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Assuming $T_1$ is the force that acts on box $1$ and $T_2$ is the force that acts on box $2$.

Exactly what causes the Tension? Why does $T_1 = T_2$?

The problem is we are told to memorize that $T_1 = T_2$ for mass less ropes, but I do not understand why, especially in an Atwoods machine.

When I think about it, I know it has something to do with the force $m_2g$ which causes the box with mass $m_1$ to rise, and vice versa. But I cannot apply physics terminology to this, or really understand whats going on.

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3 Answers 3

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Imagine a finite segment of the rope, say on the left side. Suppose the tension at the top of the segment is $T_t$ and the tension at the bottom of the segment is $T_b$. Then the segment of rope feels a force $T_t$ up and $T_b$ down, or a net force $T_t - T_b$ up. Since the rope is massless there is no gravitational force so that $T_t - T_b$ is the net force on the rope.

Newtons law says that the acceleration of the rope is $\dfrac{T_t - T_b}{m}$ upward, where $m$ is the mass of the rope segment. However, since $m$ is zero this acceleration must be infinite, which is a problem.

What is really going on is that if there were a net force up, the rope segment would immediately accelerate upward. This upward movement would relax the tension in the upper part of the rope ($T_t$ decreases) and increase the tension in the lower part of the rope ($T_b$ increases). This will continue until $T_t$ equals $T_b$ and there is no net force on the segment of rope. This happens throughout the rope as it becomes taut. Since the rope is massless, this process happens very fast so it can always be assumed that $T_t$ is equal to $T_b$.

Since the segment of rope could have been anywhere, this means that any two points on the rope must have the same tension.

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  • $\begingroup$ "This upward movement would relax the tension in the upper part of the rope (Tt decreases) and increase the tension in the lower part of the rope (Tb increases)".Can you explain this further please ? $\endgroup$ Feb 22, 2018 at 15:25
  • $\begingroup$ @AbhinavDhawan Imaging you are pulling on a rope that is fixed on the other end in order to create tension. Now if you move your end of the rope towards other end, the tension goes away. In my case, the small segment of rope was pulling the rope above it down in order to create a tension, so if this small segment of rope moves up, the tension goes away. $\endgroup$ Feb 22, 2018 at 16:06
  • $\begingroup$ @BrianMoths How can we explain this for a segment which is in contact with pulley, where we will also have a normal force? $\endgroup$ Jul 1, 2021 at 3:03
  • $\begingroup$ @BrianMoths "Now if you move your end of the rope towards other end, the tension goes away (from comments)," this would be true because we are not applying any force on the rope right? As in, there is no external pull on the rope and hence there is no pull on that rope due to the other part of the rope. But in the case of your answer, it seems to not validate it because as the rope moves up there still is a force that is pulling it down. Can you further express this point please? $\endgroup$
    – Floatoss
    Nov 24, 2021 at 16:50
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Consider two cases.

CASE 1: Without friction

The pulley is totally friction less. Consider segment A. Here, it is acted upon by its weight, upward tension force(T2) exerted by the segment of rope above it and a downward tension force(T1) exerted by the segment below it. Since, it accelerates downwards, ma=mg+T1-T2 Now, if the rope is mass less, mg=0, also infinitesimal force is required to accelerate that mass less segment, thus ma=0 and T1=T2. Following the same argument, you can tell downward tension in segment B(T3)=T1=T2. Since, the pulley is friction less, no additional friction force act on segment B and the upward tension force(T4)=T3. In this way, you can tell, tension is same along all the segments of the rope.

CASE 2: With friction

The same argument follows for the segment A as in the previous case. Ultimately, you can show downward tension force in segment B(T3)=T2=T1

But, its different for the upward tension force in the case of segment B. Here, friction acts in segment B. ma=tangential component of mg+T3-T4-friction Since m=0, T3=T4+friction. Hence T3=/=T4. So, in all the segments which are in contact with the pulley, tension aren't equal at both the ends. But as soon as the segment loses contact with the pulley, there you go, tension is equal at both the ends.

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The assumption that $T1$ and $T2$ are the same comes down to the approximation $m_{rope} = 0$. At the end, we are trying to make sense of an assumption that is not trivial to interpret physically.

Besides Brian's great answer I would like to add that the assumption $m_{rope} = 0$ leads to a force balance (FBD) about the rope that suggests the rope is "feeling" the same force at its extremities. So as a reaction, the rope will apply a tension $T$ at each object. For further details see my answer on another post: https://physics.stackexchange.com/a/387934/46464

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