How can pulleys or levers multiply forces?


by NANDHU001
Tags: forces, levers, multiply, pulleys
NANDHU001
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#1
Feb8-12, 07:32 AM
P: 22
Can anybody please tell me how pulleys and levers multiply forces without using ideas such as input power(or work)= output power(or work), torque etc. I can't understand the basic mechanism behind it. Please check out my doubt attached to this thread as a bmp image.
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JHamm
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#2
Feb8-12, 09:30 AM
P: 389
I like to think of it like this, take a wrench and grab it a distance [itex]d[/itex] from the nut so that [itex]d << l [/itex] where [itex]l[/itex] is the length of the wrench. If you push down (or across depending on how you're picturing it) a very small amount then the wrench has been displaced by an angle [itex]\phi[/itex] and the net displacement of the wrench from its original position is the sum of the changes in position of each part, since each part moves a distance [itex]r \phi[/itex] where [itex]r[/itex] is the distance from the nut and is summed over the length of the wrench so that we have
[tex] D = \int_0^l r \phi dr = \frac{\phi}{2}l^2 [/tex]
Where [itex]D[/itex] is our total displacement.

Now if you try this again at another point [itex]d_2 >> d_1[/itex] the angle is much much less which means your total displacement is also much less. So small movements a greater distance from the fulcrum should require less force because you end up displacing less mass per "push".
NANDHU001
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#3
Feb9-12, 10:09 AM
P: 22
Thanks JHamm, but I still don't feel comfortable with that idea, may be I didn't understand it well enough.

Why should displacing less mass (also how does mass get lesser) per push make it easier, isn't it acceleration of mass that is to be considered? Also try to clear my doubt attached with the thread in the beginning.

Consider a spherical mass attached to one end of a rigid rod whose other end is pivoted and is the axis,(mass of rod is negligible)(the whole setup is like a simple pedulum). Now why sholud the same force applied at different distances from the mass have different effect.
If we consider the instantaneous acceleration of the mass due to the application of a force, the path can be treated as a small linear displacement and then isn't it like a linear force acting on a body making it move in a straight line, then why should the point of application of force matter in the other case.
Please don't explain it using torque and work as I don't get the logic in such explanations.

Also consider clearing my doubt posted as a bmp file.

bootsnbraces
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#4
Feb9-12, 02:23 PM
P: 13

How can pulleys or levers multiply forces?


Ok try it with the pulleys idea, if you have a small gear driving a large gear )lets say twice the number of teeth) for one revoloution of the smaller gear- say 10teeth only half a revoloution of the larger gear (20th in this case) can take place as the teeth lock each other together.
The force you have used to turn the first gear 360 degrees has only been used to turn the larger gear 180 degrees simple

The work youve done is the same throughout the system but in the second gear instead of having say a 200watt motors input divided by 360degrees you have 200 watts divided by 180 degrees giving 0.5watt per degree and 1watt per degree. Therefore the larger second gear can drive a load requiring 1 watt per degree of rotation to move it but it can only move it at half the rate of the first gear.

I know technically my watts per deg system doesnt work but its the easiest way i can think of explaining it
Is that any use?
rcgldr
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#5
Feb9-12, 03:39 PM
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Quote Quote by NANDHU001 View Post
Please check out my doubt attached to this thread as a bmp image.
The issue in the case of your image is that force is related to the rate of acceleration, and the rate of acceleration for the block on the left is 1/2 the rate of acceleration of the block on the right.
zoobyshoe
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#6
Feb9-12, 04:00 PM
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Quote Quote by NANDHU001 View Post
Now why sholud the same force applied at different distances from the mass have different effect.
In your diagram, once the small mass hits the longer lever arm it moves the longer lever arm at twice the speed for twice the distance than the shorter arm moves. Whatever distance and speed the large mass is moved by the lever arm, the smaller mass moves double that, since it is acting on the lever at twice the distance from the fulcrum.

Mechanical advantage permits the movement of a large weight by a small one, but only because the large weight does not move as far or as fast as the small one. There is always a trade off: a small force acting faster over a longer distance can move a larger weight, but only because the larger weight moves more slowly and over a smaller distance.

Consider your lever with no masses at all. Any motion of the arms requires that the longer arm move twice the distance and speed the shorter arm moves. This ratio of distance and speed is locked in to the ratio of the length of the arms.

Everything is conserved as usual: energy, mass, momentum.
NANDHU001
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#7
Feb10-12, 10:28 AM
P: 22
Are levers useful only when force of gravity acts on the loads ,that is do they actully help only in raising a large weight using a foce less than its weight(force due to gravity). Are they useful in the absence of a gravitational force ?
James_Harford
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#8
Feb10-12, 11:18 AM
P: 67
Quote Quote by NANDHU001 View Post
Are levers useful only when force of gravity acts on the loads ,that is do they actully help only in raising a large weight using a foce less than its weight(force due to gravity). Are they useful in the absence of a gravitational force ?
Levers are useful in both cases, so any explanation about how a lever works has nothing to do with gravity. It is only one of many ways to apply a force to a lever.

Based on your questions, perhaps you are more comfortable with F = dp/dt, than F= dE/dx? As definitions of force, they are equivalent, but the advantage of using F = dE/dx to answer your question is that we can dispense with the relatively complicated details of accelerating masses or earth's gravity.

So I would ask myself, what is it about F = dE/dx that I need to better understand? Because once you do, the answer to your question will be trivial.

You may get a somewhat intuitive feel for F = dE/dx, by considering the difference of force involved in climbing a hill up its steep side versus its gently sloping side.

- Regards
zoobyshoe
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#9
Feb10-12, 02:28 PM
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Quote Quote by NANDHU001 View Post
Are levers useful only when force of gravity acts on the loads ,that is do they actully help only in raising a large weight using a foce less than its weight(force due to gravity). Are they useful in the absence of a gravitational force ?
I get the feeling you don't realize how many different manifestations of the lever there are. There are three classes of lever, not just the first class lever you depicted, and there are so many man made (tools, machines) and natural configurations of (arms, legs, wings) levers that it's safe to say the lever is ubiquitous. Any lever that works on earth regardless of gravity would obviously also work in space. For example: a pair of pliers.

All machines are elaborations of the simple machines and all simple machines are useful because they provide mechanical advantage. Every space vehicle ever launched has had purely mechanical elements, as does every computer, CD player, printer, scanner, and any other device you can think of that has nothing to do with lifting weights against gravity.
NANDHU001
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#10
Feb11-12, 04:36 AM
P: 22
Can anyone explain the working of pulleys and levers by using momentum tranfer alone and some kinematics without using ideas of work,energy etc?
zoobyshoe
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#11
Feb11-12, 05:19 AM
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Quote Quote by NANDHU001 View Post
Can anyone explain the working of pulleys and levers by using momentum tranfer alone and some kinematics without using ideas of work,energy etc?
On your diagram you write:

"By Newton's third law since m exerts a force F on lever arm 2d it should experience a reaction force F itself reducing its momentum by F x t. Doesn't this violate law of conservation of momentum."

Can you explain this apparent violation of conservation of momentum you're worried about more clearly?
NANDHU001
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#12
Feb11-12, 11:02 AM
P: 22
What I meant is that the initial momentum was contributed by 'm' alone with 'M' at rest,
later when 'm' loses a momentum of 'F x t' the mass 'M' gains a momentum of ' 2xFxt ' (as force is double at 2d than that at d,force is now 2F). To conserve momentum when 'm' loses a momentum of 'Fxt' , 'M' should gain the same (ie, F x t) but here it ' is ' 2xFxt.

Thanks zoobyshoe for your interest.
zoobyshoe
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#13
Feb11-12, 05:25 PM
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Quote Quote by NANDHU001 View Post
What I meant is that the initial momentum was contributed by 'm' alone with 'M' at rest,
later when 'm' loses a momentum of 'F x t' the mass 'M' gains a momentum of ' 2xFxt ' (as force is double at 2d than that at d,force is now 2F). To conserve momentum when 'm' loses a momentum of 'Fxt' , 'M' should gain the same (ie, F x t) but here it ' is ' 2xFxt.

Thanks zoobyshoe for your interest.
You are welcome. Part of the problem, at least, may be that you are using the formula for impulse and calling it momentum.

If you stick to one concept at a time, you should be able to see there's no violation. If you are worried that conservation of momentum may have been violated, then momentum is the quantity to account for, mass x velocity. All the momentum is initially in the smaller moving mass, as you say, then some, or all of it is transferred to the larger mass.

Let's suppose this to be a one dimensional elastic collision. The small mass has some momentum composed of its mass times its velocity. The large mass is at rest.

When the small mass hits the lever at 2d, the initial mv of the small mass, channeled through the lever, acts on the larger mass as though it (the small mass) had twice the mass it actually has, BUT it also acts on the larger mass as if it (the small mass) only had 1/2 the velocity it actually has.

If the initial momentum of the small mass were, for instance, composed of 2kg 's of mass by 12 m/s of velocity = 24 kg*m/s of momentum, acting through the lever would increase its effective mass to 4kg, BUT it would cut its velocity in half to 6 m/s. Therefore, it would be acting on the larger mass at rest with a momentum of 24 kg*m/s, exactly the same momentum it had before it hit the lever.
cshum00
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#14
Feb11-12, 09:08 PM
P: 213
Well, think of it this way. You have to make a stack of 20 books. (1) You can either put one book on top of another one at the time. (2) Or put put five books at the time. In the first case, it will take you more time but it will be easy because 1 book is not heavy. In the second case, you will finish faster but it will be more difficult because 5 books at the time is heavier.

Now, i use your lever picture as the example. On the left side, the beam is shorter while the right side is larger. On the larger beam, it will require you to move more space to complete a angle and you only have to push with little force at the time. However, on the smaller beam; you will complete the angle faster but it will require to use more force at the time.

Another way to explain it is using fractions and ratios. Angle is actually a ratio of two numbers:[tex]\theta = \frac{arc}{radius}[/tex]The arc or total space displaced divided by the radius. And asume you moved an angle of 2 on the lever you get the following ratio:[tex]\frac{2m}{1m} = \theta_1 = \theta_2 = \frac{4m}{2m}[/tex]On the smaller beam, you only have to move 2m. But on the larger beam you have to move 4m, twice more the space. The amount of effort put in both is the same. The difference is that in one, you put all at once which is more difficult and requires more force. The other one requires little force at the time but longer time to add up the amount of effort.
NANDHU001
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#15
Apr30-12, 10:48 AM
P: 22
I see that in a pulley system if the distance of application of force is halved at the load end, the force gets doubled at that end. But even then the acceleration it produces at the load end in the absence of resistive forces such as gravity,friction etc is not what we expect from a=F/m. ie, even if 1N is multiplied to 2N at the load end the load of mass 1 kg is not accelerated by 2m/s*s in the absence of resistive forces. Why is it so?
Bob S
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#16
Apr30-12, 11:20 AM
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In the block and tackle picture (courtesy Wikipedia), all of the ropes supporting the load have equal tension. ADD up the number of ropes supporting the load. The SUM of the number of ropes, MULTIPLIED by the force of the person pulling the rope, is the weight of the load.
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chingel
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#17
Apr30-12, 02:25 PM
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In the original question, lets say both of the masses weigh 1 kg, mass A is 1 m from the fulcrum and mass B is 2 m from the fulcrum. Since they are connected by a lever, any force applied by mass B gets felt twice as strongly by mass A. Lets say we make the forces apply for a second and the average force felt by mass B is 1 N, then the average force felt by mass A is 2N. Change of momentum is force times time, for mass A it is 2, for mass B it is 1. One seems to have gotten more momentum and I'm pretty sure it would actually get twice the momentum. But how? Well, to apply a force through such a lever, you need a solidly attached fulcrum, so that you have something to push off of. The total force felt by the fulcrum is 3N. You can think of it as the fulcrum applying a 3N force to the rod. Since one mass in closer to the application point of the force, it feels more force. So in essence the earth gives more momentum to mass A, because of a lever advantage.

I don't know why levers multiply force. With the ropes, the explanation that all the ropes are at the same tension is interesting. But with a regular lever, how would it work? The work principle was taken into use because it was noticed that you can multiply force while force times distance stays constant. But the lever doesn't know that, it just transfers forces and how exactly does the force transfer such as to multiply it?


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