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While I can get the right answer for these problems by memorizing how to do it, I want to understand it conceptually.

There is a (massless) pulley with a rope through it. A person sits (in a bucket/chair/box) and pulls down on the other end of the rope to lift herself up. The question is how much force she needs to pull down with to lift herself at constant velocity.

From practice, I know the answer is half the force of her weight. This is because the pulling force is equal to the tension. When she pulls on the rope, it creates tension throughout the rope. The tension pulls her up on both sides, so the total upward force is twice the tension (2T). Then, you subtract her weight (mg) from the upward force to get 2T - mg. Since velocity is constant, acceleration = 0, so according to Newton’s Second Law the net force is 0. Thus, you set the equation to 0 and solve to get T = mg/2. I do understand why the answer ends up being half the weight, and why the tension pulls on both ends of the rope. (I’ve clarified the question to focus on what I’m actually confused about.)

My question is, why is the tension equal to the pulling force applied by the person? If an object hangs from a rope, the tension in the rope is equal to that object’s weight. The person is, after all, hanging from the rope as she pulls it. So as she pulls herself up, why does the tension only equal her pulling force, and not include her weight? Doesn’t the person’s weight strain the rope more?

I would really appreciate any and all help. Thank you.

Qmechanic
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1 Answers1

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This is a simple analysis;

Take an object, and hang it from a pulley with two ropes coming to the object.What is the force on the pulley? The weight of the object, you could measure with a proper scale. This means that , all other things being equal, the tension on each rope is half the weight of the object, by construction.

Now if it is a girl being the object and pulling on one of the ropes , the algebra should be the same.

anna v
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