Everyone, welcome back. So in this practice problem, we have these two stacked blocks on top of each other in this diagram. Basically, we have block A that's sort of resting on top of block B, but I'm pulling block B to the right with a force of 45 Newtons. There's a twist to this one because instead of the whole system moving, what happens is that block A on the top is actually tied to the wall with some kind of a rope. We're supposed to figure out what the tension force is on block A. We are given some masses and some coefficients of friction. So let's just jump right in and stick to our steps. We're going to draw free body diagrams and choose our direction of positive, like we've done a bunch of times before.
Let's start out with block A. The free body diagram is going to be the weight force, so this is going to be mAg, and there's also a normal force because it's resting on top of B. In other words, that's the normal force from B on A. There's a tension force that's holding it to the wall, this T over here. If these were the only three forces acting, the block would accelerate in this direction. However, that doesn't make any sense because block A is on top of block B. As you pull block B to the right, why would block A go to the left? That makes no sense. So there must be some kind of a force that's sort of keeping it from moving off to the right, and that force is actually friction. As you pull B, you're going to cause these two surfaces to slip. So block A needs a force to prevent that slipping, which is going to be a force this way, that's the force of friction trying to prevent that motion.
There are no other forces on block A. Let's take a look at block B, which is a little bit more complicated. We start off with the weight force, which is going to be mBg. There's also an action-reaction pair from the normal force; B pushes up on A, so that means A has to push down on B. This is going to be the normal force from A on B. That means the normal force that's on B from the ground has to support both of those things, which is going to be NB. Those are all the vertical forces. For the horizontal forces, we know that we're starting off with an applied force of 45 Newtons. There are actually two forces we have to consider on the left; block A on B is going to exert a friction force in this direction. This is going to be the friction force of B on A. But that means, because of action-reaction, there has to be another force that's acting from A on B. So, these are basically just the action-reaction pair of the friction force between the two surfaces.
Now, there's actually one more friction force to consider. We're told here that the coefficient of kinetic friction between all of the surfaces is 0.2. Thus, there's another friction force, which we'll denote as FB, and it's really the force on block B from the ground. So there are two friction forces here. Now that we know that these frictions are all kinetic frictions, let's move on to the third step. We're going to write F = m a starting with block A and look at forces in the X-axis. The equation here turns out to be FBAk - T = mA × a. However, given that the rope is holding it, the acceleration is actually zero, meaning FBAk = T. With this, knowing that the friction force is μ k × NBA, and that NBA is equal to mAg, the weight force, the tension force can be found by substituting these values. Plugging in the numbers, T equals 0.2 × 5 kg × 9.8 m/s², which calculates to 9.8 Newtons. So, the tension force keeping block A attached to the wall is 9.8 Newtons.
Thank you for watching, and I'll see you in the next tutorial.