# Why is the net work done in a pulley-string system zero?

In any pulley system, where the pulleys and strings are massless and frictionless, why is the net work done by Tension zero?

In the systems you describe, each string connects always two masses. The tension force exerted on these two masses by the string is equal in magnitude and opposite in direction with respect to the displacement. Hence, the work done by each string on the two masses attached have opposite sign. As a consequence, if you sum up all contributions from each string and each mass, the net work done by tension is zero.

• Yes that's right, but I'm not talking only about one rope system. What about bigger systems where there are 4-5 pulleys and 2-3 ropes, in that case how can we prove that the net work done will be zero? Commented Jul 4, 2015 at 12:34
• I was talking about the general system, with many ropes. Any string connects two masses. So you can say that the tension of the string has opposite direction on each of the two masses attached, and so on. I edited the answer to make this point more explicit. Commented Jul 4, 2015 at 12:37
• Oh, so is it that when we say net work done is zero, then we mean that net work done by each rope individually is zero and so the total net work done is also zero? Commented Jul 4, 2015 at 12:41
• Just saw your edit..that clears my doubt, thanks a lot! Commented Jul 4, 2015 at 12:42

In a two mass dependent motion systems with pulleys, tension displacement ratios are inverse of each other. That is, given blocks A and B, where A is attached to a certain number $n$ of cables by pulleys, and B is attached to $m$ number of cables. The ratio of forces would be $\frac{n}{m}$ and the ratio of displacements would be $\frac{m}{n}$. From here you might be able to guess that they could be equal, but to be sure, here's some work you can look at. HA, PUNS!

Dependent motion in pulley systems tells us that $${\Delta}s_B = -{\Delta}s_A \frac{n}{m}$$ Work on A is simply $${\int_0^{x_A}}\vec{F_A}\cdot d\vec{x_A} = W_A = T n{\Delta}s_A$$ compared to the work on B, $${\int_0^{x_B}}\vec{F_B}\cdot d\vec{x_B} = T m{\Delta}s_B =$$ $$T m(-{\Delta}s_A \frac{n}{m}) =$$ $$W_B = -T n {\Delta}s_A = -W_A$$ and $${\Sigma}{\int}\vec{F}\cdot d\vec{x_i} = 0$$

You can then generalize this to a multiple pulley system by saying the work of a pulley sub-system is equivalent to $W_A$ and the other sub-system is equivalent to $W_B$.

• Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. Commented Mar 13, 2016 at 21:57

You will have to go back to the equation F = m a. If the mass of the string is 0, then F = 0. So yea, the net force of the tension by the string is 0.

To add on, in order for the net tension of the string to be zero, the tension everywhere has to be equal to cancel out.

It is a consequence of total length and tension being constant throughout the taut rope. Here is how it extends to work done:

Let two ends of a segment of a taut rope be at $$x_1$$ and $$x_2$$. Then, $$d|x_1-x_2|=d\sqrt{(x_1-x_2)\cdot(x_1-x_2)}=\frac{1}{2}\frac{2\cdot(dx_1-dx_2)\cdot(x_1-x_2)}{|x_1-x_2|}=dx_1\cdot n + dx_2 \cdot (-n)$$ where $$n$$ is unit vector along the rope segment. The sum represents dot product of displacement of ends of rope to the opposite of unit vector of tension at each point. Multiplying by constant value of tension and summing up for all segments of the rope, the LHS will become tension multiplied by total change in length of rope and the RHS will become the negative work done. LHS being zero, the work done by rope is also zero.