# Questions about statics, mechanics, pulleys

Gold Member

## Main Question or Discussion Point

1) Does the radius of a single, fixated frictionless pulley affect the force I have to apply on one side in order to reach equilibrium with a constant load on the other? Or, will I always have to apply the same amount of force on the other side in order to reach equilibrium, regardless of the radius of the pulley?

And on that same note, would the number of fixated pulley affect the force I'd have to apply?

2) Is my formula for pulleys correct?:

x = number of mobile pulleys/0.5

T (tension of rope) = mg/x

3) If a motorcycle has two wheels, then it has two normal forces, right? and if it's standing on a flat surface with no forces acting on it other than gravity then mg = N1+N2... And if it has 3 wheels, it has 3 normal forces... mg = N1+N2+N3... so on and so forth...right?

4) Under what conditions do I draw more than 1 Free Body diagram to a static equilibrium problem?

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1) Does the radius of a single, fixated frictionless pulley affect the force I have to apply on one side in order to reach equilibrium with a constant load on the other? Or, will I always have to apply the same amount of force on the other side in order to reach equilibrium, regardless of the radius of the pulley?
Remember it is not the pulley that provides the lifting force, it is the tension in the rope, which you supply. If the rope is continuous, the size the of pulley does not matter.

And on that same note, would the number of fixated pulley affect the force I'd have to apply?
In the case of one continuous rope ran through multiple pulleys on the load side, then yes--the rope has the same tension all along its length, and you have multiple sections of the rope lifting the load so the force you must supply is less.

If, instead you simply have one rope attached to multiple ropes, each on a single, fixed pulley, then you still have to supply the full weight of the load.

2) Is my formula for pulleys correct?:

x = number of mobile pulleys/0.5

T (tension of rope) = mg/x
I believe that would be true if the rope is attached at the fixed pulley, but it is better just to think of it as "how many lengths of rope are lifting the load?"

3) If a motorcycle has two wheels, then it has two normal forces, right? and if it's standing on a flat surface with no forces acting on it other than gravity then mg = N1+N2... And if it has 3 wheels, it has 3 normal forces... mg = N1+N2+N3... so on and so forth...right?
yup! The next step is to write the moment equations.

4) Under what conditions do I draw more than 1 Free Body diagram to a static equilibrium problem?
Never! Actually, it is a matter of semantics, some people say "draw a FBD for each particle", while others say "draw your FBD, and clearly indicate the forces on each individual particle/object." The point is to get all the forces attached to the correct body, in the correct place, in the correct direction. For example, for your motorcycle, there is only one object--the multiple wheels simply provide multiple places for the ground to hold it up.

Randy

Gold Member

Hi, and thanks for the reply... :)

Remember it is not the pulley that provides the lifting force, it is the tension in the rope, which you supply. If the rope is continuous, the size the of pulley does not matter.
Ah..makes sense.

In the case of one continuous rope ran through multiple pulleys on the load side, then yes--the rope has the same tension all along its length, and you have multiple sections of the rope lifting the load so the force you must supply is less.
Really?
So....how do you calculate how do fixated pulleys reduce the weight? I thought that fixated pulleys are rather irrelevant to the reduction of force regardless of how many there are. Don't they just increase the length of the pulley but don't really affect the weight? I just can't imagine something being easier to lift because I got more fixed pulleys on the ceiling. We're talking about a single rope and subsequent fixed pulleys, yes?

I believe that would be true if the rope is attached at the fixed pulley, but it is better just to think of it as "how many lengths of rope are lifting the load?"
Then I might make do with that. :)

Never! Actually, it is a matter of semantics, some people say "draw a FBD for each particle", while others say "draw your FBD, and clearly indicate the forces on each individual particle/object." The point is to get all the forces attached to the correct body, in the correct place, in the correct direction. For example, for your motorcycle, there is only one object--the multiple wheels simply provide multiple places for the ground to hold it up.
Our lecturer rather confused me then. He says it's a must in some places. And also, in Hibbeler solution manuals he sometimes break diagrams to 2, 3, and even 4 parts at times! Does it give them a better introspection to the question? Doesn't it help to sorta "break things down"? I'm not sure if I'd be able to solve those problems without splitting the diagrams-- which I have been doing, but rather unsure as to how it's really helping me. If I have more than 1 object that isn't a pulley/wall/support I just automatically break it down...

Really?
So....how do you calculate how do fixated pulleys reduce the weight? I thought that fixated pulleys <snip>
Ahh, I should have been more clear there. I was leading into the next question, where you have multiple moving pulleys along with the fixed pulleys. You are absolutely correct that if you only have one attachment to the load, it doesn't matter how many fixed pulleys there are--still only one section of rope doing the lifting.

Our lecturer rather confused me then. He says it's a must in some places. And also, in Hibbeler solution manuals <snip>
Well, like I said its a matter of semantics. If you find it easier to focus on each component individually then feel free to split them up and consider them separate FBDs--especially if that is the way your book presents it. I should admit, I didn't actually take statics--I jumped straight into dynamics--so Hibbeler's method may be the more popular way to teach statics. Plus, teaching methods do have a way of changing over 20 years . . .

My dynamics prof always drilled it into us that no matter how many pins, wires, girders etc there were, it is still only ONE system, hence one FBD--that has as many unique components as degrees of freedom in the problem. That viewpoint is certainly more consistent with the way physicists tackle problems, which is probably part of why I ended up being a physicist ;)

Best luck,
Randy

Gold Member

Ahh, I should have been more clear there. I was leading into the next question, where you have multiple moving pulleys along with the fixed pulleys. You are absolutely correct that if you only have one attachment to the load, it doesn't matter how many fixed pulleys there are--still only one section of rope doing the lifting.
Ah good, makes sense :)

Well, like I said its a matter of semantics. If you find it easier to focus on each component individually then feel free to split them up and consider them separate FBDs--especially if that is the way your book presents it. I should admit, I didn't actually take statics--I jumped straight into dynamics--so Hibbeler's method may be the more popular way to teach statics. Plus, teaching methods do have a way of changing over 20 years . . .

My dynamics prof always drilled it into us that no matter how many pins, wires, girders etc there were, it is still only ONE system, hence one FBD--that has as many unique components as degrees of freedom in the problem. That viewpoint is certainly more consistent with the way physicists tackle problems, which is probably part of why I ended up being a physicist ;)
Ah...I envy you!

Anyway...many thanks for clarifying this for me