The rest of this chapter consists of sample problems. You should pore over them, learning not just their particular answers but, instead, the procedures they show for attacking a problem. Especially important is knowing how to translate a sketch of a situation into a free-body diagram with appropriate axes, so that Newton’s laws can be applied. We begin with a sample problem that is worked out in exhaustive detail, using a question-and-answer format.
Figure 5-14 shows a block S (the sliding block) with mass M = 3.3 kg. The block is free to move along a horizontal frictionless surface and connected, by a cord that wraps over a frictionless pulley, to a second block H (the hanging block), with mass m = 2.1 kg. The cord and pulley have negligible masses compared to the blocks (they are “massless”). The hanging block H falls as the sliding block S accelerates to the right. Find (a) the acceleration of block S, (b) the acceleration of the block H, and (c) the tension in the cord.
Q What is this problem all about?
You are given two bodies—sliding block and hanging block—but must also consider Earth, which pulls on both bodies. (Without Earth, nothing would happen here.) A total of five forces act on the blocks, as shown in Fig. 5-15:
Fig. 5-14 A block S of mass M is connected to a block H of mass m by a cord that wraps over a pulley.
Fig. 5-15 The forces acting on the two blocks of Fig. 5-14.
1. The cord pulls to the right on sliding block S with a force of magnitude T.
2. The cord pulls upward on hanging block H with a force of the same magnitude T. This upward force keeps block H from falling freely.
3. Earth pulls down on block S with the gravitational force 
, which has a magnitude equal to Mg.
4. Earth pulls down on block H with the gravitational force 
, which has a magnitude equal to mg.
5. The table pushes up on block S with a normal force 
.
There is another thing you should note. We assume that the cord does not stretch, so that if block H falls 1 mm in a certain time, block S moves 1 mm to the right in that same time. This means that the blocks move together and their accelerations have the same magnitude a.
Q How do I classify this problem? Should it suggest a particular law of physics to me?
Yes. Forces, masses, and accelerations are involved, and they should suggest Newton’s second law of motion, 
. That is our starting Key Idea.
Q If I apply Newton’s second law to this problem, to which body should I apply it?
We focus on two bodies, the sliding block and the hanging block. Although they are extended objects (they are not points), we can still treat each block as a particle because every part of it moves in exactly the same way. A second Key Idea is to apply Newton’s second law separately to each block.
Q What about the pulley?
We cannot represent the pulley as a particle because different parts of it move in different ways. When we discuss rotation, we shall deal with pulleys in detail. Meanwhile, we eliminate the pulley from consideration by assuming its mass to be negligible compared with the masses of the two blocks. Its only function is to change the cord’s orientation.
Q OK. Now how do I apply to the sliding block?
Represent block S as a particle of mass M and draw all the forces that act on it, as in Fig. 5-16a. This is the block’s free-body diagram. Next, draw a set of axes. It makes sense to draw the x axis parallel to the table, in the direction in which the block moves.
Q Thanks, but you still haven’t told me how to apply to the sliding block. All you’ve done is explain how to draw a free-body diagram.
You are right, and here’s the third Key Idea: The expression 
is a vector equation, so we can write it as three component equations:
in which Fnet,x, Fnet,y, and Fnet,z are the components of the net force along the three axes. Now we apply each component equation to its corresponding direction. Because block S does not accelerate vertically, Fnet,y = May becomes
Thus in the y direction, the magnitude of the normal force is equal to the magnitude of the gravitational force.
No force acts in the z direction, which is perpendicular to the page.
In the x direction, there is only one force component, which is T. Thus, Fnet,x = Max becomes
This equation contains two unknowns, T and a; so we cannot yet solve it. Recall, however, that we have not said anything about the hanging block.
Q I agree. How do I apply to the hanging block?
We apply it just as we did for block S: Draw a free-body diagram for block H, as in Fig. 5-16b. Then apply in component form. This time, because the acceleration is along the y axis, we use the y part of Eq. 5-16 (Fnet,y = may) to write
We can now substitute mg for FgH and −a for ay (negative because block H accelerates in the negative direction of the y axis). We find
Now note that Eqs. 5-17 and 5-18 are simultaneous equations with the same two unknowns, T and a. Subtracting these equations eliminates T. Then solving for a yields
Fig. 5-16 (a) A free-body diagram for block S of Fig. 5-14. (b) A free-body diagram for block H of Fig. 5-14.
Substituting this result into Eq. 5-17 yields
Putting in the numbers gives, for these two quantities,
Q The problem is now solved, right?
That’s a fair question, but the problem is not really finished until we have examined the results to see whether they make sense. (If you made these calculations on the job, wouldn’t you want to see whether they made sense before you turned them in?)
Look first at Eq. 5-19. Note that it is dimensionally correct and that the acceleration a will always be less than g. This is as it must be, because the hanging block is not in free fall. The cord pulls upward on it.
Look now at Eq. 5-20, which we can rewrite in the form
In this form, it is easier to see that this equation is also dimensionally correct, because both T and mg have dimensions of forces. Equation 5-21 also lets us see that the tension in the cord is always less than mg, and thus is always less than the gravitational force on the hanging block. That is a comforting thought because, if T were greater than mg, the hanging block would accelerate upward.
We can also check the results by studying special cases, in which we can guess what the answers must be. A simple example is to put g = 0, as if the experiment were carried out in interstellar space. We know that in that case, the blocks would not move from rest, there would be no forces on the ends of the cord, and so there would be no tension in the cord. Do the formulas predict this? Yes, they do. If you put g = 0 in Eqs. 5-19 and 5-20, you find a = 0 and T = 0. Two more special cases you might try are M = 0 and m → ∞.
In Fig. 5-17a, a block B of mass M = 15.0 kg hangs by a cord from a knot K of mass mK, which hangs from a ceiling by means of two other cords. The cords have negligible mass, and the magnitude of the gravitational force on the knot is negligible compared to the gravitational force on the block. What are the tensions in the three cords?
Fig. 5-17 (a) A block of mass M hangs from three cords by means of a knot K. (b) A free-body diagram for the block. (c) A free-body diagram for the knot.
Solution: Let’s start with the block because it has only one attached cord. The free-body diagram in Fig. 5-17b shows the forces on the block: gravitational force 
(magnitude Mg) and force T3 from the attached cord. A Key Idea is that we can relate these forces to acceleration via Newton’s second law (). Because the forces are both vertical, we choose the vertical component version of the law (Fnet,y = may) and write
Substituting Mg for Fg and 0 for the block’s acceleration ay, we find
This means that the two forces on the block are in equilibrium. Substituting for M (= 15.0 kg) and g and solving for T3 yield
We next consider knot K in the free-body diagram of Fig. 5-17c, where the negligible gravitational force on the knot is not included. The Key Idea here is that we can relate the three other forces acting on the knot to the acceleration of the knot via Newton’s second law () by writing
Substituting 0 for the knot’s acceleration 
yields
which means that the three forces on the knot are in equilibrium. Although we know both magnitude and angle for 
, we know only the angles and not the magnitudes for 
and 
. With unknowns in two vectors, we cannot solve Eq. 5-22 for or directly on a vector-capable calculator.
Instead we rewrite Eq. 5-22 in terms of components along the x and y axes. For the x axis, we have
which, using the given data, yields
(For the first term, we have two choices, either the one shown or the equivalent T1 cos 152°, where 152° is the angle from the positive direction of the x axis.)
Similarly, for the y axis we rewrite Eq. 5-22 as
Substituting our previous result for T3 then gives us
We cannot solve Eq. 5-23 or Eq. 5-24 separately because each contains two unknowns, but we can solve them simultaneously because they contain the same two unknowns. Doing so (either by substitution, by adding or subtracting the equations appropriately, or by using the equation-solving capability of a calculator), we discover
In Fig. 5-18a, a cord holds a 15 kg block stationary on a frictionless plane inclined at angle θ = 27°.
(a) What are the magnitudes of the force 
on the block from the cord and the normal force on the block from the plane?
Solution: Those two forces and the gravitational force on the block are shown in the block’s free-body diagram of Fig. 5-18b. Only these three forces act on the block. A Key Idea is that we can relate them to the block’s acceleration via Newton’s second law (), which we write as
Substituting 0 for the block’s acceleration 
yields
which says that the three forces are in equilibrium.
With two unknown vectors in Eq. 5-25, we cannot solve it for either vector directly on a vector-capable calculator. So, we must rewrite it in terms of components. We use a coordinate system with its x axis parallel to the plane, as shown in Fig. 5-18b; then and line up with the axes, making their components easy to find. To find the components of , we first note that the angle θ of the plane is also the angle between the y axis and the direction of (Fig. 5-18c). Component Fgx is then −Fg sin θ, which is equal to −mg sin θ, and component Fgy is then −Fg cos θ, which is equal to −mg cos θ.
Now, writing the x component version of Eq. 5-25 yields
from which
Similarly, for the y axis, Eq. 5-25 yields
(b) We now cut the cord. Does the block accelerate as it slides down the inclined plane? If so, what is its acceleration?
Solution: Cutting the cord removes force from the block. Along the y axis, the normal force and component Fgy are still in equilibrium. However, along the x axis, only force component Fgx acts on the block; because it is directed down the plane (along the x axis), that component must cause the block to accelerate down the plane. Our Key Idea here is that we can relate Fgx to the acceleration a that it produces with Newton’s second law written for x components (Fnet,x = max). We get
Fig. 5-18 (a) A block of mass m held stationary by a cord. (b) A free-body diagram for the block. (c) The x and y components of . The y component is directly into the plane. The x component is down the plane but can be drawn as part of a triangle (both ways are shown).
which gives us
Substituting known data then yields
The magnitude of this acceleration a is less than the magnitude 9.8 m/s2 of the free-fall acceleration because only a component of (the component directed down the plane) is producing acceleration a.
CHECK POINT 7 In the figure, horizontal force 
is applied to a block on a ramp. (a) Is the component of that is perpendicular to the ramp F cos θ or F sin θ? (b) Does the presence of increase or decrease the magnitude of the normal force on the block from the ramp?
In Fig. 5-19a, a passenger of mass m = 72.2 kg stands on a platform scale in an elevator cab. We are concerned with the scale readings when the cab is stationary and when it is moving up or down.
(a) Find a general solution for the scale reading, whatever the vertical motion of the cab.
Solution: One Key Idea here is that the reading is equal to the magnitude of the normal force on the passenger from the scale. The only other force acting on the passenger is the gravitational force , as shown in the free-body diagram of Fig. 5-19b.
A second Key Idea is that we can relate the forces on the passenger to his acceleration with Newton’s second law (). However, recall that we can use this law only in an inertial frame. If the cab accelerates, then it is not an inertial frame. So we choose the ground to be our inertial frame and make any measure of the passenger’s acceleration relative to it.
Because the two forces on the passenger and his acceleration are all directed vertically, along the y axis in Fig. 5-19b, we can use Newton’s second law written for y components (Fnet,y = may) to get
This tells us that the scale reading, which is equal to FN, depends on the vertical acceleration. Substituting mg for Fg gives us
for any choice of acceleration a.
(b) What does the scale read if the cab is stationary or moving upward at a constant 0.50 m/s?
Solution: The Key Idea here is that for any constant velocity (zero or otherwise), the acceleration a of the passenger is zero. Substituting this and other known values into Eq. 5-28, we find
This is the weight of the passenger and is equal to the magnitude Fg of the gravitational force on him.
(c) What does the scale read if the cab accelerates upward at 3.20 m/s2 and downward at 3.20 m/s2?
Fig. 5-19 (a) A passenger stands on a platform scale that indicates either his weight or his apparent weight. (b) The free-body diagram for the passenger, showing the normal force on him from the scale and the gravitational force .
For an upward acceleration (either the cab’s upward speed is increasing or its downward speed is decreasing), the scale reading is greater than the passenger’s weight. That reading is a measurement of an apparent weight, because it is made in a noninertial frame. For a downward acceleration (either decreasing upward speed or increasing downward speed), the scale reading is less than the passenger’s weight.
(d) During the upward acceleration in part (c), what is the magnitude Fnet of the net force on the passenger, and what is the magnitude ap,cab of his acceleration as measured in the frame of the cab? Does 
Solution: One Key Idea here is that the magnitude Fg of the gravitational force on the passenger does not depend on the motion of the passenger or the cab; so, from part (b), Fg is 708 N. From part (c), the magnitude FN of the normal force on the passenger during the upward acceleration is the 939 N reading on the scale. Thus, the net force on the passenger is
during the upward acceleration. However, his acceleration ap,cab relative to the frame of the cab is zero. Thus, in the noninertial frame of the accelerating cab, Fnet is not equal to map,cab, and Newton’s second law does not hold.
CHECK POINT 8 In this sample problem, what does the scale read if the elevator cable breaks so that the cab falls freely; that is, what is the apparent weight of the passenger in free fall?
In Fig. 5-20a, a constant horizontal force 
of magnitude 20 N is applied to block A of mass mA = 4.0 kg, which pushes against block B of mass mB = 6.0 kg. The blocks slide over a frictionless surface, along an x axis.
(a) What is the acceleration of the blocks?
Solution: We shall first examine a solution with a serious error, then a dead-end solution, and then a successful solution.
Serious Error: Because force is applied directly to block A, we use Newton’s second law to relate that force to the acceleration of block A. Because the motion is along the x axis, we use that law for x components (Fnet,x = max), writing it as
However, this is seriously wrong because is not the only horizontal force acting on block A. There is also the force 
from block B (Fig. 5-20b).
Dead-End Solution: Let us now include force by writing, again for the x axis,
(We use the minus sign to include the direction of .) Because FAB is a second unknown, we cannot solve this equation for a.
Successful Solution: The Key Idea here is that, because of the direction in which force is applied, the two blocks form a rigidly connected system. We can relate the net force on the system to the acceleration of the system with Newton’s second law. Here, once again for the x axis, we can write that law as
where now we properly apply to the system with total
Fig. 5-20 (a) A constant horizontal force is applied to block A, which pushes against block B. (b) Two horizontal forces act on block A: and the force from block B. (c) Only one horizontal force acts on block B: force 
from block A.
mass mA + mB. Solving for a and substituting known values, we find
Thus, the acceleration of the system and of each block is in the positive direction of the x axis and has the magnitude 2.0 m/s2.
(b) What is the (horizontal) force on block B from block A (Fig. 5-20c)?
Solution: The Key Idea here is that we can relate the net force on block B to the block’s acceleration with Newton’s second law. Here we can write that law, still for components along the x axis, as
which, with known values, gives
Thus, force is in the positive direction of the x axis and has a magnitude of 12 N.
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