[spoiler]Physics QBANK#74 (work, pulley)

[FatherTime-PhD]

For this question....
So I get the solution (kind of). Yes, we're using pulleys so maybe we use less force, but that force is just applied over a larger distance so that in the end, total work done is the same.

BUT this is how I thought of it ... would really appreciate it if someone could tell me where exactly my flaw in thinking is:

I wanted to try to find the y component of the force because I knew that component would equal 40N if I assumed the the mass would be lifted at a constant velocity (a=0).... then somehow use SOHCAHTOA to get F in terms of Fy and then just multiply it by the distance to get work?

Although what also doesn't make sense to me is that the F in the y direction is a downwards vector which doesn't make sense to me because you need to use a upwards vector F to pull up the box?

Probably multiple issues with how I'm thinking lol ... any feedback is appreciated.

---

Members don't see this ad.

 

[sapientnarwhal]

Your flaw is the assumption that the y component would be 40N. If this were true, then F = sqrt(40^2 + Fx^2) ≥ 40N. If F ≥ 40N, then the force exerted on the object via tension is ≥ 80N.
More specifically, you were incorrect in assuming that only the y-component is involved in work. Recall that work is F*d in the direction of movement. If the cable moves -60* to the horizontal, then you would use F and d in this direction.

Note that the tension forces for each cable section do not directly oppose each other. There exists a tension force (T1) opposing the pull force exerted by the ceiling, a Tforce (T2) opposing the Fg on the object, another Tforce (T3) opposing the Fg on the object, a Tforce (T4) opposing the pull force exerted by the second pulley (essentially the ceiling bc pulley 2 is attached to the ceiling), another Tforce (T5) opposing the pull force of pulley 2, and lastly a Tforce (T6) opposing the pull force F shown in the diagram.

The key to understanding this pulley (under assumptions of massless cable, pulleys, and no friction) is that each of these tension forces are identical in magnitude. Because we are considering the forces concerned with the object, we can isolate T2 and T3. If we wish to suspend the object, T2 + T3 = 40N. Therefore T6 = 20N and F = 20N (to avoid confusion, F + T6 = 0 so F = -T6). Fy is 20sin(60), but this value alone is useless because we only care about the force F in the direction of cable elongation.

You can also work backwards. If the object is lifted by 5m, then cable section 1 loses 5m, section 2 loses 5m of length, and section 3 gains 10m of cable.
Work done = 4*5*10 = F*10 = 200. Therefore F = 20N.