# For a massless pulley moving upwards with acceleration, is the upward force equal to the downward force?

Imagine a massless and frictionless pulley with two weights hanging either side of the pulley by a massless string.

Like this except not attached to a ceiling

Rather than being fixed to a ceiling, the pulley is being pulled upward by an external force F, with the weights and string still attached.

Due to Newton's 2nd Law,

$\Sigma F_y=F-2T=ma$,

where $T$ is the tension in the string on either side of the pulley and $a$ is the vertical acceleration of the pulley.

Clearly, since there is a net upward force, the pulley itself will accelerate upwards.

But because the $m=0$,

$F-2T=0$.

Does this not then suggest that the pulley has a constant velocity?

• What do you mean by "not attached to a ceiling?" Are you saying that the whole setup is just free-falling straight down? If it's not attached to a ceiling, then what object is exerting the force F on the pulley? – Ben Crowell Oct 22 '14 at 15:09
• Clearly, since there is a net upward force, the pulley itself will accelerate upwards. Huh? Why doesn't the whole apparatus just drop? – Ben Crowell Oct 22 '14 at 15:19
• Someone's pulling the pulley upward by a hook of some sort... – ODP Oct 22 '14 at 15:27
• Quite clear by "a force pulls the pulley upward". – ODP Oct 22 '14 at 15:27
• Good homework question in my opinion. Effort has been put in, and the OP is asking about a conceptual interpretation of their derived equation. – BMS Oct 22 '14 at 17:47

In the equation $F_{net}=ma$, normally we would assume that $F_{net}=0$ implies $a=0$ on the right-hand side. However, for a massless object, we can satisfy the equation by having $F_{net}=0$, $m=0$, and $a\ne0$. In reality, of course, the pulley is not massless, so $m$ is small, $a$ is some nonzero number, and $F_{net}$ is small.