# The force exerted on a pulley in a atwood machine

There was a question on my test where I had to find the force exerted on pulley by the string in an atwood machine. 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 $$\frac{g}{7}$$. 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?

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
$$\displaystyle \frac {4 \times \frac g 7 - 3 \times \frac g 7}{4+3} = \frac g {49}$$
$$\displaystyle P = 7g - \frac {7g} {49} = \frac {48}{7}g$$
• 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. Mar 25, 2021 at 17:48