A 25.0-kg box of textbooks rests on a loading ramp that makes an ...

Question

A 25.0-kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (b) At this angle, find the acceleration once the box has begun to move.

Accepted Answer

Hey everyone in this problem, We have a 13 kg wooden block that rests on a ramp. Okay, the block is going to slide down the incline when the minimum incline plane angle is theta. The friction coefficients are the static friction coefficient 0. and kinetic friction coefficient 0.29. And were asked to determine the value of theta and then compute the acceleration at this angle once the block begins moving. Okay, alright, so let's just draw a little picture of what's going on here, we have this ramp inclined at some angle feta, we have our block, 13 kilograms. Okay, it's rusting there and then at some point when we compute the acceleration, it's going to be sliding downwards, let's go ahead and draw a free body diagram related to this block. Okay, so we have our block drawn here, we have the normal force which always points up, and when we say up, I mean perpendicular to the surface. Okay, so it's going to be pointing up in this direction, Okay, perpendicular to the surface that the block is on, this is our normal force. We have a weight pointing down always. And then in this case we have the friction force and the friction force is going to try to resist the motion. Okay, so that block is gonna be sliding down the ramp kind of to the right, which means that the friction force is going to be acting this way to the left. All right, there are two ways that we can kind of go about this. Okay, we can choose our coordinate system to be X and Y. Regular up and down vertical and horizontal. Okay, So that the weight would be in that coordinate system and then F and N. We would have to pick components from or we can choose our coordinate system to be tilted just like the incline. Okay. Like this. So we have X and y in our coordinate system and we're gonna go ahead and do that. And then all we have to do is decompose this weight. W so we can decompose it into the weight in the Y direction and the weight in the extraction. Alright, so that's our free body diagram. Now the first thing we're asked to find is the value of feta right before this block begins to me. Okay, so that minimum angle where that block is gonna start to slide. Now we know that this is gonna be data critical. Okay, stack Okay, so this is our critical angle. We have static friction because the block is at rest. This is just before it's going to start to slide. And recall that this angle theta is given by tan inverse of um U. S. A. The coefficient of static friction. Well this is gonna be tan inverse of. We're told the coefficient of static friction is 0.41. So 0.41. And this gives us an angle of 22.3°. Okay and so the value of theta. That minimum incline plane angle for the block to slide is going to be 22. degrees. Okay. And now we're asked to find the acceleration of the block when we look at the block, when we're talking about its acceleration as it slides down that ramp, that's gonna be in the X direction. Okay. Based on our coordinate system. So we want to look at the sum of the forces in that X direction. Okay, so let's give ourselves some more room. I want to try to leave the free body diagram up. We have the sum of the forces in the X. Direction. Okay? And we know from Newton's Law we have the sum of the forces in the X direction is going to be the mass times the acceleration. Okay, We'll write the acceleration A. X. In the X. Direction. Alright, so what forces do we have in the extraction? Well, in the positive X direction we have this weight W. X. And then in the opposite direction we have the friction force that's resisting case we're gonna have negative F. And in this case the block is sliding. So we're gonna be talking about kinetic friction. So we have F. Sub K. Okay, and on the right hand side we still have a mess times the acceleration in the X direction. Alright, so our weight in the X direction when we know that our weight W is equal to the mass times gravitational acceleration. Mg. Okay, so our way in the X direction is going to be M G M sign of data, our friction force or kinetic friction force recall that this is equal to mu k coefficient of kinetic friction times a normal end. We know the mass M were given in the problem. Gravitational acceleration is a constant. We know we know theta because that's the theta value we just found we're told the U. K. And the problem but we don't know is N. Okay, we're trying to find a X. We don't know yet. So let's just go over to the right side here and see if we can figure out. Okay. And we'll see that N In our free body diagram is in the Y direction. Ok. That's a Y component. So, let's look at the sum of the forces in the Y direction. Ok. Just like in the X direction that some of the forces in the Y direction. Newton's law tells us is equal to mass times acceleration in the Y direction and this is going to be equal to zero. Okay, we have no acceleration in the Y direction because we have no movement in the Y direction. All right, so some of the forces in the positive Y direction, we have the normal force and and in the negative Y direction we have the weight W. And the Y component of it and these are equal to zero. Okay, that means that our normal force end is just going to be equal to w Y. The weight. The Y component of the weight which is equal to M G. Course of theta. Alright, so now we can come back to our equation in the X direction. Okay, Because we know what end is now. So our equation becomes MG sign of theater minus mu K. The normal which we found to be M G cosign Theta is equal to M A X. All right now we have Mass here in every term. You can eliminate that. Let's just go ahead and plug in all of the numbers so we can see how this works out um with the units and everything. So we're gonna have 13 kg G is 9.8 m per second squared. We have sine of the angle 22.3 degrees minus mu K. We're told. And the problem is 0.29 Hey Times M 13 kg G 9.8 m/s squared Times. Close of the angle 22.3° is equal to 13 kg times the acceleration. A. X. All right. On the left hand side we have the units kilogram meter per second squared. Okay, we get 14.16 kilogram meter per second square. Okay, on the right hand side we have 13 kg times a. X. And when we divide by 13 kg we're going to get AX. Is equal to the unit of kilogram cancels were left with meters per second squared, which is a unit we want for acceleration. We get 1.89 m per second squared. Ok. And we can approximate that to 1. meters per second squared. If we go back up to our answer choices, we see that we have answer A that critical angle. Theta we found was 22.3 degrees and the acceleration once the block is moving is 1.9 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

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A 25.0-kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (b) At this angle, find the acceleration once the box has begun to move.

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