In FIGURE CP7.54, find an expression for the acceleration of m₁. The pulleys are massless and frictionless.

Hint: Think carefully about the acceleration constraint.

Question

In FIGURE CP7.54, find an expression for the acceleration of m₁. The pulleys are massless and frictionless.

Hint: Think carefully about the acceleration constraint.

Accepted Answer

Everyone in this problem, we're asked to consider an arrangement of massless and frictionless pulley used to connect a bag weighing 4 kg to a block of 1.5 kg through a massless string as shown in the figure. OK. So we have this 4 kg bag hanging two pulley connecting it to this block with 1.5 kg that is resting on a frictionless surface. And we're asked to determine the magnitude of the acceleration of the 1.5 kg block. We have four answer choices all in meters per second squared. Option A 4.9 option B 7.8 option C 9.8 and option D 6.5. So we're looking for the acceleration of the block that's resting on the surface. We know that the acceleration is related to the forces through Newton's second law. OK. So let's go ahead and draw some free body diagrams and think about the forces we have acting in this case. All right. So we're gonna start with the big. OK. So we have our 4 kg bag and we have the force of gravity acting downwards. We're gonna call that FG for the bag, we have two forces acting upwards. OK? We have two tensions because this string comes down around this pulley and then back. So we have two tensions acting on this. We're gonna call those T one and T two. All right. So that's the bag. So this is the bag and then we have the block. And what does the free body diagram look like for the block? Well, it's resting on a surface. So we know we're gonna have a normal force pointing upwards. We have a force of gravity pointing downwards. So this is gonna be FG block and in this case, we have a tension force pointing to the left. OK. That's where the string is. And we're gonna call this tension force T three. All right. So we have our free body diagrams. We're gonna choose to the left and down to be our positive directions. This 4 kg bag heavier than a 1.5 kg bag. It's hanging. So we kind of expect that things are gonna be moving downwards until. So we're gonna pick that as a positive direction, but we're asked for the magnitude of the acceleration. So you can choose any direction you want to be positive as long as you're keeping it consistent throughout the problem. Now, if we look at these tension forces, we have T one, we have T two and we have T three seems like a lot to figure out. But we have a single string connecting these, this block to this bag. OK. So all three of those tensions are from the same string. So they must be equal and we have to have that consistent tension force throughout. But we don't have any breaking of this string. OK. So what we have is that T one is equal to T two is equal to T three. And we're gonna call all of them just T, OK. That's our tension force. And it's the same for all three of them. All right. So let's go ahead and try to write an equation for the block. OK. We wanna find the acceleration of the block. So let's go ahead and write an equation that gives us the acceleration of the block. And we're called the sum of the forces. OK? And in this case, in the X direction for our block is gonna be equal to the mass of the block multiplied by the acceleration in the X direction of the block. And we've chosen the X direction because again, that block is resting on a surface. So it's not moving up or down in the wide direction. The acceleration is going to be in the X direction in that horizontal direction. Yeah, for the sum of the forces, what forces do we have acting in the extraction? Well, the only one we have is the tension and T acting in the positive direction. So we get that the tension T is going to be equal to the mass of the block multiplied by the acceleration in the X direction of the block. Hey, and we're actually, we're just gonna call that a block. We're gonna drop that X because that's the only direction it's accelerating in. All right. So we want to find this acceleration, we know the mass of the block, but we don't know the tension team. OK. So in order to find the tension team, we are going to need or sorry, in order to find the acceleration, we're gonna need to find the tension. OK. Now, because these tensions are all the same, we can write an equation for the bag. OK. And relate those tensions and try to figure out what that tension T is using the bag information in order to substitute it it. So let's move over to the bag. We have the, the sum of the forces in the right direction for the bag, it's going to be equal to. OK. We have these two tensions acting in the negative direction and we have the force of gravity of the bag acting in the positive direction. OK. So on the left hand side, we have the force of gravity due to the back minus the first tension minus the second tension. And this is all equal to the mass of the bag multiplied by the acceleration of the bag K due to Newton's second law math of the bag acceleration of the bag. All right. So what can we do here? OK. How can we figure this out? We have two equations but we have three unknowns. OK? We don't know the acceleration of the block. We don't know the acceleration of the bag and we don't know the 10 to what we can do now is figure out our acceleration constraint. OK. So we want to find a relationship between the acceleration of the bank and the acceleration of the block in order to simplify this. And we're gonna do this by looking at our diagram. OK. If this block moves by a distance of D, so you can imagine it moving to the left a distance of D, then the bank is gonna move by a distance of one half of D. OK? Because this string leading to the bag is looped around the pulley. OK? So if this top block moves a distance of D, the bag is only gonna move by half of D again because of the way it's looped around this bully. Now the distance and the acceleration are proportional. OK? We recall it the distance and the acceleration have that second derivative. OK? Or second integral relationship depending on which way you're going. And so those two are proportional. OK. So if the distance is one half, then the acceleration is also going to be one half. OK. So our acceleration constraint is that the acceleration of the bag, it's only gonna be one half of the acceleration of the block. All right. Now, let's use this in our questions. Now, the first equation, the one for the block we end up with the tension T is going to be equal to OK. The mass of the block multiplied by the acceleration to, to the, we're not gonna change it. We want the acceleration due to a block. So we're not gonna do anything different there. We're gonna call this equation one and on the other side, we have the force of gravity due the bag, which we can write as the mass of the bag multiplied by the acceleration due to gravity G minus two, multiplied by the tension T gonna be equal to the mass of the ba multiplied by one half the acceleration due to the block. OK? And we're gonna call this equation too. Now, equation one is an equation for T. So what we're gonna do is we're gonna substitute equation one into equation two that will eliminate that T term. The only known is gonna be the acceleration due to the block and we can finally solve. All right. So when we do that, we get the mass of the bag multiplied by the acceleration due to gravity minus two, multiplied by the mass of the block multiplied by the acceleration of the block is equal to the mass of a bag multiplied by one half the acceleration of the block. And again, the acceleration of the block that is what we're looking for. And that is the only unknown. Now, we know all of these masses, we know the acceleration due to gravity. So let's simplify and isolate the acceleration of a block first, then we're gonna substitute in our numbers if you prefer to substitute the numbers in right now and then rearrange, you can do that as well either is fine. All right. So we're gonna move all of our acceleration of the block terms to the right hand side. And what we get when we factor is that the acceleration of the block multiplied by two mass of the block plus one half mass of the bag is going to be equal to the mass of the bag multiplied by the acceleration due to gravity G and we can divide by these masses. We get that the acceleration of the block is the mass of the bag multiplied by the acceleration due to gravity G divided by two multiplied by the mass block plus one half multiplied by the mass of the bag. So now we have our equation, our expression for the acceleration of the block that we were looking for, we can substitute in our values. And so we get that the acceleration of the block is going to be equal to 4 kg multiplied by 9. m per second squared divided by two multiplied by 1.5 kg plus one half multiplied by 4 kg. If we simplify this and work this out on our calculator. We get that the acceleration of this block is going to be 7.84 m per second squared. All right. So let's go back just to a quick refresh of what we did. OK. What we did is we drew our free body diagrams and wrote on our equations using Newton's second law for the some of the forces for both the bank and the block. OK. We wanted the acceleration of the block but in order to do that, we needed to solve for the tension team to solve for the tension. T we had to find the relationship between the acceleration of the bag in the block. Use that in our equation so that we had two equations and two a notes substitute one of our equations into the other and then solve for that acceleration that we were interested in. So be round to two significant digits. We can see that our answer is going to correspond with answer choice B 7.8 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

In FIGURE CP7.54, find an expression for the acceleration of m₁. The pulleys are massless and frictionless. Hint: Think carefully about the acceleration constraint.

Verified Solution

Key Concepts

Newton's Second Law

Acceleration Constraint

Massless and Frictionless Pulleys