In the image above, assume that the person is lifting the 12.5 kg weight at a constant velocity and the pulleys are frictionless. In this case, what would the pulling force be?
Here's what I've tried:
I assumed that the resultant force on the pulley A would be equal to the tension T. Since the velocity is constant, the acceleration is 0.
T – 125 = 12.5 * 0
=> T = 125 N
So the resultant force on the pulley A is 125 N.
Here's the part I'm unsure about:
If the resultant force on the pulley A is 125 N, then should the tension on both sides of pulley B also be 125 N? And if so, would the tension on both sides of pulley C also be 125 N (which would result in a pulling force of 125 N)? Or would the resultant force from pulley A be divided somehow?
Best Answer
You're missing a few things, which is why your answer isn't as you expect. I'll solve this two ways - using tensions (as you did), and using conservation of energy. We will see that they give the same answer.
Working using tension.
Your big omission here is that, from the point of view of pulley A, there are * two * downward forces on it (other than its own weight which we ignore), not one.
Look at the rope that starts at the ground, runs round pulley A, and attaches to the weight. That rope has tension in it. When the weight is just about to lift more, the tension pulling it up must be just about /more than 125 N (taking g as 10 for simplicity, actually it's closer to 9.8).
That tension is in all of that rope. So pulley A experiences that tension * twice* - once as a tension on the left side, where the ground is 'trying' to pull down the pulley, with tension 125N, and again on the right side where the weight is also pulling downward, creating a tension in the right half of the rope, of 125N. At the top of A, these two tensions in the bottom rope are equal and opposite (and both pull horizontally!), so they cancel. But vertically, pulley A experiences 250N downward because of them.
As A is in equilibrium, it must also have a balancing force upward of 250N as well, so the tension in the upper rope must be 250N. So this is the force you must exert to hold it stationary.
Answer - pull with force 250N.
(Comment on tensions at B and C: At pulley B, the tension at the left of B is 250N, so the tension on the right of B is also 250N. That means the tension on the left of C is 250N, so the tension on the right of C is 250N, so we are pulling with a force of 250N. In effect, the rope from A-B-C-human is one rope, with the same tension throughout (as it's not moving internally!), so the tension at A is the force applied by the human at the right. Looked at another way, B and C change the direction of the rope, and the direction of the tension in it, but they don't change the size of that tension)
Working using energy.
You can also solve this using conservation of energy (assuming no friction).
Suppose we pull the rope by some small amount, call it 'h'. By looking at how the pulleys are arranged, we can confirm that the weight would be lifted by twice that amount, or 2h.
(Example: suppose there was 50cm of rope each side of pulley A, so the total rope length around A is 100cm. Imagine that we pull the rope at C by 1cm. Then A will rise by 1cm. So there will now be 51cm of rope on the left side of A, so there must be 49cm on the right side. So the weight is 2cm off the ground, so it has risen by 2cm)
The work done is calculated as the force used x the distance the force worked.
Work done by weight (= potential energy gained by weight): the change in energy when an object moves against gravity is given by a well known formula: (mass) x (g) x (distance moved). The mass is 12.5 kg, gravity is say 10 m/s2 and the distance raised is 2h. Total potential energy gained = 12.5 x 10 x 2h = 250h.
Work done by human: the work done is whatever force they had to use (call it F), times the distance that force acted through (which was 'h'). So work done is Fh.
Assuming no friction, the work done equals the potential energy gained (human did work, weight gained altitude, no energy was gained or lost by the system as a whole). So 250h = Fh. So F=250.
The force used was 250 N, same result.