Determine a formula for the acceleration of the system shown i...

Question

Determine a formula for the acceleration of the system shown in Fig. 4–49 (see Problem 55) if the cord has a non-negligible mass m_C. Specify in terms of ℓ_A and ℓ_B , the lengths of cord from the respective masses to the pulley. (The total cord length is ℓ_A + ℓ_B.)

Diagram of a pulley system with two masses, mA on a slope and mB hanging, illustrating vertical forces and acceleration.

Accepted Answer

Hi, everyone. Let's take a look at this practice while dealing with Newton's laws. During this problem, we have a wooden block of mass M one and it's placed on a horizontal frictionless table and it's connected um by a cord over a pulley to a hanging nylon block of mass M two. The cord that is connecting the system, it has a mass of MC and that's the cord that passes over the pulley. The chord is divided into two segments with length L one on the table side and length L two on the hanging side. So the total length of chord is L equals L one plus L two. The question wants us to drive an equation uh to determine the acceleration a of the system. We're given an image of what was described here in the question. We're also given four choices and we'll actually come back to those choices at the end once we've done our calculation. So let's begin by looking at a free by diagram and we're gonna do things a little bit differently. So we're going to look at a free by diagram for the combination of M one and the length, accord L one that is on the table side of the court, we're gonna look at the forces acting on that system. So if I look at that system, um I can draw here actually on my diagram with the free body diagram, I'm going to draw the tension which is acting to the right. So it's gonna be the force of tension and that's the only force that I have acting on that in the X direction. Now do have forces in the Y direction, but they all cancel out. So they're not going to contribute to our acceleration. We also define a coordinate system so that the positive X direction is going to the right. So what I'm going to do is write down the Newton's second law for this combination of the mass and the chord. And in doing so I'll have the net force, the X direction and that's just gonna be equal to the force of tension that's going in the positive X direction. And this is gonna be equal to the total mass of the system. So it's gonna be M one plus the mass of the cord that is on the table side of the pulley. So what I'm going to do is take my total mass of the chord that's gonna be MC and multiply it by the fractional length that I have on the table side of the pulley. So that's gonna be L one divided by L. So it's gonna be the fraction of the mass that is on that side. And so I'm gonna multiply um that quantity. So the M one plus MC L one divided by L that's all gonna get multiplied by our acceleration A. And this is as far as I can go with this equation. But what it does tell me is what my um tension force is equal to in terms of the masses and the length of the board and the acceleration, I'm gonna do something similar for mass M two and the length of the cord that's hanging vertically, L two. So I'm gonna treat that as a single system. So it's gonna have two forces acting on it. It's gonna have the tension force pulling up, it'll be ft and that's the same tension force that's in the horizontal direction. And we also have the weight going downward here. It's gonna be the combined weight due to the mass of M two and the mass of the cord that's hanging. So I'll have M two plus MC multiplied by L two over L and that whole quantity gets multiplied by G. So here we've done the same sort of trick that we did um the horizontal direction for the length of the chord L one. We've taken the mass of the chord and multiplied by the fractional length of a chord that is hanging vertically. And so when I add those two masses together and multiply it by G that's give me the total weight pulling downward. So here I can write down Newton's second law um in the Y direction or this combination of the M two and L two. But first, before we do that, let's define a quad and system where the positive Y direction is going downward. The reason why we define it downward is now both of our accelerations will be in the positive direction. So I already done to the second law or this uh second mass or combination. So I'll have the net force in the Y direction is equal to the weight. So it's gonna be M two plus MC multiplied by L two divided by L. And that whole quantity gets multiplied by G and then we'll have the 10 force go in the negative Y direction. So I'll have minus the force attention FT and that's gonna be equal to the total mass, which again is the M two plus MC L two divided by L and that mass gets multiplied by the acceleration A. So now what I can do is replace the force of tension here with the expression that I found in the horizontal direction in terms of the M one MC L one and the acceleration. So I'll rewrite my um net force in the Y direction. And that's gonna be M two L MC L two over L all multiplied by G. That term did not change, then I'll have minus and then I'll have the substitution for the force of 10. So I have a quantity of M one plus MC L one divided by L. And that whole quantity is multiplied by A and that has to equal the quantity M two plus MC L two divided by L and that quantity is multiplied by A. So at this point, I want to um collect my accelerations on one side of the equation. And so I'm gonna move the second term on the left hand side of the equation to the right side of the equation. So my first term on the left side is going to remain and it's going to be, remain unchanged. So I'll just have the quantity of M two plus MC L two divided by L multiplied by G. And that's gonna be equal to. And here I'm gonna factor on a of both of those terms. So I'm just gonna have the quantity of M one plus M two plus M two, sorry L mt. And here I can collect the terms or MC. So I'm gonna have MC multiplying L one plus L two all divided by L and all those masses get multiplied by the acceleration A. So I can simplify things a little bit L one plus L two equals L. That's what I was given here in our um problem. So these two terms actually cancel out and I can now offer the acceleration A. So A is gonna be equal to the quantity M two plus MC L two over L, that whole quantity is multiplied by G divided by M one plus M two plus MC. And so that is the acceleration that I was looking for. So that's my answer. And if I compare that to the answers that I was given, this corresponds to answer C, the other answers um either have, are missing the L terms. So it's missing the factor of L two over L. So that's true for um an answer choices A and B and answer D is just A is equal to G which is not going to be true. Your acceleration is not gonna be equal to the acceleration due to gravity because you have that additional mass that's being pulled in the horizontal direction on the table. So just to sort of recap real quick, we actually um used Newton's Second law here twice for two different free body diagrams. But here the trick was to combine the mass of the chord in with your free body diagram. And then we have two equations with two unknowns that we can solve simultaneously to get the acceleration. So I hope that this has been useful and I'll see you in the next video.

Determine a formula for the acceleration of the system shown in Fig. 4–49 (see Problem 55) if the cord has a non-negligible mass m_C. Specify in terms of ℓ_A and ℓ_B , the lengths of cord from the respective masses to the pulley. (The total cord length is ℓ_A + ℓ_B.)

Key Concepts

Newton's Second Law

Tension in a Cord

Kinematics of a Pulley System

Watch next

Determine a formula for the acceleration of the system shown in Fig. 4–49 (see Problem 55) if the cord has a non-negligible mass mC. Specify in terms of ℓA and ℓB , the lengths of cord from the respective masses to the pulley. (The total cord length is ℓA + ℓB.)

Key Concepts

Newton's Second Law

Tension in a Cord

Kinematics of a Pulley System

Watch next

Determine a formula for the acceleration of the system shown in Fig. 4–49 (see Problem 55) if the cord has a non-negligible mass m_C. Specify in terms of ℓ_A and ℓ_B , the lengths of cord from the respective masses to the pulley. (The total cord length is ℓ_A + ℓ_B.)