# Atwood machine: force on pulley

In an Atwood machine where we assume the string is massless, I understand that the tension in the string is constant throughout. However, I'm having a little trouble imagining the forces on the pulley itself. I'm not sure why the pulley feels $$2T$$ force downwards, where $$T$$ is the tension in the string.

If both the pulley and the string are massless then the experiences the twice the tension force in the downward direction and if pulley is not massless and has some mass $$m$$ then it will experience two forces namely $$mg$$ and $$2T$$ .

In the picture the string applies a force $$T$$ on both the blocks and in return the two blocks apply a force $$T$$ on both the string in the downward direction. Since the string is fixed on the pulley this means that there must be a force $$F$$ acting on the string to balance the $$2T$$ force on the string as a whole and this force is provided by the pulley in contact with the string as shown in the figure below :

[ Here $$F_{sp}$$ represent the force on string by the pulley , $$F_{ps}$$ represent the force on pulley by the string , $$T_{sb}$$ represent the tension force on the string by the block .]

Now since the pulley applies $$2T$$ force on the string in contact in the upward direction , so the string in contact also applies $$2T$$ force on the pulley but in opposite directions i.e. in downward direction. If the pulley is hinged then the downward $$2T$$ force is balanced by the force applied by the hinge on the pulley in the upward direction.)

The following diagram explains how pulley feel $$2T$$ force downward:

If the pulley is massless and frictionless, the normal force upward on the free-body diagram of the pulley should be equal to the total downward force since it does not move vertically (keep in mind that an idealized case, tension of the string is same on the both side of the pulley).

Because the rope pulls at both sides of the pulley downwards, hence the total force exerted on the pulley is 2T.