# Movable pulley system problem

#### Neasiac

Hello,

I'd be grateful if someone could check my solution to the following problem on movable pulleys. Forces with pulleys tend to confuse me. Excuse if terminology is a bit wonky, I'm translating this from another language.

Steel cube with side of length a = 0.2 m and density r = 7800 kg*m^-3 lies at the bottom of a container submerged underwater. The container has depth of h = 1 m. We're pulling the cube with a movable pulley system (see image bellow), so that the bottom face of the cube is 1 m above the water surface. To pull the rope without any body, we need force of F_t = 55 N (because of friction and stuff). What force F do we need to pull the rope with, while:

a) Entire cube is underwater?

b) Entire cube is above the water?

The pulley system looks like this (ignore the 200kg - this is not actually the image from the textbook, it's a random one I found online, but the pulley system looks the same)

My solution:
From dimensions and density, we can get mass of the cube => m = r * a^3 = 62.4 kg

Drawing free body diagram, we can determine that without friction, the force needed to pull the cube using the pulley system would be 4 times lower than the force weighting the cube down. To this force, however, we need to add F_t. Therefore we get:

b)
F = F_gravity / 4 + F_t = (m * g) / 4 + F_t = 207.88 N

a)
Same as b), but uptrust from the water is helping us, lowering the force which weights the cube down.
F = (F_gravity - F_uptrust) / 4 + F_t = (m * g - waterDensity * g * a^3) / 4 + F_t = 188.28 N

Information about depth was not needed.

#### Woody

I haven't checked in detail,
but the basic ideas look correct.