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Old Oct 26th 2017, 12:50 PM   #1
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Movable pulley system problem


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:

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

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.
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Old Oct 26th 2017, 02:41 PM   #2
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Location: England
Posts: 473
I haven't checked in detail,
but the basic ideas look correct.
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