Blocks of mass m₁ and m₂ are connected by a massless string that passes over the pulley in FIGURE P12.64. The pulley turns on frictionless bearings. Mass m₁ slides on a horizontal, frictionless surface. Mass m₂ is released while the blocks are at rest. a. Assume the pulley is massless. Find the acceleration of m₁ and the tension in the string. This is a Chapter 7 review problem.

Question

Diagram showing a pulley system with 3.5 kg and 3.0 kg blocks for a physics problem on acceleration.

Accepted Answer

Hey, everyone in this problem, a box of mass three kg is attached to an iron ball with a mass of 3.5 kg through a massless string passing over a frictionless pulley. We're told to consider the system is initially at rest. When that iron ball is suddenly released, we're asked to determine how much tension is developed in the string and also the acceleration experienced by the box. We're given four answer choices options A through D each with a different combination of accelerations and tensions. And we have two possible accelerations, 5.3 m per second squared or 4.5 m per second squared and two possible tensions. 14 newtons and 16 newtons. We've shown our diagram here. So we have our iron ball of 3.5 kg attached through this pulley k, that iron ball is hanging attached to this pulley. And we have our block of three kg resting on a surface. We're asked to determine the tension and the acceleration. Let's go ahead and start with free body diagrams and we know that tension is a force. So we're gonna be working with forces here. So for the iron ball we have the tension acting upwards. We're gonna call that t we have the force of gravity acting downwards. And we're gonna use a subscript I to indicate that we're talking about the iron ball. So we have F G I pointing downwards for the box. Uh We have a normal force pointing upwards and we're gonna use subscript B to indicate we're talking about the box. So we have N B pointing upwards, we have the force of gravity pointing downwards, the tension force pointing to the left. And you'll notice that I've used the same tension teeth in both of these three body diagrams. And that's because the tension is going to be the same, gonna be in a different direction. The magnitude of that tension is going to be the same and this is one single string if the tension was different in either parts of these strings and you could get that string breaking or something like that. OK. So the tension is going to be equal. And that's really key to this problem. We're gonna take up into the right to be our positive directions. And what we want to calculate is first that tension and then the acceleration. So let's start with our iron bowl if we look at the sum of the forces in the wide direction of the iron ball, and Newton's second law tells us that this is gonna be equal to the mass of that iron ball multiplied by its acceleration. Well, with some of the forces in the wide direction, that's gonna be the tension T minus the force of gravity of the iron ball. And because the attention is acting in the positive direction, the force of gravity is acting in the negative direction and this is equal to the mass M I multiplied by the acceleration A I. Now because we know these tensions are equal in our other two equations we're actually gonna solve for the tension. And what you'll notice is that we have the tension T that's unknown. But we also have this acceleration that is unknown. And we wanna calculate both of those. We know that the tensions are equal. We've already said that. So let's start by solving for the tension T. In this equation, we get that the tension T is gonna be equal to the mass M I multiplied by the acceleration A I plus the force of gravity which recall is equal to the mass multiplied by the acceleration due to gravity G. And we're gonna call this equation one. Now again, we have two unknowns here, the tension and the acceleration A I. So what we wanna do is write an equation for our second object which, which is that box, so that we can have two equations with two unknowns and substitute them in. So we have the, the sum of the forces in this case in the X direction of the box. And we're looking at the X direction because that's where the tension acts is gonna be equal to the mass of that box multiplied by the acceleration of the box. We have the tension acting to the left which is on our negative direction. And so we get negative teeth is equal to the mass of the box multiplied by the acceleration of the box. And this is gonna be equation two. Yeah. So on the left, we have T is equal to something on the right, we have negative T is equal to something again, the tension is the same because we have a massless string and a frictionless pulley. And so we're gonna substitute equation one into equation two. When we do this, we're not gonna have the tension in our equation anymore. So we're actually gonna be solving for the acceleration first, but we can come back to the tension and you could do this in either order whether you wanna do the acceleration first or the tension first. We've just chosen to write our equations in terms of that tension T, all right. So if we substitute these in, we get the M I multiplied by A I four S M I multiplied by G is equal to negative M B multiplied by A B. OK. So we've just written the ne negative on the right hand side instead of on the left hand side just to make it a little bit simpler because we have only one term on the right hand side. Now, what you'll notice is we have two accelerations. We have A I and we have A B. So how can we solve for the acceleration? All the good news is that these accelerations will be the same as well and the magnitude of the acceleration will be the same and again, like the tension they're acting in different directions. OK. The iron ball is gonna be falling accelerating downwards, the block or the box is gonna be moving to the left. So accelerating to the left, however, the magnitude will be the same. OK. And this is also really important in this problem. And so we have that A I is gonna be equal to A B and we're just gonna call both of those A, that's gonna be our acceleration, whether we're talking about the iron ball or the box, it's the same thing. We call it A. So we can write that M I multiplied by A plus M I multiplied by G is equal to negative M B multiplied by A we want to solve for A. OK. That's part of the question solving for the acceleration. And now that's the only unknown we have, we know the masses and we know the acceleration due to gravity. So let's move our M I G term to the right hand side, we're gonna move our M B A term to the left hand side and we're gonna factor out that A as well. On the left hand side, we have a multiplied by M I plus M B. And on the right hand side, we have negative M I G. It tells us that the acceleration A is gonna be equal to negative M I G divided by M I M B. And you can substitute in your values at any point. I'm choosing to substitute them at the very end. So we're not getting any round off error. But if you want to substitute the values in and write these in terms of numbers all the way through before that final calculation, you can do that as well. OK. So our acceleration A is going to be negative the mass of the iron ball 3.5 kg multiplied by the acceleration due to gravity 9.8 m per second squared, all divided by the sum of the mass three kg plus 3.5 kg. And this gives us an acceleration a of negative five 0.2 seven meters per second squared. If we go to our answer choices, the accelerations are positive, this is based on the direction we've chosen. What we're really interested in here is that magnitude of the acceleration. So we have 5.2769 m per second squared, which is going round a 5.3 m per second squared. So we can eliminate option B and C because they do not have that same acceleration. And now let's move to the tension. And recall from equation two, OK, we could substitute into equation one or 22 is a little bit simpler. So let's use equation two. So from two, we had the, the negative tension is equal to the mass of the box multiplied by the acceleration. So the tension is gonna be negative and B three kg multiplied by the acceleration negative 5.2769 m per second squared. And we get a tension t of 15.8307 newtons approximately. OK. So now we have our acceleration, we have our attention and we can choose the correct answer. Looking back at our answer choices, we've already eliminated B and C because they had the incorrect acceleration, we found a 10 of 15.83 newtons rounding this to the nearest noon. We get the correct answer is option D. Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Newton's Second Law

Tension in a String

Free Body Diagram