# Force and tension in a pulley

So imagine a pulley (frictionless) with a massless string. On each side are 2 weights 1 of 10 g and 1 of 20g. In the free body force diagrams for the 2nd body we say that $$m(g+a)= t$$ But shouldn't the left hand side be greater than the right one considering the force acting downwards is more than the tension? Can someone please explain this part?

• One each side is *1 weight not 2 Aug 22 '19 at 16:17
• You are just writing $m$, without distinguishing which mass it is. Aug 22 '19 at 16:19
• The pulley is frictionless, but is it massless? Aug 22 '19 at 17:09
• Show us the free body diagram. Aug 22 '19 at 17:24
• What left-hand-side are you referring to? The $m(g+a)$ term? Aug 22 '19 at 17:33

For any body, you can always assume that the net force is equal to the net acceleration.

$$F_{net} = ma$$

In your problem, there are two forces acting on the mass: gravity and tension from the rope.

$$F_g + F_t = ma$$

And of course the force of gravity is given my the mass and local value of g.

$$mg + F_t = ma$$ $$F_t = m(a-g)$$

This will hold if you are consistent in your coordinate choices. Now, if you want to assume positive displacement/velocity/acceleration is in the upward direction, then $$g$$ would be a negative number.

So you could reduce some things by swapping the sign of both $$g$$ and the sign in front.

$$F_t = m(a-(-9.8\text{m/s}))$$ $$F_t = m(a+9.8\text{m/s})$$ or the way it was written originally (so the left-hand side still refers to the same term) $$m(a+9.8\text{m/s}) = F_t$$

Therefore your first equation is correct only if the $$g$$ in the equation isn't really a coordinate-consistent value of $$g$$, but the positive magnitude value.

shouldn't the left hand side be greater than the right one considering the force acting downwards is more than the tension?

But the left hand side is not only the force of gravity. It's also the acceleration term. So that value must be something that reduces the magnitude of left hand side to be equal to the tension.

• I think it would be better to say if up is positive then $F_g=-mg$ rather than saying $g$ is negative. Aug 22 '19 at 18:02

With a pulley, I like to wrap the coordinate axis around the pulley with the cord, with the direction of motion being positive. Then: Mg – t = Ma and t – mg = ma