Atwood machine with forces, not masses

[CaptainP]

I am having difficulty reconciling this in my head. Imagine two Atwood machines (example). In both machines, the pulley is massless, there is no friction, but the rope is not massless, though it is inextensible. The rope and pulley in each machine are identical. The first is a traditional Atwood machine, with a 10N block hanging on the left of the pulley and a 50N block hanging on the right of the pulley. The second Atwood machine also has a 10N block on the left, but on the right there is a constant applied force of 50N downward.

I am not sure whether the 10N block will accelerate at different rates in each Atwood machine. On the one hand I could see an argument that the block will accelerate faster in the second one simply because there is less overall mass to accelerate with identical forces. However, if this were the case, it would imply that the tension in the rope is greater in the second Atwood machine, and I cannot figure out how this would happen when both the mass in the first machine and the applied force in the second machine are exerting the same constant downward force. Help?

 

[PhanthomJay]

I am having difficulty reconciling this in my head. Imagine two Atwood machines (example). In both machines, the pulley is massless, there is no friction, but the rope is not massless, though it is inextensible.

Is it given that the rope is not massless? If so, you don't state its mass, so let us proceed assuming its mass is negligible.

The rope and pulley in each machine are identical. The first is a traditional Atwood machine, with a 10N block hanging on the left of the pulley and a 50N block hanging on the right of the pulley. The second Atwood machine also has a 10N block on the left, but on the right there is a constant applied force of 50N downward.

I am not sure whether the 10N block will accelerate at different rates in each Atwood machine. On the one hand I could see an argument that the block will accelerate faster in the second one simply because there is less overall mass to accelerate with identical forces.

yes...

However, if this were the case, it would imply that the tension in the rope is greater in the second Atwood machine,

yes..

and I cannot figure out how this would happen when both the mass in the first machine and the applied force in the second machine are exerting the same constant downward force. Help?

Draw free body diagrams for each case. Although the downward applied force is the same, the NET downward force is not.

 

[Nugatory]

However, if this were the case, it would imply that the tension in the rope is greater in the second Atwood machine, and I cannot figure out how this would happen when both the mass in the first machine and the applied force in the second machine are exerting the same constant downward force.

You're right, the tension in the rope is greater in the second case.

In the case with the falling 50N weight, there's not 50N tension in the rope - if there were, the 50N weight wouldn't be falling, as the tension would exactly balance gravity. The acceleration can never exceed g, so that puts an upper bound on the possible tension.

In the case where you're applying a constant 50N force to the rope, the acceleration can exceed g; indeed the only way of keeping 50N in the rope when it's accelerating a 10N weight is to have an acceleration much greater than g.

 

[CaptainP]

In the case with the falling 50N weight, there's not 50N tension in the rope - if there were, the 50N weight wouldn't be falling, as the tension would exactly balance gravity. The acceleration can never exceed g, so that puts an upper bound on the possible tension.

I buy that, but...

In the case where you're applying a constant 50N force to the rope, the acceleration can exceed g; indeed the only way of keeping 50N in the rope when it's accelerating a 10N weight is to have an acceleration much greater than g.

Why does the rope necessarily have a 50N tension in the second machine? Does the weight on the left not contribute to the tension? Ex: what if it was a 20N weight, would it change the tension in the second Atwood machine? Or what if it was a 1N weight, or what if there was no weight on the left side at all?

 

[PhanthomJay]

It is given that the tension is 50 N in case 2. If the applied force was 20 , the rope tension would be 20.if there was no mass on the left and the rope was massless, the acceleration would be infinite in case 2 and g (free fall) in case 1.

 

[CaptainP]

It is given that the tension is 50 N in case 2. If the applied force was 20 , the rope tension would be 20.

K but why is this considered given? Why is the only determinant of tension the applied force? Why does the weight on the left not impact the tension at all?

 

[PhanthomJay]

K but why is this considered given? Why is the only determinant of tension the applied force? Why does the weight on the left not impact the tension at all?

Why should it? You can apply any force you want. If you applied 10 N , the mass on the left would not accelerate. Apply 50 N and it accelerates. Apply 1000 N and it accelerates at a faster rate.
on the

 

[Nugatory]

Why does the rope necessarily have a 50N tension in the second machine? Does the weight on the left not contribute to the tension? Ex: what if it was a 20N weight, would it change the tension in the second Atwood machine? Or what if it was a 1N weight, or what if there was no weight on the left side at all?

If you're exerting 50N of force on the end of the rope, then by Newton's laws (equal and opposite force) the rope is exerting 50N on you and therefore the tension in the rope is 50N.

You can visualize this by imagining that there's a spring (nice linear $F=kx$ spring) between your hand and the end of the rope. To exert 50N of force, the spring has to be stretched by some amount and it has to stay stretched by that amount (if it's allowed to contract, you're no longer applying 50N). Keeping it stretched requires 50N of tension in the rope.

It also requires that if the end of the rope is accelerating towards you, you must accelerate your hand on the other end of the spring as well to keep the length of the spring constant at 50N of tension. Put a lighter weight on the other end and you end up with higher acceleration.

 

[PhanthomJay]

Or you might want to consider a free body diagram at the right end of the rope, assuming, for ease of calculation, a massless particle at the right end of the rope, and you apply a force F to it. There are 2 forces acting on it, F_applied acting down and T acting up. Using Newtons 2nd Law
F_net = ma
F - T = ma
F - T = (0)a
F - T = 0
F = T