Work done on or by any object?

[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.

 

[sophiecentaur]

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.

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?

 

[Momento]

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?

Not really sure where you're going at here?

 

[Doc Al]

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.

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

Example:

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

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.

 

[A.T.]

If work is done on or by a object. Is the force perpendicular to the displacement?

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

only the component of the force parallel to the application point velocity is doing work

 

[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.)

 

[Doc Al]

I know that W = F x D

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

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°.)

 

[sophiecentaur]

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

 

[Momento]

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°.)

Thank you!

 

[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 :-)

 

[dipole]

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 :-)

Less competition for me!

 

[Momento]

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 :-)

Less competition for me!

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.

 

[sophiecentaur]

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.

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?

 

[Momento]

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?

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

 

[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!

 

[mikeph]

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

 

[tiny-tim]

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"

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

 

[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.

 

[CWatters]

Please accept my apologies. No offense intended.

 

[Darwin123]

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.)

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.