Hey I need help figuring out the free body diagram for a block and tackle system. Problem will be in the comments.

- anonymous

- katieb

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- anonymous

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- anonymous

I've worked with pulley systems before, but I'm not sure how to go about this one.

- anonymous

take the boy as the system....forces acting on him are 1) tension by rope 2) normal force exerted by the beam he is sitting on

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- anonymous

+ 3) the weight of the boy downwards

- anonymous

So the whole pully system doesn't affect the problem? I'm sure it does...

- anonymous

can u explain what u meant?

- anonymous

Sorry I'm really bad at explaining myself. In order to determine the force he's exerting we need to figure out how much force is required by this pulley system to lift 120 lbs right?

- anonymous

So perhaps the question is rather what are all the relevant free body diagrams in this problem.

- anonymous

since it moves at a constant velocity, i guess the force exerted by him = tension in the rope

- anonymous

yeah that is true.

- anonymous

and u are right,this is all about taking free body diagrams and finding unknowns

- anonymous

So how can I take the free body diagrams at the pulleys? the boy's is relatively simple as you mentioned.

- anonymous

as long as the pulley is massless, u neednt take it into consideration

- anonymous

I mean the whole mess of ropes around the pulleys

- anonymous

next you can take the beam as the free body, and the forces acting on it,,,1) normal force from the boy downwards, 2) weight downwards 3) tension upwards,,,,,equate to zero

- anonymous

the tension in the rope goes around the top pulley so the tension in the rope from the top to the bottom pulley is the same as the tension in the rope....does the tackle system provide a mechanical advantage of 2?

- anonymous

or is it an advantage of 4 since the rope goes around the bottom pulley twice?

- anonymous

the pulley thing is rather complex,,,i guess the advantage is 4,,,perhaps @LastDayWork may help u better,,,expert at physics

- anonymous

@LastDayWork is it 4? It would seem so looking at the picture, since the weight of the boy/beam system is 120lb, the boy only needs to pull with 1/4 of the force since 4 ropes are supporting that weight no?

- LastDayWork

Is there something below the diagram; because I can't clearly makeout how the rope is attached to the beam (in which the boy is sitting).

- anonymous

No. but I don't think it matters, since they just say the system and ignore the angle at which the rope is.

- anonymous

Assuming that the tension in the hook in the bottom is 120lbs, and the tension in the other rope is whatever pull the boy is applying, wouldn't the pull be a quarter of the weight? not sure.

- LastDayWork

I am not considering the angle; its just that the tone of the question implies that the boy is lifting the beam along with himself (hence the combined weight).

- LastDayWork

^^ But I cant see how

- anonymous

yeah, but doesn't that mean we can pretend the boy/bar system is just a box of weight 120lb? I think the bar slides up the vertical column, but that doesn't affect the problem.

- LastDayWork

A box climbing a rope ?? :P

- anonymous

Well I have seen physics problems that say to work with perfectly spherical elephants whose mass can be ignored soo....why not?

- JoannaBlackwelder

It looks like to me that the problem states that 120 lbs is the combined weight.

- anonymous

exactly. from the picture we need to determine the force exerted by the boy. I'm thinking that the tackle system provides a mechanical advantage of4 but i'm not sure.

- anonymous

Since the weight is supported by 4 ropes that lead to the top pulley, and the rope the boy is pulling on is equal to one of these ropes, then the force he uses to pull is a quarter of the weight right? or do I have faulty reasoning?

- LastDayWork

As far as I can understand -
There is some sort of spring attached to the hook of upper pulley (above the picture).
By pulling the rope, the boy decreases the distance between the pulleys; and as the lower pulley can't move the (hypothetical) spring will get stretched to account for the displacement of upper pulley.
There is a slag in the rope below the point where the boy is holding it (implies no tension).
So the question is how is the boy lifting himself??

- anonymous

Ok this is getting ridiculously complicated. This shouldn't be that convoluted....

- wolfe8

Wouldn't it be the same as the weight of him and the beam? (Is he lifting the beam up as well?)

- LastDayWork

Can he??

- anonymous

yeah he is. However, the force he's using isn't equal to the weight of the system. The pulleys provide mechanical advantage. I just looked up a double block and tackle system and yeah the advantage is 4, so I guess he pulls with a force of 30 lbs.

- wolfe8

The pulley just allows complete transfer of force though. Since we assume the weight of the pulley is negligible, it does not contribute to the force through torque as a pulley with mass would.

- anonymous

exactly. Well I think I have my answer. Thanks everyone for taking the time to break your heads on this one.

- anonymous

It's not a simple pulley. It's a double tackle

- wolfe8

Whoah. Didn't notice that. That's what happens when you jump into a question. Well good luck. Sorry I couldn't help more.

- anonymous

a single pulley would transfer, so he would pull with the same force as the combined weight. If it were a single block and tackle, the rope would loop once around the bottom ,providing a mechanical advantage of 2, etc. Thanks anyways Wolfe :)

- anonymous

good night everyone! Thanks.

- Vincent-Lyon.Fr

If somebody else were pulling the rope and not the boy himself, I would divide the total weight by 4, as there are 4 ropes supporting him.
But as the boy is pulling the rope himself, there is a 5th tension acting on him, so I would divide the total weight by 5.

- LastDayWork

@Vincent-Lyon.Fr
How can the boy pull himself (along with the beam) ??

- Vincent-Lyon.Fr

Well, I understand he is holding and pulling the rope with his hands.

- LastDayWork

That way he can pull himself but not the beam (why would question give us combined weight).

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