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Let's suppose we have two people Alice and Bob in a vacuum. The are connected by a rope.

At first, Alice pulls Bob with force $F$ and Bob only clings to the rope. Afterwards, the experiment is repeated with Bob pulling with force $F$ and Alice only clinging to the rope.

The two situations seem different because in one case, Alice seems to be the active part and in the other case, Bob seems to be the active part. My question is: Is there a way to describe this difference in physical terms?

It can't be the force, because due to the third law $F_{Alice \rightarrow Bob}$ = - $F_{Bob \rightarrow Alice}$ in both cases. Also the work done by the active person seems to be equal to the work done by the passive person because both the forces and the displacements are equal in magnitude. Is this correct? Does it take exactly the same amount of energy to pull as it takes to being pulled?

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    $\begingroup$ No, the signs of the works are opposite. When you are passively pulled you move the same way as the rope’s tension force on you. When you actively pull a rope in front of you, you move your arms back as the tension force on you pushes forward. $\endgroup$ – knzhou Apr 25 '18 at 14:45
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    $\begingroup$ If Alice pulls, Alice does positive work on the rope (so the work does negative work on Alice), and the rope does positive work on Bob. The net result is a transfer of energy from Alice to Bob. $\endgroup$ – knzhou Apr 25 '18 at 14:46
  • $\begingroup$ What does "vacuum" have to do with anything? Did you mean to say, "free fall?" $\endgroup$ – Solomon Slow Apr 25 '18 at 16:26
  • $\begingroup$ @knzhou: Thanks for catching this! It makes sense because if we tie the rope around Bob instead of having him cling, all the energy needs to come from Alice. I think I got sidetracked because in the situation of the OP, Bob does need energy to cling to the rope. But having read your comment, I now think this is much less than the work Alice is doing. $\endgroup$ – Marc Apr 25 '18 at 18:24
  • $\begingroup$ I was just thinking about the tug of war and I think it improved my understanding. I'll share it here both for others and for finding further misconceptions in my thinking. In the tug of war, work is done by the net force, the magnitude of which is equal to the magnitude of the reaction force of the ground on the winning party minus the magnitude of the tension in the rope. But in this case, the (macroscopic) work is only a small part of the chemical energy which is supplied by the parties. Almost all of the calories burned are needed to keep up the tension in the rope. $\endgroup$ – Marc Apr 25 '18 at 19:04
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If the rope has mass, the person pulling will experience a larger force than the one clinging. However, let's assume the rope is massless (and also that Alice and Bob are initially at rest). In that case, as you say, the force magnitudes are identical. So, kinematically there is no difference between puller and clinger. The tension in the rope determines the acceleration of both Alice and Bob, independent of who is pulling. There is a physical difference though: the puller needs to do work, expending energy to reel in rope. The clinger expends no energy.

The tension does some work on Alice and some on Bob. The ratio of the amounts or work is in inverse proportion to their masses: if they weigh the same, the tension force does the same amount of work on each of them. If Bob is twice as heavy, half as much work is done on him as on Alice. The point though, is where the energy to do this work comes from. And the answer is from chemical energy: the puller metabolizes food they have eaten in order to reel in rope. Energy required is equal to the force exerted multiplied by the amount of rope reeled in. That is equal to the combined kinetic energies of Alice and Bob after the pulling is complete, and also (once you account for the efficiency of converting food energy to muscle motion) to the amount of calories burned by the puller.

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  • $\begingroup$ Thanks for your clear answer. I think I am going to accept your answer soon, but for now I'm still thinking about these things and have summarized my current understanding in a comment above. $\endgroup$ – Marc Apr 25 '18 at 19:07
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The paradox arises because it's easy to think that the rope runs between the centres of mass (CM) of A and B (or, more precisely, one of its ends is stationary wrt A's CM, and the other end wrt B's CM). But when being actively hauled by Alice, the rope moves towards her CM. In fact the rope moves with B's CM. We can think of the rope as part of B, an extension of B, if you like. [wrt = with respect to.]

So in the frame of reference of the centre of mass of the AB system, B starts to move in the direction of the force exerted on her by A, and so has work done on him. A's hauling hand also moves in this direction (even allowing for the movement of her centre of mass in the opposite direction), but exerts a force in the same direction, and therefore does work.

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Consider an "arm" as massless in the force diagram shown below with force $F_{\rm 1a}$ as the force on mass $m_1$ due to the arm etc.

All the forces have the same constant magnitude, $F$.

enter image description here

In the lower diagram the arm contracts by a distance of $x_1+x_2$.

The mass $m_1$ moves a distance $x_1$ and the mass $m_2$ moves a distance $x_2$.

The work done on mass $m_1$ is $\vec F_{\rm 1a}\cdot \vec x_1 = +Fx_1$ and the work done on mass $m_2$ is $\vec F_{\rm 2a}\cdot \vec x_2=+Fx_2$.

The total work done on the two masses is $+F(x_1+x_2)$.

The work done on the arm is $\vec F_{\rm a1} \cdot \vec x_1 +\vec F_{\rm a2} \cdot \vec x_2= -F(x_1+x_2)$ and so the work done by the arm is $+F(x_1+x_2)$ which come from the chemical reactions occurring in the arm.

Note that it does not matter to which mass the arm is attached.

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If my understanding of this experiment is correct, a person A that pulls the rope remains stationary and a person B, that clings to the rope, moves.

We should assume that person A holds on to something or relies on the static friction - otherwise, he or she will move as well.

Person B will accelerate and gain some kinetic energy all due to the work performed by person A, unless person B is helping somehow to move things along.

Ignoring the intricacies of muscle contractions, this scenario, from the forces and energy prospective, should not be any different than the scenario when person A is pulling a sled: person A is applying the force and spending energy, which goes toward increasing the kinetic energy of the sled and overcoming friction.

Of course, for person B to move, the tension force has to be greater than the effective static friction, while, for person A to remain stationary, the effective static friction has to be greater than the tension.

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protected by Qmechanic Apr 26 '18 at 13:50

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