What is the difference in torque produced by a wrench versus a belt?

[joeyh1234]

Hello All

I have recently been trying to get my head around torque and think I am getting there.

but I am struggling to understand the different of using a wrench to move an object i.e a bolt and using a belt to move something ok i.e an alternator.

I don't get is it the entire lenth of the belt or just the distonce from the belt runner and the belt, which determines the amount of torque produced by a belt.

i have done some research and at the moment the maths seems to go over my head massively

so if someone could explain it easier that would be great

thanks
joe

 

[A.T.]

Here a nice video:

https://www.youtube.com/watch?v=Sm4pV3xyJRE

I don't get is it the entire lenth of the belt or just the distonce from the belt runner and the belt, which determines the amount of torque produced by a belt.

See here:
http://www.engineeringtoolbox.com/belt-transmission-power-efficciency-d_1378.html

 

[joeyh1234]

hi many thanks ill give this one a watch when i get home from work.

 

[SalfordPhysics]

A very nice video A.T.

 

[joeyh1234]

out of curiosity as i can't watch it just yet, does it explain belts and torque??

 

[jaydnul]

For a wrench, the further you are along the handle, the more torque you'll get. Think of it like this. $W=Fd$. The work being done is always the same, you're turning the bolt. So as the radius increases, so too does the arc length, requiring less continuos force to satisfy the relationship.

 

[joeyh1234]

hi thanks for this, so are you saying the bigger the radius of the wheel which the belt is attached to, essentially the less force required.?
i do appreciate your response and apologies if my questioning is idiotic...

 

[jaydnul]

With belts or gears, the wheels are different sizes. Your work will always be the same to turn a gear 360 degrees. If you attach a bigger gear to the one that you are trying to turn, then one full rotation of the big gear will translate to many rotations in the original gear, which means more work. In other words, to turn your original gear 360 degrees,it will take only a fraction of a turn on the bigger gear, but more force is required because $W=Fd$.

 

[joeyh1234]

hey thanks for the response...i think my question was poorly constructed.
What i want to understand/know is when you have an alternator as a for instance, which is turned using a belt. i am getting a little confused with is it the total length of the belt, is it the size of the initial belt disc or ? which determines how much force is excerted on the alternator... I understand the belts and gears bit by the way i am trying to figure out how much torque needs to go through the initial disc to turn the alternator?

thnaks
joe

 

[A.T.]

i am trying to figure out how much torque needs to go through the initial disc to turn the alternator?

That depends on the internal resistance of the alternator.

 

[jaydnul]

Theoretically, the length of the belt wouldn't matter, assuming it is rigid. The bigger the first gear is relative to the alternator, the more force it will apply, but for a shorter interval of time (assuming the same energy). The torque on both wheels would be the same.

 

[sophiecentaur]

Here a nice video:





See here:
http://www.engineeringtoolbox.com/belt-transmission-power-efficciency-d_1378.html

I don't like that video simulation much. (Simulations should always be looked on as possibly suspect; see Tom and Jerry). It shows the balance returning to the horizontal position but does not explain why it does. It is not just because the weights are the same; it is because the beam is suspended below the pivot and there is a restoring moment due to this when the beam is displaced. If the pivot were in the middle of the beam, the beam would stay in any position (imagine weights, diametrically apart on a cycle wheel) or it would be completely unstable if the pivot were under the beam. (Stable / neutral / unstable equilibrium - as I learned at School)

 

[Nathanael]

It is not just because the weights are the same; it is because the beam is suspended below the pivot and there is a restoring moment due to this when the beam is displaced.

The video explains this (at about 1:30)

 

[sophiecentaur]

The video explains this (at about 1:30)

That was remiss of me. I missed that bit.

 

[Okefenokee]

You say that you understand the concept of torque on a wrench correct?

Now imagine that you tie a rope on the end of the wrench and pull with a perpendicular force, meaning that the rope is at a right angle to the wrench. Isn't that sort of like a belt on a pulley? Think about it. If the belt can't slip then it's not much different from pulling a rope tied to a lever. Does it matter how long the rope is? What happens if you tie the rope in the middle of the wrench? You get less torque because you have less leverage, correct? If the wrench is 100 meters long then pulling on the rope would really torque that bolt.

Now imagine that you're doing the reverse. Let's say that you are twisting the bolt that the wrench is on in order to lift a weight tied to the end of the wrench. If the wrench is long then your wrists would have work really hard to lift the weight, right? If the wrench is really short then it's not so bad.

So, on the load side, a big pulley means that the load receives a big torque. On the supply side the opposite is true. A big pulley means that that a smaller torque is supplied.

 

[sophiecentaur]

There are dozens of web sites with help with the 'Principle of Moments'. Once you get the idea that the distance that counts, in calculating torque, is the so-called perpendicular distance from the pivot to the line of action of the force and not just the length of the pole, . This link shows it well.

When you are dealing with a belt on a pulley, the belt always leaves along a tangent to the circle so it is the radius of the pulley that is always the perpendicular distance or "lever arm", as the Hyperphysics page calls it (one less thing to work out when you deal with pulleys rather than levers-at-an-angle).