The 1.0 kg physics book in FIGURE P7.40 is connected by a string to a 500 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are μₛ = 0.50 and μₖ = 0.20.

b. At the highest point, does the book stick to the slope, or does it slide back down?

Question

The 1.0 kg physics book in FIGURE P7.40 is connected by a string to a 500 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are μₛ = 0.50 and μₖ = 0.20.

b. At the highest point, does the book stick to the slope, or does it slide back down?

Accepted Answer

Everyone in this problem, we have a figure showing a block of mass 2 kg attached to another block weighing 750 g through a massless string. The coefficient of static and dynamic friction between the block and the incline are 0.45 and 0.25 respectively. And the block is on an incline surface. And we're shown in the diagram that, that makes an 18 degree angle with the horizontal, it's pulled to the top of the plane. And at this point, we're asked to determine whether the block will remain steady or move downwards on the incline plane. OK. So we have four answer choices. It's gonna remain steady. Option B it's gonna move upwards. Option C, it's gonna move downwards. And option D is that we can't actually determine this. So let's get started on this problem. And we have our diagram, we're gonna give yourself some room to work. And the first thing we're gonna do is draw a free body diagram and think about the forces we have acted. OK. So for our 2 kg block, that's our block on the incline, it's resting on a surface. So we have the normal force that's gonna point perpendicular to that incline. We have the force of gravity pointing straight down which we can break up into its X and Y components on this tilted axis. Yes, we have FGX pointing to the right Fgy pointing downwards and then we have a force of friction that's gonna point to the left. OK. It's gonna oppose the motion. So if there was no friction at all, this block would slide down that incline to the right. And so the friction opposing that motion points to the left. Now for the block that's hanging, we have two forces to consider. We have the force of gravity acting downwards. We have the tension force from that string acting upwards. And in our first diagram, we missed, we have the tension force as well pointing to the right in the x direction for that 2 kg block on the incline. All right. So we have our free body diagrams. We have all of the forces written down. Now, what we want to think about. OK, is the static friction. OK. If the sum of all of the other direction forces is enough to overcome static friction, then our block will slide. If it's not enough to overcome static friction, our block will remain stationary. All right. So let's look at this maximum static friction force. OK. We're gonna call it Fs Max and the maximum static friction force recall is gonna be equal to us. Multiplied by N. So the coefficient of static friction multiplied by the normal force. All right, we know the coefficient of static friction. But what about this normal force? Well, if we go back to our free body diagram, OK, we're gonna say that this upwards direction perpendicular to the incline is our positive direction and to the right parallel to that incline is our positive X direction. OK. So we're gonna take that frame of reference and we can see that the sum of the forces in the Y direction on this tilted axis is going to be zero. OK. This block is not moving upwards or downwards relative to the incline, it's staying resting on that incline. OK. So if we look at the sum of the forces and let me do this in another color. If I look at the sum of the forces in the wide direction for this block, OK. We know that they're gonna be equal to zero. What forces do we have? Well, we have the normal force in the positive direction. We have the force of gravity, Y component in the negative direction. So we have N minus FGY is gonna be equal to zero. That tells us that this normal force is going to be equal to the Y component of the force of gravity. Now, the force of gravity is given by the mass multiplied by the acceleration due to gravity G. And because we're looking at the white component OK. We need to break our force of gravity up. We know that the angle with the vertical is going to be degrees because that's the angle that we make with the incline. OK. So related to this angle, we're looking at the adjacent side. So we wanna be using cosine, so we have MG multiplied by cosine of 18 degrees. All right. So if we get back to our maximum static friction equation, now we can write it as a coefficient of static friction, us multiplied by M multiplied by G multiplied by cosine of 18. OK. And we are dealing with the upper block here. So when I say m the mass, we're talking about the upper block. All right. So substituting in our values, the coefficient of static friction we're told is 0. the mass is 2 kg. The acceleration due to gravity 9.8 m per second squared, then we have cosine of 18 degrees. And when we work this out on our calculator, we can get a maximum static friction force of about 8.39 S. All right. So what this tells us is that the maximum static friction force is 8.39 nodes. If the other forces acting in the X direction in the opposite direction, trying to get this block to move are greater than this, then that block is gonna move. If they're less than that, then static friction is gonna be greater and this block is gonna remain at rest. So let's look at those other forces. OK? So the other forces acting in the extra direction, what are they? Well, we have the some of the forces other. OK. In the X direction is gonna be the tension T plus the force of gravity in the X direction. If we go up to our rebo diagram, we can see those two forces pointing to the right now the force of gravity in the extraction, we can calculate. We know that. But the tension, we're gonna need to find another way. The good news is that tension is the same tension that the lower block experiences because they're connected with the same string. So we can go over to the lower block, right, calculate the attention and then come back. So there's a lot of jumping around here. But the concept is still simple. The concept is the same is are the forces that are trying to pull the block down greater than the maximum static friction force or not. So for our lower block, we're looking at the sum of the forces in the wide direction. That's the only direction we have forces acting. And we know this is gonna be equal to zero if our block is not moving. OK. At the very beginning of this problem, our block is not moving. That moment is what we're looking at. Hm. All right. So we have our attention minus the force of gravity, FG that's gonna be equal to zero. That tells us that the tension is gonna be equal to the force of gravity, which we know is the mass multiplied by the acceleration due to gravity. And here when we're talking about mass, we're talking about the lower block. And so our tension is going to be equal to 0. kg. OK. We're told that the mass is 750 g. We want to use our standard unit of kilograms. So we divide by 1000 and that's multiplied by 9.8 m per second squared for attention of 7.35 news. All right. So we have this tension now and we can use that in our sum of the X direction forces. So we go back up to this equation. OK. Now the sum of our other forces is gonna be 7.35 newtons who FGX which is gonna be the mats. OK? 2 kg multiplied by the acceleration due to gravity 9.8. Let me do this where we have more room, we are running out of space. So the some of the other forces, all right, is that tension 7.35 nos who lost the force of gravity, which is the mass of 2 kg multiplied by the acceleration due to gravity 9.8 m per second squared multiplied by sine of 18 degrees. OK. When we were looking at the Y component. It was cosine for the X component it's going or yeah, for the X component it's going to be sine. All right, we work all this out and we get this other forces with a magnitude of 13.4 nodes. OK. All right. So what we wanna do is compare this value to the maximum static friction force. And we can see that the other forces, the forces acting downwards in that direction to the right are greater than the force of static friction, right? So the other forces or greater then the maximum static friction force which tells us that the block is going to and those forces are gonna be enough to overcome static friction. That block is gonna start to move and then we're gonna be dealing with kinetic friction. All right. So going back up to the top, we figured out that this block is going to move. OK. So option A remains steady. Nope. Um we can't determine it so we can eliminate. Option D the question now is move upwards or move downwards. And we already know the answer to that, hey, these forces are acting downwards. It's the force of gravity and that tension that are gonna pull this block to the right. So it's going to move downwards. And the correct answer here is option C Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Friction

Kinematics

Energy Conservation