# Utterly lost - internal torques

1. Jan 29, 2006

### kajalove

It'd be best if I start with example.
Screwdriver helps us by having a handle with a large radius that provides a mechanical advantage in turning a blade with a smaller radius. So whatever force exerted on screwdriver's handle by hand, same force is exerted on screw. But torque on a handle is different from torque on a blade?

Next question
If you pull a body with certain force it will start accelerating and exert equal but oposite force on you.
Is there something similar with torques? If I make an object rotate will I experience same kind of torque caused by that object?

This question also relates to internal torques, since the sum of internal torques in a system is zero. If me and an object I made rotate comprise a system, then all internal torques should sum to zero. Will there be torque on my body with same magnitude? I don't get it!

thanx

2. Jan 30, 2006

### Hootenanny

Staff Emeritus
This statement is incorrect. It is the torque which remains constant, but because this torque is exerted on an object with a smaller radius (the screw) the force is increased, thus giving a mechanical advantage.

3. Jan 30, 2006

### FredGarvin

Absolutely. Look at a helicopter. There's a reason why there is either a tail rotor or counter-rotating rotors...The torque the main rotor disk requires to turn requires an equal and opposite torque to keep the aircraft from spinning in the opposite direction to the main rotor. That torque must be countered which is why tail rotors are specifically called "counter torque" or "anti-torque" rotors.

With the same magnitude as what? If the system you are in is not experiencing an angular acceleration, then you will have no NET torque, i.e. all torques will have an equal and opposite. They will still be there, but they will effectively cancel each other out. Again, think of the helicopter example. Look at a helicopter hovering. For all intents and purposes, you have two torques; the torque created by the engines to turn the main rotor and the torque counteracting the main engine torque. The two torques are still there, but they negate each other to give a zero NET torque.

4. Jan 31, 2006

### kajalove

Torques really can get confusing:grumpy:

Why would torques be the same ?Is there some proof you can show me (mathematical or vector)?

And why would force be greater? I do know that when lever arm is smaller there needs to be greater force for same torque, but if you only applied force 10 N where did additional force come from?

This is what is confusing me the most.How can torques negate eachother? I know how torques negate eachother when it comes to wheels ( where you apply two forces equal in size and direction one on left side and other on right side of the wheel ->that way torques have opposite directions and cancel eachother out). Those are relatively easy to understand. But what if me and an object that together comprise isolated system are floating somewhere in space and I exert torque on object and in turn object exerts on me same force I applied to it. But what torque will I experience?
If object has an axis around which it rotates then we know the magnitude of torque I caused (if we know the force I applied and the lever), but where is my axis (around which I would rotate)? For object to apply same torque on me then perpendicular distance from mine axis of rotation to the line of action of the force applied by object would need to be the same as distance from object's axis of rotation to the line of action of the force exerted on that object?

5. Jan 31, 2006

### Hootenanny

Staff Emeritus
6. Jan 31, 2006

### Hootenanny

Staff Emeritus
Think about Newton's 3rd law as proof.

7. Feb 1, 2006

### kajalove

I'm sorry mate but I still don't get it. I would really need some thorough explanation

8. Feb 1, 2006

### Hootenanny

Staff Emeritus
Forget about the screwdriver, helicopter and internal torques for the moment. The most simple example of a torque is a seesaw. Imagine you have a 2 meter long plank pivoted about its center, with a mass of 5kg 0.5m from the pivot. If your pushing on the end of the other side how hard would you have to push to keep the plank level?

Use the formula $$\tau = Fd$$ where $\tau$ is torque, $F$ is force and $d$ is perpendicular displacement from the pivot.

9. Feb 2, 2006

### kajalove

that's easy.Both clockwise and counterclockwise torques would have to be equal in order to balance eachother out

T = F * d = 50N * 0.5m = 25 N m

F = T / d = 25 N

I would have to push with force 25 N.

10. Feb 3, 2006

### Hootenanny

Staff Emeritus
Yes now apply this formula to a screwdriver. This is called equating torques. Basically what you are saying is "If I apply a torque on the handle then the same torque will be applied at the screwdriver head (because the screwdriver doesn't deform); but beacuse the radius of the screwhead is smaller than that of the handle (and therefore the distance of the acting force from the axis of rotation) therefore the force applied by the screwhead is larger than that of the handle". Hope that helps.

$$F_{handle} R_{handle} = \tau = F_{head} R_{head}$$

Last edited: Feb 3, 2006
11. Feb 3, 2006

### kajalove

Now I know better how torques negate eachother when it comes to screwdrivers and similar .
But the following I still don't understand:

12. Feb 4, 2006

### Hootenanny

Staff Emeritus
The applied torques would certainly be equal but the perpendicular force you experenced would indeed depend on your axis of rotation which would depend on the position and angle at which the force was applied.

13. Feb 4, 2006

### kajalove

I would think applied forces (due to newton's third law) would be equal while torque would depend on my axis of rotation.

could you explain it in a bit more detail since I haven't got a clue?

14. Feb 5, 2006

### Hootenanny

Staff Emeritus
I apologise, I said it the wrong way round. The forces would be equal but the torques would indeed be different, which would depends on the angle and the axis of rotation. Sorry my mistake.