🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning



Numerade Educator



Problem 39 Medium Difficulty

ssm A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?


447 $\mathrm{N}$
241 $\mathrm{N}$


You must be signed in to discuss.

Video Transcript

in this question. There are four forces acting on the creek, the weight force, the normal force. That is the force exerted by the floor on the great, the horizontal force that we will use to move the crate and the frictional force that tries to make the creates not moved. And we are interested in the horizontal movement off. Great. So we begin this question by applying Newton's second law to the horizontal axis. So let me choose my referential as being that everything that is pointing to the right, it's positive and everything that is pointing to the left is negative and you'll call the horizontal axis X, then Newton's second law. When apply it on that axis gives the following the net force is there close to the mass off the great times acceleration off the Crete. But the net force acting in the crate all that horizontal access is equals to the pliant force minus the frictional force. And these easy close to the mask off The great times acceleration. We want to calculate their force f So f is it goes to the mass off the create times acceleration of the great plus that frictional forces, but note that in both situations we are interested that create is not accelerating. It's either just starting to move, so it's not yet moving. So that's elation is it goes to zero or it is sliding with that constant velocity. So again, the acceleration is the question zero therefore, a easy close to zero, and these thousands that we apply it forced easy close to the frictional force of both situations. So on the first situation, then the force is equal to the frictional force that acts on the great when it's just starting to move. So if it's just starting to move, we have to use the static friction co officials to calculate the frictional force so the frictional force is equals to the static frictional coefficient serve you as times the normal force that is acting on the crate. But note that the normal force is equal to the weight force because they create on the vertical axis is not moving. It's standing still on the floor, meaning that the normal is the coast to the weight force. Therefore, we have that the frictional force is in close to the mute static times the weight force but the weight forced is the coast mass off the great times acceleration of gravity. Then the problem gives us the following values. So the static friction coefficient is equals to 0.76 The mass off the crate is the Costa 60 kilograms and we will consider that grabbing tonight Gravitational acceleration to be equals to 9.8 meters per second squared then the force that is neccessary for just starting to me of the great Izzy Kohstuh 0.726 times 16 times 9.8 which is approximately 447 Newtons. No. On the second item we do the same thing, the same calculation. But now we will use the kinetic friction local officials because of the second situation that create is already moving. So in the second situation, the necessary force is it goes to the frictional force in this magnetic situation. So this is it goes to the kinetic friction local officials times the weight force which is close to the mass off the great times Acceleration of gravity on the problem tells us that magnetic frictional coefficient is it close to 0.4 to 1. Therefore the neccessary. Forced to keep the great moving Waas it's moving is equals to 0.421 times 60 times 9.8, which is approximately 204 to 1 noodles. So note that keep moving. The crate is easier than making the great start moving. This is true for almost everything. You try to move.