Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
- 133 Practice Problem 5.8: ose you wet down your driveway prior to delivery, so that u,-0.30 and p = What horizontal force is required to start the crate moving? What force is red to keep it sliding at constant velocity? Answers: 150 N, 125 N.

- 133 Practice Problem 5.8: ose you wet down your driveway prior to delivery, so that u,-0.30 and p = What horizontal force is required to start the crate moving? What force is red to keep it sliding at constant velocity? Answers: 150 N, 125 N.

icon
Related questions
Question
The second image has background info used to help solve the practice problem.
and ade ip- en alt W S EXAMPLE 5.8 Delivering the goods In this example we will calculate the coufficients of static and kinede friction using Equations 5.1 and 5.2 along with Newton's second law of motion. Suppose that a delivery company has just unloaded a 500 N crate full of home exercise equipment in your level driveway. You find that to get it started moving toward your garage, you have to pull with a horizontal force of magnitude 230 N. Once the crate "breaks loose" and starts to move, you can keep it moving at constant velocity with only 200 N of force. What are the coef- ficients of static and kinetic friction? SOLUTION SET UP As shown in Figure 5.16, the forces acting on the crate are its weight (magnitude w), the force applied by the rope (magnitude 7), and the normal and frictional components (magnitudes n and f, respec- tively) of the contact force the driveway surface exerts on the crate. We draw free-body diagrams for the crate at the instant it starts to move (when the static-friction force is maximum) and while it is in motion. SOLVE Whether the crate is moving or not, the friction force opposes your pull because it tends to prevent the crate from sliding relative to the surface. An instant before the crate starts to move, the static-friction force has its maximum possible value, fs,max= n. The state of rest and the state of motion with constant velocity are both equilibrium con- ditions. Remember that w, n, and f are the magnitudes of the forces; some of the components have negative signs. For example, the magni- tude of the weight is 500 N, but its y component is -500 N. We write Newton's second law in component form, F, = 0 and ΣF, = 0, for the crate. Then EF, = n + (-w) = n 500 N = 0, EF=T + (-fs) = 230 Nfs = 0, fs, max = μgn (motion about to start), fs, max μg = n 230 N 500 N = 0.46. n = 500 N. fs = 230 N, After the crate starts to move, the friction force becomes kinetic friction (Figure 5.16c). The crate moves with constant velocity, so it is still in equilibrium. Thus, the vector sum of the forces is still zero, and we have ΣF, = n + (-w) = n - 500 N = 0, ΣF, = T + (-fk) = 200 Nfk = 0, (moving). fk = μkn n = 500 N, fk = 200 N, (a) Pulling a crate A FIGURE 5.16 s,max T-230 N -X 15 (b) Free-body diagram for crate just before it starts to move pk = The coefficient of kinetic friction is fk w=500 N n 200 N 500 N Video Tutor Solution Y fkT-200 N -X = 0.40. w=500 N (c) Free-body diagram for crate moving at constant speed REFLECT There is no vertical motion, so ay = 0 and n = w. It is easier to keep the crate moving than to start it moving from rest because kg. Note, however, that just as the crate begins to move, the ac- celeration is nonzero while the velocity changes from zero to its final, constant value. Practice Problem: Suppose you wet down your driveway prior to delivery, so that s = 0.30 and k = 0.25. What horizontal force is required to start the crate moving? What force is required to keep it sliding at constant velocity? Answers: 150 N, 125 N.
Transcribed Image Text:and ade ip- en alt W S EXAMPLE 5.8 Delivering the goods In this example we will calculate the coufficients of static and kinede friction using Equations 5.1 and 5.2 along with Newton's second law of motion. Suppose that a delivery company has just unloaded a 500 N crate full of home exercise equipment in your level driveway. You find that to get it started moving toward your garage, you have to pull with a horizontal force of magnitude 230 N. Once the crate "breaks loose" and starts to move, you can keep it moving at constant velocity with only 200 N of force. What are the coef- ficients of static and kinetic friction? SOLUTION SET UP As shown in Figure 5.16, the forces acting on the crate are its weight (magnitude w), the force applied by the rope (magnitude 7), and the normal and frictional components (magnitudes n and f, respec- tively) of the contact force the driveway surface exerts on the crate. We draw free-body diagrams for the crate at the instant it starts to move (when the static-friction force is maximum) and while it is in motion. SOLVE Whether the crate is moving or not, the friction force opposes your pull because it tends to prevent the crate from sliding relative to the surface. An instant before the crate starts to move, the static-friction force has its maximum possible value, fs,max= n. The state of rest and the state of motion with constant velocity are both equilibrium con- ditions. Remember that w, n, and f are the magnitudes of the forces; some of the components have negative signs. For example, the magni- tude of the weight is 500 N, but its y component is -500 N. We write Newton's second law in component form, F, = 0 and ΣF, = 0, for the crate. Then EF, = n + (-w) = n 500 N = 0, EF=T + (-fs) = 230 Nfs = 0, fs, max = μgn (motion about to start), fs, max μg = n 230 N 500 N = 0.46. n = 500 N. fs = 230 N, After the crate starts to move, the friction force becomes kinetic friction (Figure 5.16c). The crate moves with constant velocity, so it is still in equilibrium. Thus, the vector sum of the forces is still zero, and we have ΣF, = n + (-w) = n - 500 N = 0, ΣF, = T + (-fk) = 200 Nfk = 0, (moving). fk = μkn n = 500 N, fk = 200 N, (a) Pulling a crate A FIGURE 5.16 s,max T-230 N -X 15 (b) Free-body diagram for crate just before it starts to move pk = The coefficient of kinetic friction is fk w=500 N n 200 N 500 N Video Tutor Solution Y fkT-200 N -X = 0.40. w=500 N (c) Free-body diagram for crate moving at constant speed REFLECT There is no vertical motion, so ay = 0 and n = w. It is easier to keep the crate moving than to start it moving from rest because kg. Note, however, that just as the crate begins to move, the ac- celeration is nonzero while the velocity changes from zero to its final, constant value. Practice Problem: Suppose you wet down your driveway prior to delivery, so that s = 0.30 and k = 0.25. What horizontal force is required to start the crate moving? What force is required to keep it sliding at constant velocity? Answers: 150 N, 125 N.
Page 133 Practice Problem 5.8: Suppose you wet down your driveway prior to delivery, so that µ-0.30 and p- 0.25. What horizontal force is required to start the crate moving? What force is required to keep it sliding at constant velocity? Answers: 150 N, 125 N.
Transcribed Image Text:Page 133 Practice Problem 5.8: Suppose you wet down your driveway prior to delivery, so that µ-0.30 and p- 0.25. What horizontal force is required to start the crate moving? What force is required to keep it sliding at constant velocity? Answers: 150 N, 125 N.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer