# Work done on or by any object?

1. Jul 9, 2012

### Momento

If work is done on or by a object. Is the force perpendicular to the displacement? For any object that is in motion or able to do work or work is being done on it.

Example:

A ball thrown up in the sky, force applied on the ball is perpendicular to the displacement.(Is this applied on all?)

Thanks.

2. Jul 9, 2012

### sophiecentaur

Perhaps you should re-think that statement. How does the ball go up of there is no force in that direction. How does it accelerate in the horizontal direction without a force in that direction too?

3. Jul 9, 2012

### Momento

Not really sure where you're going at here?

4. Jul 9, 2012

### Staff: Mentor

In order for a force to do work, it must have a component parallel to the displacement.

What force are you talking about? The force of the hand that launches the ball? If so, that force is in the same direction as the displacement, assuming the ball is thrown vertically upward.

5. Jul 9, 2012

### A.T.

http://en.wikipedia.org/wiki/Work_(physics)

6. Jul 9, 2012

### Momento

I know that W = F x D

Where force has to be perpendicular with distance is that correct?

(Sorry I didn't really get all you're points so... I used bad example at the wrong time.)

7. Jul 9, 2012

### Staff: Mentor

No, force has to be parallel to the displacement (at least in part).

Work is the scalar product of two vectors:
$$W = \vec{F}\cdot\vec{D} = FD\cos\theta$$
Where θ is the angle between the two vectors. If they are perpendicular, that scalar product will be zero. (Cosθ = 0 when θ = 90°.)

8. Jul 9, 2012

### sophiecentaur

I think memento is confusing work with moments, perhaps. That could explain what's emerged from his memory: right formula by wrong context. :-)

9. Jul 9, 2012

### Momento

Thank you!

10. Jul 9, 2012

### CWatters

Amazing. Never seen someone that needs to use scalar products of vectors to understand that in order to throw a ball UP the force on the ball has to be UP and not for example sideways. More time in the garden and less on the xbox perhaps :-)

11. Jul 9, 2012

### dipole

Less competition for me!

12. Jul 25, 2012

### Momento

It's also AMAZING to find rude people on this site who would join a discussion just to reply in a very childish manner. "AMAZING"

Thanks to everyone who help me out with this just needed some clarification.

13. Jul 25, 2012

### sophiecentaur

I do think he has a point, though. It can hardly be surprising that it takes an UP force to have an effect in the UP direction. I know one can't always rely on 'common sense' in Physics - the Greeks (ancient ones, that is) fell over when they tried that - but, nevertheless . . . .
"Rude" is hardly the description, though. Have you never seen REAL RUDE on the internet?

14. Jul 25, 2012

### Momento

I have, but in a forum like this I doubted to find such manner.

15. Jul 25, 2012

### sophiecentaur

Well . . . . with respect, if you had looked up the definition of work before you asked your question (thousands of possible links) you would have not asked it. Forums like this are supposed to entertain the people who answer the questions as well as the questioners and I don't think it's too much to ask that people do some of their own research before asking their questions. You know - just make a bit of an effort!!

16. Jul 25, 2012

### mikeph

I don't see any need to be rude though.

17. Jul 25, 2012

### tiny-tim

Momento, CWatters was making the friendly and helpful point that understanding a physics formula doesn't just mean understanding the maths, it also means understanding the reality

something which you could have checked just by actually throwing something in (for example) the garden, and watching it

18. Jul 25, 2012

### sophiecentaur

But I see a great need to put a bit of effort into one's questions. That, also, is a matter of courtesy.
Anyway - why be so thin skinned about it? We delivered a useful answer - even though it may not have been well deserved.

19. Jul 26, 2012

### CWatters

Please accept my apologies. No offense intended.

20. Jul 28, 2012

### Darwin123

You are interpreting the symbol "×" for cross product in vector multiplication. The symbol "×" is not always used for cross product. The symbol "×" is often used for scalar multiplication (multiplication of two numbers). That is how it is introduced.
Work is the inner product between the two vectors, force and displacement. Other people on this forum have described the inner product.
Just for later reference, the formula that you used is valid for torque. W is the torque, F is the force vector, D is the position vector relative to a fulcrum point, and "×" is the cross product. I bring this up only because I have seen some writers refer to ω as torque.