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There's this very simple problem which I've never quite been able to wrap my head around, partly because different instructors have given me different answers and, further, even looking at worked problems online there seems to be no consensus.

Suppose we have a "massless" rope that is pulled at both ends by forces of equal magnitude. What is the tension in the rope?


For example, I've recently been given a problem along the lines of:

Two masses, $m_1$ and $m_2$, are given equal charges of $+Q$ and are tied together by a massless rope of length $l$. What is the tension in the rope?

(Notice that, although the masses are implied to be different, by an energy analysis there is no kinetic energy in the system thus the forces at the ends must be of equal magnitude).

According to some, we should treat the rope as being "fixed" at one end, and then the tension in the rope is given by the magnitude of just one of the forces. According to others, we should consider both forces.

In past exams, I've had points docked due to both approaches, and in each case the instructor is adamant that their approach is correct.

To me, it seems somewhat intuitive that we should consider both forces.

I'm past introductory mechanics now, but these problems still come up from time to time in some form or another.

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The confusion, I think, comes from the fact that many people think about the tension as the resultant force on the rope, which is incorrect.

Since the rope is stationary, the resultant force must be zero. One body is pulling on one side, and another is pulling on the other side, but in the opposite direction. These two vectors add up to zero.

Tension is the magnitude of the force by which the rope acts on the body at one of the ends. Or another way, if you were to cut up the rope into two pieces (say part A and B), then tension would be the magnitude of the force by which A acts on B (and by Newton's 3rd law B acts on A).

Bottom line: if a body is acted upon by force with magnitude F and is held stationary by a rope, the tension in the rope must be F to counteract the force.

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  • $\begingroup$ This is the correct answer. Tension = force need to keep a cut segment in equilibrium. $\endgroup$ – JAlex Apr 8 at 20:08
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A "massless" rope cannot ever have unequal forces at each end. If it did it would accelerate. But due to the insignificant mass, it would accelerate quickly enough that the forces would equalize.

Draw a free-body diagram of the small portion at the very end of the rope. Assuming the rope is stationary, all the forces must sum to zero. From one direction is the external force. From the other direction is the force due to tension in the rope. These forces must be equal and opposite.

According to some, we should treat the rope as being "fixed" at one end, and then the tension in the rope is given by the magnitude of just one of the forces. According to others, we should consider both forces.

Do you have a link to one or the other?

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