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Old Apr 3rd 2010, 11:59 AM   #1
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Lifting Yourself With a Pulley

I'm having trouble explaining in forces why my intuitive answer is correct.

The problem is a rope through a single pulley attached to the ceiling. A person sits in a chair (system mass M) which is attached to one end of the rope, and pulls down on the other rope to lift himself. I need to explain mathematically how much force he must apply to remain in position and also come up with a formula for how much force to apply to achieve upward acceleration a.

From experience, I know that you will have to apply half of your weight to achieve zero acceleration. This is because by Newton's third, for every dF of force delivered downwards by your muscles on the rope, an equal force pushes you up, decreasing the effective "weight" of the chair person system. So when you apply a force equal to half the weight of the system, you effectively cancel your weight. Beyond that (I'm a little less certain here) every bit of force you apply will only count once towards the tension in the rope.

I must be missing forces, because I can't put this in terms that make mathematical sense. What equation can I set up that will justify F(applied) = F(weight)/2 when tension = weight? How do I show that the effective weight of the system is equal to the actual weight minus the 2x(applied force) but only until the two are equal?
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Old Apr 30th 2010, 11:01 AM   #2
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you can try this...
draw a free body diagram of man.assume he is sitting on a chair and the chair is attached to one end of the pulley. the other end of the rope, is in the hand of the man himself.
the man exerts a force T on the rope, by newton 3 law rope will exert an equal force on the man ie. T upwards. since pulley is assumed massless, tension on both sides of the pulley is equal. so the tension acting on the other end, ie. the one attached to the chair is also T.
NOW take (man+chair) system, this experiences net force(T+T) upwards and earth exerts downward gravity Mg. so force balance gives
2T=mg ,ie. T=mg/2
for the next part, the same concept will yield M(g+a)/2
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Old Oct 12th 2013, 02:46 AM   #3
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Unhappy action and reaction on same body

hey,
i have a doubt about this.
is there any chance to happen action-reaction on one body.
you are pulling rope, pulley changes your force direction towards you, and same force is going to apply on you. i.e action given by you is reacted upon you.
plz explain
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Old Oct 12th 2013, 03:18 PM   #4
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Exclamation

Originally Posted by siddhartha5432 View Post
hey,
i have a doubt about this.
is there any chance to happen action-reaction on one body.
you are pulling rope, pulley changes your force direction towards you, and same force is going to apply on you. i.e action given by you is reacted upon you.
plz explain
Newton 3 action- reaction force pairs always act on different objects. The man pulls down on the rope , and thus the rope pulls up on the man, away from him. No way do these forces act on the same object. When 2 equal and opposite forces act on the same object such that the net force is 0, then that is an application of his first law. Do not confuse the two laws.
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Old Oct 14th 2013, 01:28 PM   #5
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Originally Posted by Mandrake View Post
Newton 3 action- reaction force pairs always act on different objects. The man pulls down on the rope , and thus the rope pulls up on the man, away from him. No way do these forces act on the same object. When 2 equal and opposite forces act on the same object such that the net force is 0, then that is an application of his first law. Do not confuse the two laws.
bro, my question is, if anyway action and reaction is going to act on same body what happens to the body?
a) net force is zero
b) body does not move
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Old Oct 14th 2013, 03:51 PM   #6
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Action-Reaction Applicability

Hi deseo,

The 3rd Law is used when systems are sub-divided. When an initial system is not solvable, it is sub-divided to become, say, A and B. Upon separation forces one-part-to-the-other arise. (Typically called action and reaction forces). The forces, "action" AND "reaction" do not apply TOGETHER on system A nor on subsystem B. Usually "reaction" applies to the (more interesting) sub system. Action applies oppositely and equally to the matter of B.

I always intended to finish this problem wherein a guy uses a gun tackle
to hoist himself. Take a look!

http://www.thermospokenhere.com/wp/0...awks_nest.html

Ask me and I'll finish it.

Good luck... TSH
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Old Oct 15th 2013, 06:11 AM   #7
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Originally Posted by siddhartha5432 View Post
bro, my question is, if anyway action and reaction is going to act on same body what happens to the body?
a) net force is zero
b) body does not move
Say there, bro, you have to be careful using the terms 'action' and 'reaction'. When applied to the man- rope- chair system in this problem, the 'action' force is the weight of the man and chair, and the 'reaction' force is the tension forces in the cable. If the system is in equilibrium such that there is no net force acting and no acceleration, then by way of newton's first law, the tension force is half the weight of the man and chair. But the man and chair are still moving, at constant speed. Remember Newton 1 applies to bodies at rest or in motion in a straight line at constant speed.

This is an application of Newton 1 for forces on the same object, not Newton 3 which applies to different objects. Newton 3 says that if the man pulls on the rope with a tension T, the rope pulls on the man with a tension T. this would be action-reaction force pairs, whereas in Newton 1, the action reaction forces have a different meaning.

Last edited by Mandrake; Oct 15th 2013 at 06:16 AM.
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Old Oct 15th 2013, 09:24 AM   #8
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Valuable Discussion Here! Thank you!

This simple event - "lifting oneself" exposes many ideas.

Every force has an "equal and opposite" force. The one is called "reaction" the other is called "action." Obviously they cannot BOTH be considered in a calculation - to do so would be to add "ZERO" to the situation.

The "gravity force" we draw as an arrow on the system FBD is a reaction. This force has an "equal and opposite force" but that force IS NOT at the ropes. The "action force of a gravity force" is represented as a force, located at Earth's center, directed toward the cm of the system and having magnitude mg.

Thank you for this discussion. I wish I had not used a gun-tackle on my problem. I'm working to finish that - Thanks again.

Jim
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