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Reach out to learn more. Join our community of over 20, industry experts and subscribe to our newsletters to receive product announcements and offers. Find the tension in each wire, neglecting the masses of the wires. The system of interest is the traffic light, and its free-body diagram is shown in Figure c. 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 d.
There are two unknowns in this problem. It is reasonable that. There are two unknowns in this equation, but substituting the expression for. We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force.
Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples. Two tugboats push on a barge at different angles Figure. The first tugboat exerts a force of. 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. Strategy The directions and magnitudes of acceleration and the applied forces are given in Figure a. We define the total force of the tugboats on the barge as. 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. We also know that. Therefore, the net external force is in the same direction as.
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. 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. Figure shows a Calculate the scale reading: a if the elevator accelerates upward at a rate of. Strategy If the scale at rest is accurate, its reading equals. Figure a 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 b.
The only forces acting on the person are his weight. 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. The scale reading in Figure a is about 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:.
Thus, the scale reading in the elevator is greater than his N lb. 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. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary. Now calculate the scale reading when the elevator accelerates downward at a rate of. 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 the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless. Two Attached Blocks Figure shows a block of mass. 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.
We draw a free-body diagram for each mass separately, as shown in Figure. 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.
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. 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,.
From these two equations, we can express a and T in terms of the masses. 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. You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.
Calculate the acceleration of the system, and the tension in the string, when the masses are. 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 ,. 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. Solve the two equations simultaneously subtract them and the result is.
The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system,. This is found by substituting the equation for acceleration in Figure a , into the equation for tension in Figure b.
Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.
Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.
A soccer player starts at rest and accelerates forward, reaching a velocity of 8. To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. Distinguish between individual forces and net force.
Ask how the magnitudes of this pair of forces compare. As evidenced by the rest position, both mg and N must be equal and opposite. They cancel. The spool undergoes no change in motion vertically.
Next, ask what horizontal forces act while the spool is at rest. Again, the answer is none, as no change in motion occurs horizontally. Ask if a force of friction is acting on the spool at rest. This may elicit some class discussion, after which all agree that friction is zero—for the time being.
So when the spool is pulled horizontally there are two horizontal forces that act on the spool; the pull F to the right, and friction f to the left, which you sketch Figure 3 , right. At this point you can tell your class there is always, in physics, more than first meets the eye.
Motion of the spool across the lecture table is more than translational. Clearly, the spool also rotates. A torque is sort of a twisting tendency that changes rotational motion; it involves force and a particular distance between the line of force and the axis of rotation.
We call this distance that can produce leverage the lever arm. Pulling the string tends to topple the spool about its point of contact with the table, which is the point about which the spool rotates when it starts to roll. Can you see that the force exerted on the left-hand spool in Figure 4 has considerably greater leverage due to the longer lever arm shown by the green dashed line?
And, for the same amount of pull, does this produce a greater torque on the left-hand spool than on the right-hand spool?
Yes to both questions, because the lever arm distance for the left spool, D 1 , is longer than that for the right spool, D 2. No friction acts on the spool while it is at rest, but when the string is pulled, a force emerges that tends to prevent motion. This is static friction. When the force of the pull is sufficient, the spool begins to roll.
Once rolling, static friction is replaced by rolling friction. If the spool were to slide rather than roll, the friction would be kinetic friction. The force of kinetic friction is less than the maximum that can be reached by static friction, and rolling friction is even weaker than static friction. What makes this demonstration a learning experience is the predictions it elicits. Predictions by students, explanations in check-your-neighbor fashion, with some time before you present your explanations, is the mark of an effective physics class.
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