# [Physics]Force with Pulleys

• Oct 18th 2009, 05:34 PM
WhoCares357
[Physics]Force with Pulleys
Quote:

In the figure, a cord runs around two massless, frictionless pulleys; a canister with mass m = 60 kg hangs from one pulley; and you exert a force F on the free end of the cord. What must be the magnitude of F if you are to lift the canister at a constant speed?

Originally I thought this question was pretty easy. Since a=0, I thought F=mg. However, the answer (588N) is wrong.
• Oct 18th 2009, 05:43 PM
WhoCares357
Actually I figured out that I had to include the tension on both sides.

For the left side: F-T=ma therefore F=T. The on the right side, mg-2T=ma. Therefore mg/2=F. I got the correct answer as 293N.

However, there is a second part that I don't understand.
Quote:

The canister is lifted 5.4 cm. During that lift, what is the work done on the canister by the weight mg of the canister? (Hint: When a cord loops around a pulley as shown, it pulls on the pulley with a net force that is twice the tension in the cord.)

I know that W=F*d. I tried setting W=2F*d, but that's not right.
• Oct 18th 2009, 05:55 PM
skeeter
Quote:

Originally Posted by WhoCares357
Originally I thought this question was pretty easy. Since a=0, I thought F=mg. However, the answer (588N) is wrong.
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you only need to pull down with a force equal to 1/2 the weight.

note that the mass is held up by two ropes on either side of the pulley, tension in both sides the same ... a mechanical advantage of 2 ... the trade-off is that you'll have to pull down 2 m to get the mass to rise 1 m.

the work done on the mass by the mass' weight is -Wd.
• Oct 18th 2009, 07:32 PM
WhoCares357
Quote:

Originally Posted by skeeter
you only need to pull down with a force equal to 1/2 the weight.

note that the mass is held up by two ropes on either side of the pulley, tension in both sides the same ... a mechanical advantage of 2 ... the trade-off is that you'll have to pull down 2 m to get the mass to rise 1 m.

the work done on the mass by the mass' weight is -Wd.

Can you explain mathematically why it is -Wd? I can't really see it from your explanation. -Wd=-Fd^2? How did you get that?
• Oct 19th 2009, 07:03 AM
skeeter
Quote:

Originally Posted by WhoCares357
Can you explain mathematically why it is -Wd? I can't really see it from your explanation. -Wd=-Fd^2? How did you get that?

work = the dot product of force and displacement

weight of the mass (the force) is W in the down direction

displacement is d in the upward direction

work = Wd cos(180) = -Wd
• Oct 19th 2009, 05:40 PM
WhoCares357
Quote:

Originally Posted by skeeter
work = the dot product of force and displacement

weight of the mass (the force) is W in the down direction

displacement is d in the upward direction

work = Wd cos(180) = -Wd

Okay, thank you.