The second image has background info used to help solve the practice problem. andadeip-enaltWSEXAMPLE 5.8 Delivering the goodsIn this example we will calculate the coufficients of static and kinede friction using Equations 5.1 and 5.2along with Newton's second law of motion. Suppose that a delivery company has just unloaded a 500 Ncrate full of home exercise equipment in your level driveway. You find that to get it started moving towardyour 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?SOLUTIONSET UP As shown in Figure 5.16, the forces acting on the crate areits 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. Wedraw 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 opposesyour pull because it tends to prevent the crate from sliding relative tothe surface. An instant before the crate starts to move, the static-frictionforce has its maximum possible value, fs,max= n. The state of restand 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 writeNewton's second law in component form, F, = 0 and ΣF, = 0, forthe crate. ThenEF, = n + (-w) = n 500 N = 0,EF=T + (-fs) = 230 Nfs = 0,fs, max = μgn(motion about to start),fs, maxμg =n230 N500 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 inequilibrium. Thus, the vector sum of the forces is still zero, and we haveΣF, = n + (-w) = n - 500 N = 0,ΣF, = T + (-fk) = 200 Nfk = 0,(moving).fk = μknn = 500 N,fk = 200 N,(a) Pulling a crateA FIGURE 5.16s,max T-230 N-X15(b) Free-body diagramfor crate just before itstarts to movepk =The coefficient of kinetic friction isfkw=500 Nn200 N500 NVideo Tutor SolutionYfkT-200 N-X= 0.40.w=500 N(c) Free-body diagramfor crate moving atconstant speedREFLECT There is no vertical motion, so ay = 0 and n = w. It iseasier to keep the crate moving than to start it moving from rest becausekg. 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 todelivery, so that s = 0.30 and k = 0.25. What horizontal force isrequired to start the crate moving? What force is required to keep itsliding 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 isrequired to keep it sliding at constant velocity? Answers: 150 N, 125 N.

Question

The second image has background info used to help solve the practice problem. andadeip-enaltWSEXAMPLE 5.8 Delivering the goodsIn this example we will calculate the coufficients of static and kinede friction using Equations 5.1 and 5.2along with Newton's second law of motion. Suppose that a delivery company has just unloaded a 500 Ncrate full of home exercise equipment in your level driveway. You find that to get it started moving towardyour garage, you have to pull with a horizontal force of magnitude 230 N. Once the crate &#34;breaks loose&#34;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?SOLUTIONSET UP As shown in Figure 5.16, the forces acting on the crate areits 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. Wedraw 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 opposesyour pull because it tends to prevent the crate from sliding relative tothe surface. An instant before the crate starts to move, the static-frictionforce has its maximum possible value, fs,max= n. The state of restand 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 writeNewton's second law in component form, F, = 0 and ΣF, = 0, forthe crate. ThenEF, = n + (-w) = n 500 N = 0,EF=T + (-fs) = 230 Nfs = 0,fs, max = μgn(motion about to start),fs, maxμg =n230 N500 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 inequilibrium. Thus, the vector sum of the forces is still zero, and we haveΣF, = n + (-w) = n - 500 N = 0,ΣF, = T + (-fk) = 200 Nfk = 0,(moving).fk = μknn = 500 N,fk = 200 N,(a) Pulling a crateA FIGURE 5.16s,max T-230 N-X15(b) Free-body diagramfor crate just before itstarts to movepk =The coefficient of kinetic friction isfkw=500 Nn200 N500 NVideo Tutor SolutionYfkT-200 N-X= 0.40.w=500 N(c) Free-body diagramfor crate moving atconstant speedREFLECT There is no vertical motion, so ay = 0 and n = w. It iseasier to keep the crate moving than to start it moving from rest becausekg. 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 todelivery, so that s = 0.30 and k = 0.25. What horizontal force isrequired to start the crate moving? What force is required to keep itsliding 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 isrequired to keep it sliding at constant velocity? Answers: 150 N, 125 N.

Accepted Answer

Answered: Image /qna-images/answer/178e2cdc-23d9-4907-83aa-8fd9e9f8fcff.jpg

- 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.