98 N - 70.7 N = 27.3 N

F x = (100 N) • cosine (45 degrees) = 70. N

The net force is the vector sum of all the forces.

F net = 50 N, right + 70.7 N, left = 20.7 N, left

The acceleration is can be found from a = F net / m :

StickMan Physics

Animated Physics Lessons

F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

                a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

               a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

            a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

             a= 4.47 m/s 2

5b. What direction would this box accelerate?

            63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

            6300 kg Elephant

            The more mass the more inertia

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

            F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

            F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

            a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

            m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

            F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

                F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

            0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

            F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

            x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

            v f = 8.58 m/s

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• Newton's Law of Motion

Newton's Third Law of Motion: Solved Problems

In this article, you learn the concepts of Newton's third law of motion by problem-solving approach. Here, we are going to find pairs of forces appearing in Newton's third law using some worked problems. 

Introduction

Consider the following everyday experiment: Imagine applying a force with a hammer to drive a nail into a wooden block. As a result, the hammer comes to a complete stop. 

What forces caused the hammer's speed to be reduced to zero? 

This is the question that Newton speculated on. 

According to Newton's idea, the nail exerts a force of the same magnitude but in the opposite direction on the hammer, causing it to rapidly come to a complete stop upon contact. 

This is the essence of Newton's third law, which states that if two objects 1 and 2 interact, the force $\vec{F}_{12}$ that object 1 exerts on object 2 is equal in magnitude but opposite in direction to the force $\vec{F}_{21}$ that object 2 exerts on object 1. \[\vec{F}_{12}=-\vec{F}_{21}\] One of the most important consequences of this law is that the forces cannot exist in isolation and always occur in pairs. 

Solved Problems:

Problem (1):  A box stands on a table. Identify all action-reaction forces between the box and the table. 

Solution : If object 1 exerts a force of $\vec{F}_{12}$ on object 2, object 2 also exerts a force $\vec{F}_{21}$ on object 1. This force is equal in magnitude and opposite in direction and can be written as $\vec{F}_{12}=-\vec{F}_{21}$. This simple relation represents Newton's third law of motion.

In this definition, one of the forces, regardless of which one, is referred to as the action force, while the other force is known as the reaction force.

When dealing with problems involving Newton's third law, the main objective is to identify pairs of forces that comply with this law.

The box sitting on a chair does not fall through due to its weight $\vec{w}$. An upward force called the normal force $\vec{n}$ is exerted on the box to balance out the downward weight force and keep the box at rest on the table. These two forces are actual external forces.

We consider this normal force as the action force. The reaction to $\vec{n}$ is the force exerted by the box on the table, denoted as $\vec{n}'$. This reaction force has equal magnitude but opposite direction to $\vec{n}$.

Thus, we can express this pair of forces as $\vec{n}=-\vec{n}'$. Since these forces are exerted on two different objects, they can be considered an action-reaction pair according to Newton's third law.

Problem (2): Identify the action-reaction force pair in each of the following situations: (a) An apple falls freely.  (b) A hammer hits a nail. (c) The blades of a helicopter push air downward.

Solution : Newton's third law states that when two objects interact, the force that object 1 applies on object 2, called $\vec{F}_{12}$ is equal in magnitude but in the opposite direction to the force that object 2 exerts on object 1, i.e., $\vec{F}_{21}$. \[\vec{F}_{12}=-\vec{F}_{21}\] In all Newton's third law problems, first of all, identify the system under study. 

(a) Here, the apple and the Earth create our system. The Earth applies a downward force on the apple, known as weight $\vec{w}$, and correspondingly, the apple exerts a force on the Earth by $\vec{w}'$. These two are third law force pairs. \[\vec{w}=-\vec{w}'\]  (b) In this case, our system is hammer-nail, two different interacting bodies. The force exerted by the hammer on the nail $\vec{F}_{hn}$ is equal in magnitude but in the opposite direction to the force exerted by the nail on the hammer, $\vec{F}_{nh}$. \[\vec{F}_{hn}=-\vec{F}_{hn}\]  (c) The blades and air form a system as a combined two interacting objects. In this case, the blades exert a force downward on the air and conversely, the air also exerts a force upward on the blades. These two forces act on different objects, not the same object, so it can be considered as Newton's third law pair force. 

Problem (3): A car is skidding to a complete stop on a horizontal road. First, identify all action-reaction pairs of forces, then show them on a free-body diagram.

Solution : The car slows down and eventually comes to a complete stop, experiencing deceleration. According to Newton's third law, action-reaction pairs of forces act between two interacting objects. In this case, the car and the road (Earth) form our system. The gravitational force on the car due to Earth is represented by $\vec{w}$, while the reaction force exerted by the car on Earth is represented by $\vec{w}'$. Therefore, these two forces act on different objects. \[\vec{w}=-\vec{w}'\] When the car is in contact with the road, an upward force is exerted by the road's surface on the car, known as the normal force ($\vec{n}$). Conversely, the car also applies a downward force on the road with the same magnitude but in the opposite direction ($-\vec{n}'$). These two forces form another pair of action-reaction forces according to Newton's third law.

Problem (4): A girl exerts a force of 40 N upward to hold a box. Identify the reaction force by determining (a) its magnitude and direction, (b) on what object it is exerted, and (c) by what object it is exerted. 

Solution : Here, three objects are interacting, the girl, the box, and the Earth. The Earth exerts a gravitational force $\vec{F}_g$ on the box. The reaction to this force is the force exerted by the box on the Earth, $\vec{F}'_g$. These two forces are acting on two different objects and are equal in magnitude and opposite in direction, so they form an action-reaction pair force. \[\vec{F}_g=-\vec{F}'_g\] On the other hand, the box is held by the girl. The girl exerts an upward force to hold it up and the reaction to this force is the force exerted by the box downward on the girl's hand. 

Problem (5): A girl stands on a spring scale while riding in an elevator accelerating at $2\,\rm m/s^2$ upward. How does Newton's third law apply to find the scale reading? 

Solution : The scale reads the value of the downward force that the girl exerts on the spring scale. By Newton's third law, the reaction to this force is the upward force exerted by the scale on the woman, known as the normal force $\vec{n}$. These two forces are depicted in the following figure. 

Applying Newton's second law gives us the magnitude of the apparent weight that the girl feels while riding in a moving elevator. To see how to solve such elevator problems, read this. 

Problem (6): A box is pushing on a rough horizontal surface at constant velocity with a force of $\vec{F}$. The friction is also applied by $-\vec{F}$. Are the pushing force and friction the action-reaction pair of forces?

Solution : It is important to remember that according to Newton's third law, force pairs appear between two interacting bodies. In this scenario, three objects are interacting: the person applying a push on the box, the box itself, and the surface over which the box is moving. 

In the person-box system, an external force $\vec{F}$ is applied to the box by the person. As a reaction to this force, the box exerts an equal and opposite force $-\vec{F}$ on the person. 

In the box-surface (Earth) system, friction $-\vec{F}$ is exerted on the box by the rough surface. Conversely, as a reaction to this frictional force, an equal and opposite force $\vec{F}$ is exerted by the box on the surface. 

As you can see, both the push force and friction are applied separately to the same object (the box). Therefore, they cannot be considered as third law pair forces.

Problem (7): A person is standing motionless on the level ground.  (a) Identify all forces acting on the person. (Hewit 49, p.120) (b) Are these forces equal in magnitude and opposite? (c) Do these forces make an action-reaction force pair? Why or why not? (d) Determine all other pairs of forces not shown here.

Solution : (a) The Earth pulls down the person by the weight force, $\vec{w}$ and the surface prevents the person from sinking into the Earth by the upward support force, called normal force $\vec{n}$.  (b) The person is motionless on the level ground, so by the second law, the net force on it must be zero which yields \begin{gather*} F_{net}=ma\\ \vec{w}+\vec{n}=0 \\ \Rightarrow \boxed{\vec{w}=-\vec{n}}\end{gather*} This expression tells us that these two above forces are equal in magnitude and opposite in direction.  (c) Every two equal and opposite forces do not constitute an action-reaction pair of forces. Newton's third law states that these forces must be applied to two different objects.  In this problem, the Earth exerts a downward force (weight) on the person, and the person also exerts an upward force on the Earth with the same magnitude.  Additionally, the Earth's surface applies an upward force (normal force) on the person, and the person exerts an equal and opposite force on the Earth. As you can see, both weight and normal forces are exerted separately between the Earth and the person and do not interact with each other. Therefore, these forces do not form a pair according to Newton's third law.

(d) In the previous section, we mentioned that the weight and normal forces are actual external forces and not paired forces. In this context, the downward weight acts as an action, with its reaction being the force that pulls the Earth towards the person. Similarly, the upward normal force acts as an action, with its reaction being the downward force exerted by the person on the surface.

Problem (8):  An astronaut performing a spacewalk with a combined mass of 165 kg applies a force of 280 N to a satellite that is freely floating in space and has a mass of 850 kg. Determine the magnitude and direction of the reaction force exerted by the satellite on the astronaut. Additionally, calculate the accelerations experienced by both the astronaut and the satellite.

Solution : The satellite and the astronaut are two distinct objects that interact with each other by the forces that make the action-reaction pair force, according to Newton's third law of motion. 

(a) The astronaut applies a force $\vec{F}_{as}$ to the satellite and, by Newton's third law, the satellite also exerts an equal force in the opposite direction on the astronaut as $-\vec{F}_{sa}$. Thus, the satellite exerts a $280\,\rm N$ force on the astronaut in the negative $x$-direction or in vector notation as \[\vec{F}_{sa}=-280\,\rm N\]  (b) The second law gives us the acceleration of an object when a force $F$ is applied to it as $\vec{a}=\frac{\vec{F}}{m}$. Therefore, the astronaut's acceleration is \[a=\frac{-280}{165}=-1.70\,\rm m/s^2\] and for the satellite can be found as \[a=\frac{280}{850}=0.33\,\rm m/s^2\] The negative indicates the direction of the acceleration, which is toward the negative $x$-axis. 

Author : Dr. Ali Nemati Published : August 27, 2023

© 2015 All rights reserved. by Physexams.com

Problems & Exercises

1.2 physical quantities and units.

The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in meters per second? (b) How many miles per hour is this?

A car is traveling at a speed of 33 m/s 33 m/s size 12{"33"" m/s"} {} . (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h 90 km/h size 12{"90"" km/h"} {} speed limit?

Show that 1 . 0 m/s = 3 . 6 km/h 1 . 0 m/s = 3 . 6 km/h size 12{1 "." 0`"m/s"=3 "." "6 km/h"} {} . Hint: Show the explicit steps involved in converting 1 . 0 m/s = 3 . 6 km/h. 1 . 0 m/s = 3 . 6 km/h. size 12{1 "." 0`"m/s"=3 "." "6 km/h"} {}

American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)

What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

The speed of sound is measured to be 342 m/s 342 m/s size 12{"342"" m/s"} {} on a certain day. What is this in km/h?

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1 s at this speed? (b) What is its speed in kilometers per million years?

(a) Refer to Table 1.3 to determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilometers per second. (b) What is this in meters per second?

1.3 Accuracy, Precision, and Significant Figures

Express your answers to problems in this section to the correct number of significant figures and proper units.

Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty. What is the uncertainty in your mass (in kilograms)?

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

(a) A car speedometer has a 5.0 % 5.0 % size 12{5.0%} {} uncertainty. What is the range of possible speeds when it reads 90 km/h 90 km/h size 12{"90"" km/h"} {} ? (b) Convert this range to miles per hour. 1 km = 0.6214 mi 1 km = 0.6214 mi size 12{"1 km" "=" "0.6214 mi"} {}

An infant’s pulse rate is measured to be 130 ± 5 130 ± 5 size 12{"130" +- 5} {} beats/min. What is the percent uncertainty in this measurement?

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y?

A can contains 375 mL of soda. How much is left after 308 mL is removed?

State how many significant figures are proper in the results of the following calculations: (a) 106 . 7 98 . 2 / 46 . 210 1 . 01 106 . 7 98 . 2 / 46 . 210 1 . 01 size 12{ left ("106" "." 7 right ) left ("98" "." 2 right )/ left ("46" "." "210" right ) left (1 "." "01" right )} {} (b) 18 . 7 2 18 . 7 2 size 12{ left ("18" "." 7 right ) rSup { size 8{2} } } {} (c) 1 . 60 × 10 − 19 3712 1 . 60 × 10 − 19 3712 size 12{ left (1 "." "60" times "10" rSup { size 8{ - "19"} } right ) left ("3712" right )} {} .

(a) How many significant figures are in the numbers 99 and 100? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

(a) If your speedometer has an uncertainty of 2 . 0 km/h 2 . 0 km/h size 12{2 "." 0" km/h"} {} at a speed of 90 km/h 90 km/h size 12{"90"" km/h"} {} , what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h 60 km/h size 12{"60"" km/h"} {} , what is the range of speeds you could be going?

(a) A person’s blood pressure is measured to be 120 ± 2 mm Hg 120 ± 2 mm Hg size 12{"120" +- 2" mm Hg"} {} . What is its percent uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg 80 mm Hg size 12{"80"" mm Hg"} {} ?

A person measures his or her heart rate by counting the number of beats in 30 s 30 s size 12{"30"" s"} {} . If 40 ± 1 40 ± 1 size 12{"40" +- 1} {} beats are counted in 30 . 0 ± 0 . 5 s 30 . 0 ± 0 . 5 s size 12{"30" "." 0 +- 0 "." 5" s"} {} , what is the heart rate and its uncertainty in beats per minute?

What is the area of a circle 3 . 102 cm 3 . 102 cm size 12{3 "." "102"" cm"} {} in diameter?

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

A marathon runner completes a 42 . 188 -km 42 . 188 -km size 12{"42" "." "188""-km"} {} course in 2 h 2 h size 12{2" h"} {} , 30 min, and 12 s 12 s size 12{"12"" s"} {} . There is an uncertainty of 25 m 25 m size 12{"25"" m"} {} in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?

The sides of a small rectangular box are measured to be 1 . 80 ± 0 . 01 cm 1 . 80 ± 0 . 01 cm size 12{1 "." "80" +- 0 "." "01"" cm"} {} , {} 2 . 05 ± 0 . 02 cm, and 3 . 1 ± 0 . 1 cm 2 . 05 ± 0 . 02 cm, and 3 . 1 ± 0 . 1 cm size 12{2 "." "05" +- 0 "." "02"" cm, and 3" "." 1 +- 0 "." "1 cm"} {} long. Calculate its volume and uncertainty in cubic centimeters.

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm = 0 . 4539 kg 1 lbm = 0 . 4539 kg size 12{1" lbm"=0 "." "4539"`"kg"} {} . (a) If there is an uncertainty of 0 . 0001 kg 0 . 0001 kg size 12{0 "." "0001"`"kg"} {} in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?

The length and width of a rectangular room are measured to be 3 . 955 ± 0 . 005 m 3 . 955 ± 0 . 005 m size 12{3 "." "955" +- 0 "." "005"" m"} {} and 3 . 050 ± 0 . 005 m 3 . 050 ± 0 . 005 m size 12{3 "." "050" +- 0 "." "005"" m"} {} . Calculate the area of the room and its uncertainty in square meters.

A car engine moves a piston with a circular cross section of 7 . 500 ± 0 . 002 cm 7 . 500 ± 0 . 002 cm size 12{7 "." "500" +- 0 "." "002"`"cm"} {} diameter a distance of 3 . 250 ± 0 . 001 cm 3 . 250 ± 0 . 001 cm size 12{3 "." "250" +- 0 "." "001"`"cm"} {} to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this volume.

1.4 Approximation

How many heartbeats are there in a lifetime?

A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?

How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10 − 22  s 10 − 22  s size 12{"10" rSup { size 8{ - "22"} } " s"} {} .)

Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10 − 27  kg 10 − 27  kg size 12{"10" rSup { size 8{ - "27"} } " kg"} {} and the mass of a bacterium is on the order of 10 − 15  kg. 10 − 15  kg. size 12{"10" rSup { size 8{ - "15"} } "kg"} {} )

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?

(a) What fraction of Earth’s diameter is the greatest ocean depth? (b) The greatest mountain height?

(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human?

Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?

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NAPS Intermediate Physics

12. solving force problems ¶, 12.1. overview ¶.

We have started with the basic definitions of momentum, and used it in Newton’s laws of motion to consider the collision of objects. Hopefully, this gives you a sense of some of the important aspects of forces – forces are due to an interaction between two objects, and over a time interval will change the momentum of each of the objects.

However, there are many more forces that we have not considered. I am sure that, after a moment’s thought, you can make a long list of forces acting on objects. A few forces we will discuss over the coming lessons are friction, buoyancy, and gravity. Because these forces can have different properties, it is important to develop a standard way of combining these forces together to find the motion of the object they act on. This recipe is what is covered in this lesson. In particular, we will look at free body diagrams as a visual way of finding the correct Newton’s 2nd law equation of a particular system.

Newton’s second law is a method

There is a standard method for solving force problems, but there is not a standard equation . Every time you have forces involved in a situation, you should start with a free body diagram and Newton’s 2nd law. But the equation you derive from Newton’s 2nd law will be different in each case! This is because the forces may be different, or the directions they act in may change. You will have to think through what is affecting the system, but always using the force equations to do so. This lesson will help to build that intuition within you.

Here are the objectives for this lesson:

Describe how to draw the free body diagram for an object with one or more forces acting on it.

Calculate the acceleration or final velocity of an object after it experiences a force over a given time interval.

Given the gravitational field at an object’s location, state the gravitational force on that object.

State the direction of the tension force acting on an object.

Define the normal force.

Describe the difference between the static frictional force and the kinetic frictional force.

State the relationship between the frictional force and the normal force.

State the maximum value of the static frictional force.

Given an object moving along a surface with friction, calculate the acceleration of the object.

12.2. Free body diagrams and Newton’s 2nd law ¶

One of the convenient facts about the Universe is that you can solve for the overall motion of a complicated system by considering only a few pieces of information. We have seen this before in previous lessons – for example, how a system of multiple objects can be thought of as a single point, moving along with the center of mass velocity of the system. We will build on this when considering forces.

Since we can simplify things quite a bit, it makes sense to represent the situation for a system by drawing only those pieces of information that are required. This leads to the idea of a free body diagram. A free body diagram (abbreviated FBD) is a picture of the forces acting on a system, a way of succinctly summarizing the forces acting on a system, a single object, or at a single point. We will use these as a thinking aid when solving the equations of Newton’s 2nd law.

Because we are eliminating any non-essential information in a free body diagram, only the force vectors, and the angles needed to find their components, will be included. This allows for the possibility that a lot of forces are acting on the system at once.

Guidelines for drawing an FBD

Draw the object as a dot; if desired, draw \(x\) and \(y\) axes with the object at the origin

Draw all forces ( not their components!) as arrows

Forces always point away from the object

Draw appropriate angles for forces next to the dot

Be careful of direction of forces

Label forces and angles symbolically (no numbers)

Do not include other vectors

No acceleration

No velocity

No net force

In Lesson 19 (work and energy) , we will add the displacement vector to FBDs, to help solve for work

Below is an example of a physical situation, and its corresponding FBD.

Fig. 12.1 A monkey hanging off of a vine, and the free body diagram at the location of its hand ¶

Here, a monkey hanging off of a vine. Obviously, the force of gravity is acting downward on the monkey, so that means there is a like downward force on the vine at point \(B\) . This bends the vine, so at the monkey’s hand, two tension forces are acting, one each to the left and right. Thus, the hand is where multiple forces are acting at once. The monkey will remain stationary only if these three forces add up to zero as vectors .

This is represented in the free body diagram to the right of the illustration. Because we are interested in maintaining equilibrium, the FBD is drawn at the point where the monkey’s hand is grabbing the vine. The three forces acting at that point are then represented as vectors, pointing away from the dot indicating the position of the hand. This FBD shows all the information needed to write down the Newton’s 2nd law equations for that position.

Next, we will formally introduce the two forces from the free body diagram above – the tension force, and the force of gravity.

12.3. The tension force ¶

If something like a rope or string (or a vine!) is attached to an object, and then pulled on, that object will experience a tension force. This force will act at the point of contact between the rope and the object, and point in the direction along the rope at this contact point. So really, this is a type of applied force, and its magnitude can change, depending on how hard the rope is pulled on. You can imagine a tug-of-war, where the winner is determined by who can pull hardest; this can change, depending on who is working the rope.

This means that there is no set value or equation for the tension force. Assuming the rope is strong enough, it can be any magnitude. This will usually be a given number, but sometimes it may be something you are solving for.

Two crewmen pull a raft through a canal lock, with their forces shown in the figure below. One crewman pulls with a force of 130. N at an angle of 34.0 \(^\circ\) relative to the forward direction of the raft. The second crewman, on the opposite side of the lock, pulls at an angle of 45.0 \(^\circ\) .

Fig. 12.2 Raft pulled along a canal ¶

With what magnitude of force \(F_2\) (in N) should the second crewman pull so that the net force of the two crewmen is only in the forward direction (if the angle is kept the same)?

What is the magnitude of the net force (in N) on the raft (assuming it is being pulled straight forward)?

Answers: 103 N; 181 N

The first crewman is a bit of a jokester, and likes to change the angle at which he pulls the raft. Help the second crewman by creating a vPython function secondAngle() . This function should return the angle (in degrees) that the second crewman should pull the raft, so that the net force on the raft is only in the forward direction, down the canal. The arguments of the function should be the scalars F1 and F2 , for the magnitudes of the two crewmen’s given forces \(F_1\) and \(F_2\) , and the angle (in degrees) Q1 that the first crewman pulls the raft.

You will need to use the inverse trigonometric functions in vPython. The functions asin() , acos() , and atan() correspond to the functions \(\sin^{-1}(), \cos^{-1}()\) , and \(\tan^{-1}()\) , respectively. These functions will return values in radians, not degrees, so you need to convert back!

Note: There are certain values of these arguments where there is no angle at which the second crewman to pull, so that the net force is in the right direction! Your function should print out Impossible situation if this is true, and return the value -1.

Answer: Here is an example function.

12.4. The gravitational force ¶

Before we talk about how to calculate the gravitational force \({\vec F}_g\) , it is important to talk first about terminology. On Earth, we frequently use the terms “mass” and “weight” interchangeably, and this is not too far off. Since the gravitational field is roughly constant everywhere on the surface of the Earth, then the mass and weight of an object will always be proportional.

But there is an important conceptual distinction that this glosses over. When we say mass , we are referring to the amount of “stuff” an object has. This is a fixed property, as long as the object is not split into pieces. Another way to say this is that mass is the measure of inertia – as we saw with Newton’s 2nd law, for the same net force, an object with larger mass will have a smaller resulting acceleration. More massive objects are harder to move.

This is different that the weight of an object. Weight is the gravitational force on an object. So, unlike mass (which is a scalar), weight is a vector quantity, with both magnitude and direction. On the Earth’s surface, the weight is always acting towards the center of the Earth, which is why we usually don’t mention it when state our weight. As we will see shortly, it depends on the object’s mass, but it also depends on the gravitational field. Since weight is a force, we will use the symbol \({\vec F}_g\) for it. I will use “weight” and “gravitational force” as synonyms, so remember that weight is a vector, not a scalar.

Back in Lesson 05 (acceleration and free fall) , the gravitational field \({\vec g}\) was described as the strength of the gravitational influence of an object on everything around it. Now that we are talking about forces, we can be more explicit in what this means. Once you know what \({\vec g}\) is at the location of an object with mass \(m\) , then the gravitational force \({\vec F}_g\) on that object is given by

Thus, this now explains something else said in Lesson 05: if gravity is the only force acting on the object, then its acceleration is the same as the gravitational field. You can see this by using Newton’s 2nd law. Since the mass \(m\) is in free fall, then the sum of the force is just \({\vec F}_g\) :

Newton’s 2nd law says that the sum of the forces on \(m\) is

so we can see that the object’s acceleration \({\vec a} = {\vec g}\) .

However, in general, there are multiple forces acting on an object, so \({\vec g}\) will not be the object’s acceleration. This is why we describe it as a measure of gravitational influence. In Lesson 21 (gravitational force) , we will see how to calculate \({\vec g}\) in general. This will include not only how to find \({\vec g}\) near the surface of the Earth, but also when there are multiple sizable objects present.

One way to lose weight is to go to another planet! Suppose an 80.0 kg person weighs themself on the Earth, then travels to the Moon, where \(g = 1.63\) m/s \(^2\) , and they weigh themselves again. How much weight (in N) did they “lose” by going to the Moon?

Answer: 654 N

Write a Python procedure surfaceWeight() that prints out the values of the weights for an object on various planets and satellites in the Solar System; remember to include the appropriate units. The procedure takes as an argument mass the mass of the object. Use the gravitational field magnitudes listed in the table below.

Answer: One possible procedure is given below.

12.5. Example: Hanging ball ¶

Let’s work through an example problem and see how to find the magnitude of two tension forces acting on an object. Here, you will see how to draw an FBD for the problem, then write down the appropriate force equations, using Newton’s 2nd law.

Here is the situation to consider. A ball of mass 100. g is attached to both the wall and ceiling, as shown. String 1 is completely horizontal and string 2 makes an angle \(\theta\) of 60.0 \(^\circ\) with the ceiling. We are going to calculate the sizes \(F_{T1}\) and \(F_{T2}\) of the tension forces acting on the ball. The physical situation is shown in the picture below.

Fig. 12.3 A ball held in place by two strings ¶

The tension \({\vec F}_{T1}\) is acting on the ball along string 1, and similarly for \({\vec F}_{T2}\) . Notice that there is also a gravitational force acting on the ball, which is why the ball’s mass will be important.

Our first step will be to draw the FBD for the ball. Why are we choosing the ball as the system to consider? Well, this is the object on which all the forces of interest are acting on. By Newton’s zeroth law (see Lesson 10 ), the forces acting directly on the ball are the two tension forces and the ball’s weight. We know enough to calculate the weight, and we want to find the tensions, so this is everything we want! Now, let’s draw the FBD for the ball. Remember to draw the arrows away from the dot representing the ball, in the same direction as the actual forces. This gives the FBD below for the ball.

Fig. 12.4 The FBD for the hanging ball ¶

Notice I have moved the angle, so that it is next to the origin of the diagram. The angle here is the same as the angle shown in the physical diagram, because of alternate interior angles.

Now that we have our FBD, we can use it to write down the Newton’s 2nd law equations for this situation. If I write this down as a vector equation, I get

This is a vector equation, so the directions are “inside” the vectors. Thus, there are no minus signs just yet; these will come in when we write down the components of each vector, which will be our next step. But first, since the ball is not moving, it is definitely not acceleration, so we can say \({\vec a} = {\vec 0}\) .

The next step is to find the components of each force, in terms of their magnitudes. This will be useful in getting equations to solve for each tension magnitude. We are still doing algebra here, so only symbols, no numbers just yet. Two of the forces are along the coordinate axes, so they are simple:

As always, the vector is written with the arrow sign, while the magnitude is written without. The directions of each force component is shown by the negative signs and the unit vectors \({\hat x}, {\hat y}\) . For the last vector \({\vec F}_{T2}\) , we need to find the \(x\) and \(y\) components, so we draw a vector diagram to help.

Fig. 12.5 Vector diagram for tension \({\vec F}_{T2}\) ¶

This gives that the second tension force vector is

These are now combined into Newton’s second law. We take the vector version of it, given above, and write down separately the \(x\) and \(y\) components of the equation. In Lesson 13 (scalar product) , we will see how to do this more formally. For now, though, this gives

Remember these net force components are equal to zero, since \({\vec a} = {\vec 0}\) . We now have two equations we can solve for the magnitude \(F_{T1}\) and \(F_{T2}\) , in terms of the angle \(\theta\) and the gravitational force \(F_g\) . But we have to be careful how we solve these! The \(x\) equation has both unknowns in it, so we need to solve the \(y\) equation first, resulting in

In the second step, we used the fact that the magnitude \(F_g\) of the ball’s weight is \(m_{ball} g\) . This equation can be substituted into the \(x\) equation to find \(F_{T1}\) .

The last step here uses the definition of \(\tan \theta\) as the ratio of \(\sin \theta\) and \(\cos \theta\) . Plugging in the givens into these equations gives

Since \(F_{T1} = F_{T2} \cos \theta\) , and \(\cos 60.0^\circ = 0.500\) , then \(F_{T1}\) is exactly half the size of \(F_{T2}\) . This would not be true for any other angle \(\theta\) .

A 124 N stop light hangs at rest from two cables as shown below, where \(\theta_1=30.0^\circ\) and \(\theta_2=50.0^\circ\) . Solve for the magnitudes of the tensions \(F_{T1}\) (left rope) and \(F_{T2}\) in the two ropes (in N). Hint: The Newton’s 2nd law equations will have both unknown magnitudes in both \(x\) and \(y\) equations, so you will have to use the techniques of simultaneous equations (as I did in the example above) to find these values.

Fig. 12.6 A hanging stoplight ¶

Answers: \(F_{T1} = 80.9\) N; \(F_{T2} = 109\) N

As an employee of the City of Newport, your job is to hang up streetlights. However, you are concerned that the ropes used are not strong enough when the streetlights are hung at certain angles.

Based on your work for the previous problem, write down equations for the tension values \(F_{T1}\) and \(F_{T2}\) in each rope, using the variables \(m\) (streetlight mass), and \(\theta_1, \theta_2\) (the two rope angles). Hint: Your equations will simplify quite a bit if you use the trigonometric identity \(\sin (\theta_1 + \theta_2) = \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2\) .

Suppose that \(\theta_1 > \theta_2\) . Which tension value will then be the larger one?

Write a vPython procedure checkTension() that checks whether the tension values in the ropes exceeds the maximum safety value. The arguments of the procedure should be m for the streetlight mass, Q1 and Q2 for the angles \(\theta_1\) and \(\theta_2\) , respectively, and F_MAX for the maximum safe value of the rope tension magnitude. If the situation is safe, have the procedure print out OK ; otherwise, the procedure should print out Unsafe situation .

Answers: The equations for the two tension magnitudes are given by \(F_{T1} = mg \cos \theta_2 / \sin(\theta_1 + \theta_2)\) and \(F_{T2} = mg \cos \theta_1 / \sin(\theta_1 + \theta_2)\) . Since \(\theta_1 > \theta_2\) implies that \(\cos \theta_1 < \cos \theta_2\) , then \(F_{T1}\) would be the larger magnitude. This makes physical sense, because whichever rope has the larger angle will be supporting more of the streetlight’s weight.

Here is a possible procedure for checking whether the situation is safe.

12.6. Normal force ¶

The best way to think about the normal force is that it is the force that prevents one object from passing through another. When you sit on a chair, or lean against a wall, these objects exert a normal force on you, so you do not fall through the chair or wall. The normal force is known as a constraint force , since it constrains objects from occupying the same location!

Notice that the normal force can change, depending on the situation. If you think of yourself leaning against the wall, with your feet still on the floor, both the wall and floor exert normal forces on you. If you change your angle relative to the horizontal (your “lean”), then the sizes of the two normal forces will shift. When you are almost vertical, the floor acts on you with the larger normal force; if you move your feet away from the wall, the angle with the horizontal will decrease, and the wall exerts a normal force with increasing size. The important point is that these forces constrain you, so that you don’t accelerate through the wall or floor!

Each of these forces prevent you from passing through the floor or wall, so they must be perpendicular to these surfaces – you don’t start sliding sideways, just because you are standing on the floor! Similarly with the wall, where you may slide down the wall, but that is due to gravity, not the force of the wall on you. These forces also point away from the surface (floor or wall).

Properties of the normal force

The normal force is a force that a surface exerts on an object. The normal force on the object always points perpendicular to (and away from) the surface. The magnitude of the normal force is whatever amount is required to prevent the object from accelerating through the surface. There is no direct relationship between the magnitudes of the normal force and the weight!

What is the correct direction for the normal force given that the object is resting on an inclined surface, as shown below?

Fig. 12.7 Find the correct direction for the normal force on a block sitting on a ramp ¶

Answer: The normal force points up and to the left.

What is the correct direction for the normal force on an object pressed against a horizontal surface?

Fig. 12.8 Find the correct direction for the normal force on the block on the ceiling ¶

Answer: The normal force points downward.

What is the correct direction for the normal force given that the object is hanging by a string above a horizontal surface?

Fig. 12.9 Find the correct direction for the normal force on the hanging block ¶

Answer: The block is not on a surface, so there is no normal force.

Let’s go through an example, to see how this works in a calculation. A 1.50 kg object is pushed up a frictionless vertical wall with a force \(F\) at an angle of 35.0 \(^\circ\) above the horizontal. Starting from rest, the block is moving with an upward velocity of 11.0 m/s after two seconds. We will find the magnitude of the applied force \(F\) , and the magnitude of the normal force \(F_N\) .

Fig. 12.10 A block pushed up the wall ¶

First, we draw an FBD for this situation. Notice that, since the normal force is away from the surface and perpendicular to it, then \({\vec F}_N\) points to the left.

Fig. 12.11 The FBD for the block pushed up the wall ¶

Writing down the sum of the forces acting on the wall, we get

Now, we find the force vector component equations along each coordinate axis. This gives

Here, \(F_x\) and \(F_y\) are the \(x\) and \(y\) components of the applied force \({\vec F}\) . The block is accelerating upwards, since its velocity is increasing in that direction. However, horizontally the block is not moving, so \(a_x = 0\) . This means we can solve first for the relation between the normal force magnitude \(F_N\) and the size of the applied force \(F\) .

This makes sense – the normal force is canceling out the piece of the applied force that is directly into the wall; it is not affected by the vertical component of the applied force. This latter part will change the acceleration in the \(y\) direction, so let’s look at that now. By definition,

and with \(v_{i, y} = 0\) , then the force equation in the \(y\) direction becomes

Solving this for the applied force magnitude \(F\) gives

and a value of \(F = 40.0\) N. Using this in the equation solved above for the normal force gives \(F_N = 32.8\) N.

A 7.85 kN telephone pole is shown in the figure. Assume that the normal force on the telephone pole is vertically upwards with a value of 21.0 kN, and the angles are \(\theta = 30.0^\circ\) and \(\phi = 40.0^\circ\) .

Fig. 12.12 A telephone pole held in place with two cables ¶

What is the magnitude of the tension \(F_{T1}\) (in kN)?

What is the magnitude of the tension \(F_{T2}\) (in kN)?

Hint : The Newton’s 2nd law equations will have both unknown magnitudes in both \(x\) and \(y\) equations, so you will have to use the techniques of simultaneous equations to find these values.

Answers: \(F_{T1} = 1.07\) kN; \(F_{T2} = 1.21\) kN

12.7. Friction ¶

12.7.1. the study of friction ¶.

If you have some discomfort with Newton’s first law of motion, it’s probably because of the widespread presence of frictional forces in our everyday lives. Whenever we roll a ball along a table, the ball will eventually stop, not because that is the “natural” motion of all objects, but because of friction between the ball and the table. This was one of the difficult mental leaps that had to be made to go from the thoughts of Aristotle and the ancient Greeks, to the modern viewpoint on physics: motion stops because there are forces present that cause this stop, not because the “natural state” of an object is always at rest. This flip between viewpoints allows us to describe in detail the forces that cause motion to change, rather than throwing up our hands and saying it’s the way things are.

Fig. 12.13 A self-portrait of Leonardo da Vinci from his notebooks ¶

This leads us into the specific force of friction as our next topic. Leonardo da Vinci was one of the first people to systematically examine the size of the frictional force. In a series of experiments, he showed that friction does not depend on the contact area between the two surfaces, and that increasing the load on an object will increase the frictional force proportionally. However, although his discoveries were written into his notebooks, these were not publicly known until hundreds of years later.

Fig. 12.14 Guillaume Amontons showing the Dauphin of France his method to transmit message by optical telegraphy ¶

da Vinci’s laws of friction were rediscovered by the French instrument maker and physicist Guillaume Amontons around 1700, and hence are often called “Amontons’ laws”. These rules were verified by Charles-Augustin de Coulomb (whom we will see later with the electrostatic force in Lesson 24 ) about eight decades later.

The frictional force is a complicated area of study, and researchers are continuing to look into its properties. This is because, not only is it interesting, but it has obvious industrial and engineering applications. We will examine only a simple model of friction here.

12.7.2. Two types of friction ¶

Suppose you are rearranging the furniture in your living room, and decide you want to move your couch. You give it a push to slide it along the carpet in the room. To make things definite, let’s make it a couch with a weight of 2.00 kN, and you start out by applying a force of 200. N to the couch. It doesn’t move, so you push harder with 400. N. It still doesn’t move, so you push with 600. N. Still doesn’t move. You activate those honed muscles of yours and push with 800. N, which is finally enough to get the couch to move. Not only does it move, but it starts accelerating across the living room! So, you push with only 600. N, and (unlike before), the couch now moves at a constant velocity.

I’m sure you can think of some situation like this, where you have to apply a force to get something to start moving. Once you do get it moving, you need less force to keep it moving. This shows that there are two separate types of frictional force. The static frictional force acts on an object when it is stationary relative to the surface it is sitting on. Notice that they may be moving together – think of a crate in the flatbed of a pickup truck, for example. The kinetic frictional force acts on an object moving relative to the surface it is on. This may be due to the object moving on a stationary surface, or the other way around. The magnitude of the kinetic frictional force is usally less than that of the static frictional force.

Once the couch in the example above starts moving, then only the kinetic frictional force is acting on it. Before that, however, your applied force is opposed by the static frictional force. Note how the type of frictional force changed! As you increased the applied force, the static frictional force matches it, at least up to a point. Beyond that maximum, though, friction was not strong enough to prevent the couch from moving, and now the kinetic friction was involved.

As mentioned above, da Vinci, Amontons, Coulomb, and others started a detailed study of the properties of the friction force. Their model of friction, described in this lesson, has two basic properties.

Properties of the frictional force

The direction of the frictional force \({\vec F}_{fr}\) is always opposing the motion of an object, or the potential motion, if the object is at rest.

The magnitude \(F_{fr}\) of the frictional force on an object on a surface is proportional to the normal force due to the surface, acting on the object. The constant of proportionality is known as the coefficient of friction, and given by the symbol \(mu\) (“mu”).

Let’s talk about these in more detail, starting with the direction. Think again to the example mentioned earlier of a crate in a flatbed truck. As the truck drives along, if the flatbed’s surface were completely smooth, the crate would have no horizontal force on it, and it would slide backwards. To keep the crate in place, the static frictional force acts opposite to this motion, in the direction the truck is moving. This is what I mean by opposing the potential motion of the object. Thus, it is actually the frictional force moving the crate forward! Think about that the next time you go for a walk.

For the magnitude, since the force of friction is always proportional to the normal force, then it makes sense to look at the ratio of the frictional and normal forces; this will be the constant \(\mu\) mentioned above. Thus, the coefficient \(\mu\) is the ratio of the two magnitudes, or

Notice that, since \(\mu\) is the ratio of two forces, it has no units! This constant is roughly the same for a given type of friction (static or kinetic) with any pair of surfaces, e.g. rubber on carpet, for someone walking in their house with shoes on. There will be a coefficient of static friction \(\mu_s\) and a coefficient of kinetic friction \(\mu_k\) for each such surface pair. These coefficients must be measured experimentally, by going into the lab and moving the two surfaces against each other.

Let’s now get specific about the relationship between the normal and friction forces. Remember the example we had above, with the couch. Specifically, the static frictional force would change, since you altered your applied force. Thus, the force of static friction will actually have a range of values, up to some maximum static frictional force. To show this, the magnitude of \(F_{fr, s}\) is given by an inequality; the static frictional force is limited by its largest magnitude. Mathematically, this is

where, as before, \(F_{fr, s}\) is the force of static friction, \(\mu_s\) is the coefficient of static friction, and \(F_N\) is the normal force. On the other hand, the kinetic frictional force is constant, so it is a straight equality:

where \(F_k\) is the force of kinetic friction, and \(\mu_k\) is the coefficient of kinetic friction.

The PhET application linked below allows you to play around with the force of friction, especially if you select the “Friction” portion of the app.

Use the “Friction” part of the PhET to answer the following questions.

Using the default values in the PhET, what is the coefficient of static friction between the crate and the ground? What is the coefficient of kinetic friction? Does it matter if you change the object you are pushing, or add mass on top of it?

Answer: The crate has a mass of 50 kg, so the gravitational force on it is 490. N. By steadily increasing the applied force when the crate is at rest, you find that the maximum static frictional force is 125 N. This gives \(\mu_s\) to be around 0.26. Once the crate is moving, there is a kinetic frictional force of 94 N, giving \(\mu_k\) to be around 0.19. These coefficients appear to be the same for all objects, which you can test by finding the necessary force to get each object to start moving.

Find the value of the unknown mass for the gift-wrapped box.

Answer: By increasing the applied force from zero, while the box is at rest, the maximum static frictional force is 125 N. When the box is moving, there is a constant kinetic frictional force of 94 N. Using the same coefficients of friction obtained in the last problem, this suggests the box has a mass of 50 kg. You can confirm this by adding objects with known mass, and verifying the frictional force changes as expected.

As another demonstration of the frictional force, let’s make a table of the applied forces on the couch. We assume that your applied force and the force of friction are the only horizontal forces.

Let the coefficient of static friction \(\mu_s\) be 0.375, and the coefficient of kinetic friction \(\mu_k\) be 0.250. Remember that the couch has a weight of 2.00 kN. If the living room floor is flat, use this to calculate the values of the maximum static frictional force, and the kinetic frictional force, and fill in the values of the frictional force in the table below.

Answer: For this problem, you first need to find the normal force of the floor acting on the couch. Since the floor is flat, the magnitude of the normal force is equal to the value of the couch’s weight, or 2.00 kN. This means the maximum magnitude of the static frictional force is \(\mu_s F_N = (0.375)(2000 \textrm{ N}) = 750.\) N, while the constant value of the kinetic frictional force is \(\mu_k F_N = (0.250)(2000 \textrm{ N}) = 500.\) N. Then, the completed table of frictional forces should look like the one given below.

After the couch starts moving, and you apply a force of 600. N on it, what is the frictional force value between the floor and the couch?

Answer: Once the couch is moving, the frictional force is kinetic, so the force of friction has a magnitude of 500. N.

12.8. An example: Multiple forces ¶

Now that we have a model of the frictional force, let’s see how it interacts with some of the other forces we have dealt with. I will work through the following example. A block with a mass of 5.00 kg is moving because of a 20.0 N applied acting at an angle of 15.0 \(^\circ\) below the horizontal. The coefficient of kinetic friction between the block and the horizontal surface is 0.200. Determine the horizontal acceleration (magnitude in m/s \(^2\) ) of the block.

Fig. 12.15 Sliding a block ¶

As always, the first step should be to draw a FBD for the block. This will include not only the frictional force, but also the gravitational, normal, and applied forces. The FBD of the block is shown below.

Fig. 12.16 FBD for block sliding on floor ¶

You might think we only need to find the horizontal forces on the block, but since the friction force depends on the normal force, we need to solve for the vertical forces as well. In addition, the downward acting applied force will change the normal force, so we need to look out for that. If we write down the net force on the block, we have

Taking the \(x\) and \(y\) components, then

We solve first for the normal force magnitude \(F_N\) , using the fact that \(a_y = 0\) (the block is not accelerating vertically). Then,

We know the block is moving, so the frictional force is kinetic; substituting the normal force into the kinetic friction force equation, the \(x\) force equation becomes

With a little rearranging, this gives the size \(a_x\) of the acceleration to be

Plugging in the given values results in an acceleration of 1.69 m/s \(^2\) to the right.

A 500. N crate starts at rest on the concrete floor of a warehouse. You exert a horizontal applied force of 350. N. The coefficient of static friction is 0.735, and the coefficient of kinetic friction is 0.580.

What is the acceleration of the crate?

What force is necessary to get the crate moving?

Answers: 0.00 m/s \(^2\) ; 368 N

An applied force \(F_{app}\) exerted horizontally starts to act on a 125 N block initially at rest. Below is shown a graph of the frictional force \(F_{fr}\) on the block from the floor, as a function of the applied force as it increases in size from zero newtons. Use this graph to find the values of the coefficients of static and kinetic friction.

Fig. 12.17 Frictional force vs. applied force graph for sliding block ¶

Answers: \(\mu_s = 0.640\) , \(\mu_k = 0.400\)

A NAPSter, crazed by the latest victory of the football team, uses a force \({\vec F_{app}}\) of magnitude 90.0 N to push a 4.00 kg block across the ceiling of their room, as shown below. The coefficient of kinetic friction between the block and the ceiling’s surface is 0.400. Find the magnitude of the block’s acceleration (in m/s \(^2\) ).

Fig. 12.18 Sliding a block on the ceiling ¶

Answer: 3.16 m/s \(^2\)

12.9. Summary ¶

We are now beginning to use force methods to solve problems. This lesson introduced this technique, and started with the tension, gravitational, normal, and frictional forces as examples to see how it works. Over the course of the next few lessons, we will add on additional forces, thus allowing us to look at more complicated situations. The basic ideas from this lesson will hold: the basic technique is to start with an FBD, and derive the appropriate force equations using Newton’s 2nd law. This should be your standard practice!

After this lesson, you should be able to:

Draw the FBD of a system for any given physical situation

Using the FBD, write down the appropriate Newton’s 2nd law equations

Find the gravitational force on an object in a gravitational field

Solve for the normal force acting on an object

Describe what type of friction is appropriate for a given situation

Explain how much static frictional force is present when other forces are applied to a stationary object

Use a knowledge of forces and Newton’s second law to write down the appropriate mathematical relationship between the forces and the object’s acceleration.

Force Examples

A force of one object has a result to push or have an interaction with another object. Here are few force examples to learn how to calculate force by applying mass and acceleration.

Force Mass Acceleration Problems with Solutions

Let us consider the problem: Find the force of an object with mass(m) as 600 kg and acceleration as 50 m/s^2

We can calculate force the using the given formula.

Substituting the values in the formula,

= 600 x 50 = 30000 N Hence, force of the object is 30000 Newtons.

Let us consider the problem: Find the mass of an object with force 200 Newtons and acceleration as 10 m/s^2

We can calculate the mass using the given formula.

= 200 / 10 = 20 Kgs Hence, mass of the object is 20 Kgs

Let us consider the problem: Find the acceleration of an object with mass(m) as 300 kg and force as 1000 N

We can calculate the acceleration using the given formula.

Acceleration:

= 1000 / 300 = 3.333 m/s^2 Hence, acceleration of the object is 3.333 m/s^2

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Work done by force – problems and solutions

1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block.

Force (F) = 20 N

Displacement (s) = 2 m

Angle (θ ) = 0

Wanted : Work (W)

W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule

2. A force F = 10 N acting on a box 1 m along a horizontal surface. The force acts at a 30 o angle as shown in figure below. Determine the work done by force F!

Force (F) = 10 N

The horizontal force (F x ) = F cos 30 o = (10)(0.5√3) = 5√3 N

Displacement (d) = 1 meter

Wanted : Work (W) ?

W = F x d = (5√3)(1) = 5√3 Joule

3. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity is 10 m/s 2 , determine the work done by the force of gravity !

Object’s mass (m) = 1 kg

Height (h) = 2 m

Acceleration due to gravity (g) = 10 m/s 2

Wanted : Work done by the force of gravity (W)

W = F d = w h = m g h

W = (1)(10)(2) = 20 Joule

W = work, F = force, d = distance, w = weight , h = height, m = mass, g = acceleration due to gravity.

4. An 1-kg object attached to a spring so it is elongated 2 cm. If acceleration due to gravity is 10 m/s 2 , determine (a) the spring constant (b) work done by spring force on object

Mass (m) = 1 kg

Elongation (x) = 2 cm = 0.02 m

Weight (w) = m g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 = 10 N

Wanted : Spring constant and work done by spring force

(a) Spring constant

Formula of Hooke’s law :

k = F / x = w / x = m g / x

k = (1)(10) / 0.02 = 10 / 0.02

k = 500 N/m

(b) work done by spring force

W = – ½ k x 2

W = – ½ (500)(0.02) 2

W = – (250)(0.0004)

W = -0.1 Joule

The minus sign indicates that the direction of spring force is opposite with the direction of object displacement.

5. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a friction force F k = 2 N. Determine the net work done on the box.

Force of kinetic friction (F k ) = 2 N

Displacement (d) = 2 m

Wanted : Net work (W net )

Work done by force F :

W 1 = F d cos 0 = (10)(2)(1) = 20 Joule

Work done by force of kinetic friction (F k ) :

W 2 = F k d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule

W net = W 1 – W 2

W net = 20 – 4

W net = 16 Joule

6 . What is the work done by force F on the block.

Force (F) = 12 Newton

Displacement (d) = 4 meters

Wanted: Work (W)

W = F d = (12 Newton)(4 meters) = 48 N m = 48 Joule

7 . A block is pushed by a force of 200 N. The block’s displacement is 2 meters. What is the work done on the block?

Force (F) = 200 Newton

Displacement (d) = 2 meters

W = (200 Newton)(2 meters)

W = 400 N m

W = 400 Joule

8 . The driver of the sedan wants to park his car exactly 0.5 m in front of the truck which is at 10 m from the sedan’s position. What is the work required by the sedan?

Displacement (d) = 10 meters – 0.5 meters = 9.5 meters

Force (F) = 50 Newton

W = (50 Newton)(9.5 meters)

W = 475 N m

W = 475 Joule

Work done by Tom and Jerry so the car can move as far as 4 meters. Forces exerted by Tom and Jerry are 50 N and 70 N.

Displacement (s) = 4 meters

Net force (F) = 50 Newton + 70 Newton = 120 Newton

W = F s = (120 Newton)(4 meters) = 480 N m = 480 Joule

10 . A driver pulling a car so the car moves as far as 1000 cm. What is the work done on the car?

Force (F) = 250 Newton

Displacement (s) = 1000 cm = 1000/100 meters = 10 meters

W = F s = (250 Newton)(10 meters) = 2500 N m = 2500 Joule

11. Based on figure below, if work done by net force is 375 Joule, determine object’s displacement.

Work (W) = 375 Joule

Net force ( ΣF) = 40 N + 10 N – 25 N = 25 Newton ( rightward )

Wanted : Displacement ( d )

The equation of work :

Object’s displacement :

d = W / F = 375 Joule / 25 Newton

d = 15 meter s

12. The activities below w hich do not do work is …

A. Push an object as far as 10 meters

B. Push a car until a move

C. Push a wall

D. Pulled a box

W = work , F = force , d = displacement

B ased on the above formula, work done by force and there is a displacement.

The correct answer is C.

1 3 . Andrew pushes an object with force of 20 N so the object moves in circular motion with a radius of 7 meters. Determine the work done by Andrew for two times circular motion .

B. 1400 Joule

C. 1540 Joule

D. 1760 Joule

If the person pushes wheelchair for two times circular motion then the person and wheelchair return to the original position, so the displacement of the person is zero.

Displacement = 0 so work = 0.

The correct answer is A.

14 . Someone push an object on the floor with force of 350 N. The floor exerts a friction force 70 N. Determine the work done by force to move the object as far as 6 meters.

The force of push (F) = 350 Newton

Friction force (F fric ) = 70 Newton

Displacement of object (s) = 6 meters

There are two forces that act on the object, the push force (F) and friction force (F fric ). The push force has the same direction as the displacement of the object because the push force does a positive work. In another hand, the friction force has the opposite direction with a displacement of the object so that the friction force does a negative work.

Work done by push force :

W = F d = (350 Newton)(6 meters) = 2100 Newton-meters = 2100 Joule

Work done by friction force :

W = – (F fric )(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule

The net work :

W net = 2100 Joule – 420 Joule

W net = 1680 Joule

15 . An object is pushed by a horizontal force of 14 Newton on a rough floor with the friction force of 10 Newton. Determine the net work of move the object as far as 8 meters.

A. 0.5 Joule

C. 32 Joule

D. 192 Joule

Push force (F) = 14 Newton

Friction force (F fric ) = 10 Newton

Displacement of object (d) = 8 meters

There are two forces that act on an object, push force (F) and friction force (F fric ).

The push force has the same direction as the displacement of the object so that the push force does a positive work. In another hand, the friction force has the opposite direction as the displacement of the object so that the friction force does a negative work.

W = F s = (14 Newton)(8 meters) = 112 Newton meters = 112 Joule

W = – (F fric )(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule

W net = 112 Joule – 80 Joule

W net = 32 Joule

16 . Determine the net work based on figure below.

B. 450 Joule

C. 600 Joule

D. 750 Joule

Work = Force (F) x displacement (d)

Work = Area of triangle 1 + area of rectangle + area of triangle 2

Work = 1/2(40-0)(3-0) + (40-0)(9-3) + 1/2(40-0)(12-9)

Work = 1/2(40)(3) + (40)(6) + 1/2(40)(3)

Work = (20)(3) + 240 + (20)(3)

Work = 60 + 240 + 60

Work = 360 Joule

17 . A piece of wood with a length of 60 cm plugged vertically into the ground. Wood hit with a 10-kg hammer from a height of 40 cm above the top of the wood. If the average resistance force of the ground is 2 x 10 3 N and the acceleration due to gravity is 10 m/s 2 , then the wood will enter entirely into the ground after…. hits.

Mass of hammer (m) = 10 kg

Acceleration due to gravity (g) = 10 m/ s 2

Weight of hammer (w) = m g = (10)(10) = 100 kg m/s 2

Displacement of hammer before hits the wood (d) = 40 cm = 0.4 meters

The resistance of wood (F) = 2 x 10 3 N = 2000 N

Length of wood (s) = 60 cm = 0.6 meters

Wanted : T he wood will enter entirely into the ground after…. hits.

Work done on the hammer when hammer moves as far as 0.4 meters is :

W = F d = w s = (100 N)(0.4 m) = 40 Nm = 40 Joule

Work done by the resistance force of the ground :

W = F d = (2000 N)(0.6 m) = 1200 Nm = 1200 Joule

T he wood will enter entirely into the ground after…. hits.

1200 Joule / 40 Joule = 30

The correct answer is D.

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• Work done by force problems and solutions
• Work-kinetic energy problems and solutions
• Work-mechanical energy principle problems and solutions
• Gravitational potential energy problems and solutions
• The potential energy of elastic spring problems and solutions
• Power problems and solutions
• Application of conservation of mechanical energy for free fall motion
• Application of conservation of mechanical energy for up and down motion in free fall motion
• Application of conservation of mechanical energy for motion on a curve surface
• Application of conservation of mechanical energy for motion on an inclined plane
• Application of conservation of mechanical energy for projectile motion

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Friction Example Problem – Physics Homework Help 3

Friction is the resistance force generated between two bodies as they move across each other. It is proportional to the force that presses the two bodies together. This diagram shows the forces acting on a block sitting on a surface.

The block is pulled down onto the surface by the force of gravity while the surface pushes back in an equal and opposite force known as the normal force: N. Note there are no horizontal forces. If a horizontal force is applied such as pushing the block to the right, the block will begin to accelerate. Experience tells us this does not always happen. If you try to push something heavy, it doesn’t always move until you’ve pushed HARD enough. There must be a force working in the opposite direction of the push to resist the motion. This force is the force of friction, F f .

Experiments have shown the magnitude of this force is dependent on the normal force. The magnitude of the friction force is directly proportional to the magnitude of the normal force. The proportionality constant between them is called the coefficient of friction, μ f . The f subscript is commonly left off, and is not unusual to see just the μ listed.

The coefficient of friction depends on two factors. The first depends on the materials the two objects are made of. It is generally easier to move 50 kg of ice on a glass surface than a 50 kg stone on sand. Each two materials have their own coefficients of friction. The second factor is whether or not the block is moving. You may have noticed it is usually easier to move a heavy object once it is already moving. This means there are two different coefficients of friction. One for when the block is stationary, μ s (static) and one for when the block begins to move, μ k (kinetic).

The static coefficient is used whenever the block is stationary. As the force pushing the block increases, it will eventually reach a point where the block is on the verge of moving. The coefficient of static friction, μ s is experimentally determined by carefully measuring the force at this point. The frictional force required to reach this point is F = -μ s N. The minus sign indicates the direction of the force. Frictional forces oppose the force trying to move the object and act in the opposite direction. Any magnitude of force less than musN, the block will not move.

When the block is moving, the coefficient of kinetic friction is used. This value is experimentally calculated by measuring the force necessary to keep the block moving at a constant velocity. This force will equal -μ k N.

Now, let’s try a friction example problem.

Example Problem:

A block weighing 200 N is pushed along a surface. If it takes 80 N to get the block moving and 40 N to keep the block moving at a constant velocity, what are the coefficients of friction μ s and μ k ?

For the coefficient of static friction, we need the force needed to get the block moving. In this case, 80 N.

From the description above:

F f = μ s N

N is equal to the weight of the block, so N = 200 N. Put these values into the formula.

80 N = μ s ·200 N or μ s = 0.4

For the coefficient of kinetic friction, the force needed to maintain a constant velocity was 40 N. Use the formula:

F f = μ k N 40 N = μ k ·200 N μ k = 0.2

The two coefficients of friction for this system are μ s = 0.4 and μ k = 0.2.

There are two important things to remember in friction homework problems. The first is the normal force N is always perpendicular to the surface. The normal force is not always ‘up’. The second is the friction force works opposite in direction to the motion of the block. Friction is a resistive force.

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