There was a question on my test where I had to find the force exerted on pulley by the string in an atwood machine. Full diagram

From the 2 masses I got: $$ 4g - T = 4a \\ T - 3g = 3a \\ a = \frac{g}{7} \\ T = \frac{24}{7}g $$

I took the 2 masses and the string as 1 system. There are only 2 forces acting on the system: the force the pulley exerts on the string ($P$) and the weight of the system ($W$). Also the system is accelerating downwards at $\frac{g}{7}$. A diagram of the string+masses system

So I tried this: $$ W - P = m \cdot a \\ 7g - P = 7 \cdot \frac{g}{7} \\ P = 6g $$

But according to this and this the force should be $2T$ (which $=\frac{48}{7}g$). Therefore, at least 1 of my assumptions must be wrong. Which one of my assumptions is wrong and why?


1 Answer 1


The centre of mass of the whole system is not accelerating at $\frac g 7$.

The $4$ kg mass is accelerating downwards at $\frac g 7$ but the $3$ kg mass is accelerating upwards at $\frac g 7$. So the acceleration of the centre of mass is

$\displaystyle \frac {4 \times \frac g 7 - 3 \times \frac g 7}{4+3} = \frac g {49}$

and so

$\displaystyle P = 7g - \frac {7g} {49} = \frac {48}{7}g$

  • $\begingroup$ Wow, how did I miss that. When I found how the centre of mass was moving (to incorrectly prove that its moving at $\frac{g}{7}$), I forgot to divide by the sum of the masses. That fully solves my problem thanks. $\endgroup$
    – TheLizzard
    Commented Mar 25, 2021 at 17:48

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