What is the acceleration of the 3.0 kg block in FIGURE CP7.55 across the frictionless table?

Hint: Think carefully about the acceleration constraint.

Question

What is the acceleration of the 3.0 kg block in FIGURE CP7.55 across the frictionless table?

Hint: Think carefully about the acceleration constraint.

Accepted Answer

Hey, everyone in this problem, we're given a figure that depicts an arrangement in which a block of 5 kg is connected to a block of 1.5 kg through massless and frictionless pulleys based with the 5 kg block is sitting on a frictionless bench and the 1.5 kg block is hanging. And we're asked to calculate the magnitude of the acceleration of the upper box. We have four answer choices here all in meters per second squared. Option A 2.3 option B six, option C 4.5 and option D 2.7 we're gonna start by drawing some free body diagrams. Let's start with the hanging block. So the 1.5 kg block, we're gonna have a force of gravity acting downwards. OK? So we have FG and we're gonna call this FGL, this is gonna be for the lower block and then we have the tension acting upwards and we're gonna call this tension T one and that's it. Those are the only forces we have acting on that lower hanging block. Now let's move to our upper block that's resting on a surface and because it's resting on a surface, we know that there's going to be a normal force acting upwards. So we have nu the normal force of the upper block again, it has a force of gravity. So the force of gravity fgu upper block, if you look at the diagram, we actually have two tensions acting. OK. The rope or the um string comes around the pulley and then back out the other side. And so this 5 kg block is subject to two tensions and we're gonna call them T two NT three. All right. Let's choose some directions. So we're gonna choose to the left and down to be positive. And we want to find the magnitude of the acceleration. Let's recall that the acceleration is related to the force through Newton's Second law. OK? So we wanna think about these forces acting and we can do that because we are a free body diagram. Now, the first thing we're gonna do is deal with these tensions and we have T one, we have T two, we have T three. Well, these are all the same string, right? It's the same string that goes up and around these pulleys. In order for that not to break, we have consistent tension throughout the same or throughout the string. And we're gonna assume that it's a nice uniform string. So what that tells us is that T one is gonna be equal to T two, gonna be equal to T three and we're gonna just call all of them T so we have tension in the string and it's the same everywhere along that string. Now, let's look at the summer forces and we're gonna start with our lower block and the summer forces actually let's start with their upper book. The summer forces in the X direction on our upper block is going to be equal to the mass of the upper block multiplied by the acceleration of the upper block. And again, we're using you to be the subscript when we're talking about the upper block. Now the force is acting in the extraction. Well, we have T two and we have T three, those are both equal to T to T. So we just have two T. OK. So two T is equal to mu multiplied by A U. And we're gonna call this equation one we can see is we have this acceleration of the upper block that we're looking for A U an equation one, but we don't know the tension yet. And so we can't directly solve. So we're gonna write now are some of forces in the wide direction of the lower. OK. Now again, this is gonna be equal to the mass of the lower block ML multiplied by the acceleration of the lower block A L, some of the forces. Well, in the positive direction, we have the force of gravity acting. So FGL in the negative direction, we have our attention T so FGL minus T is equal to the mass ML multiplied by the acceleration A L. Now, we have a bit of a problem in our first equation. We have two unknowns A U and T. In our second equation, we also have two unknowns T and A L. So we have two equations with three unknowns. That's too many to solve for what we can do though is recall our acceleration constraint and we need to determine what the acceleration constraint is. That's gonna be a relationship between the acceleration of the upper block, the acceleration of the lower block, we can find a relationship between those two and write one in terms of the other. Then we'll just have two equations and two unknowns which we do know how to solve. OK. So let's take a look at our diagram. Now you can imagine that if the lower block moves by a distance of D, then the upper block is gonna be moved by a distance of D divided by two. Maybe because the string goes up and around that pulley. Now, distance and acceleration are proportional. I recall that they're related through the integral or the derivative depending on which direction you're going. And so distance and acceleration are proportional. That means that if the distance is only moving by half, then the acceleration will do the same. Mm So let's write this up. We have an acceleration constraint that tells us that the acceleration of the upper block is gonna be half the acceleration of the lower block. And again, we figured that out because if the upper block moves or sorry, the lower block moves by a distance deep, the upper block moves by a distance of half. Ok. So the acceleration is going to be and we can write this as well as the acceleration of the lower block is two times the acceleration of the upper. Yeah. All right. Now, the upper block acceleration, the magnitude of the acceleration of the upper block is what we are looking for. OK. So we wanna leave a UN R equation. So we are going to substitute A L equals two A U into equation two. And we're gonna rearrange and what we're gonna have is that the tension T we're gonna call this equation two star. We have that the tension T is gonna be the force of gravity of the lower block FGL minus ML multiplied by two A E. OK. So what we've done is we've taken equation two, we've substituted A L is equal to two A U. And then we've just rearranged to solve for T and we're gonna call that equation to star. And now we have equation one, we have equation two star, they each have two, we or sorry, we have two equations with two unknowns T and A U. Now we can really solve this thing. OK? So that acceleration constraint is key in these problems. So let's go ahead and do it. We're gonna substitute equation two star into equation one. We have two multiplied by FGL minus ML multiplied by two A U is equal to mu multiplied by A U. Let's expand and solve for A U isolate it before we substitute in our values. If you wanted to substitute in your values at this point, you can go ahead and do that as well. So expanding, we have two multiplied by FGO minus four, multiplied by Mlau is equal to MU A. You, we're gonna add four ML multiplied by A U to both sides. And then we're gonna factor out that A U value. So we have a U multiplied by MU plus four. ML is equal to two. We have the force of gravity of the lower block and we're called that the force of gravity is given by the mass. So ML multiplied by the acceleration due to gravity G. OK. So two MLG on the right hand side, one more step to isolate for A U, we need to divide by the M uh Yeah, to divide by mu plus four MF. So we have A U is equal to two multiplied by ML, multiplied by G divided by mu plus four ML. Now we just need to substitute in our values. We were given the mass of both blocks. We know the acceleration due to gravity. We have everything we need to solve. So we get that the acceleration of the upper block is going to be equal to two multiplied by the mass of the lower block. 1.5 kg multiplied by the accelerate or by the yeah, the acceleration due to gravity 9.8 m per second squared divided by the mass of the upper block 5 kg co four multiplied by the mass of the lower block 1.5 kg. When we work this out, when we get A U is going to be 29.4 kilogram meters per second squared divided by 11 kilograms. The unit of kilograms will divide out and we're gonna be left with an acceleration of about 2.67 27 repeated meters per second squared. And that is the magnitude of the acceleration of the upper block that we were looking for. So let's do a quick recap here. We wanted to find the acceleration. We drew out our free body diagrams and use Newton's second law to relate the acceleration to the forces. We have one string connecting all of these blocks. And so the tension was the equal throughout that entire string. We ended up with two equations and three unknowns. And in order to simplify it down to something we can solve, we had to look at an acceleration constraint. Once we did that, we could reduce it down to two equations, two unknowns and solve for the acceleration. Now comparing this to our answer choices, rounding to one decimal place. We can see that the correct answer is going to be option D 2.7 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

What is the acceleration of the 3.0 kg block in FIGURE CP7.55 across the frictionless table? Hint: Think carefully about the acceleration constraint.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Frictionless Surface

Acceleration Constraint