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From Work (Physics) - Wikipedia:

In physics, a force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force.

According to the mathematical formula of work: $$W = F . S = FS Cos \theta$$, even if the displacement is not along the direction of force (except in the perpendicular direction), work will not be zero.

Then, a force can do work even if, when acting, there is a displacement of the point of application in any direction w.r.t force except 90 degree. But Wikipedia seems to claim (in the above statement) work to be done only in the case of $\theta$ equal to zero.

I have got confused. Where am I wrong? Or is there a problem in wikipedia statement?

Edit: According to above Wikipedia statement, is force not said to do work, if the displacement is not in the direction of force?

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  • $\begingroup$ At the end of this section from Wikipedia the work done $W$ is equated to the dot product of the force $\vec F$ and the displacement $\vec s$ $W =Fs\cos \theta$ where $\theta$ if the angle between the force and the displacement. en.m.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation $\endgroup$ – Farcher Mar 31 '17 at 19:02
  • $\begingroup$ @Farcher: Thank you for the comment. I am not understanding your stand. I seem to be not clear. Do you disagree with the above Wikipedia statement? $\endgroup$ – Immortal Player Mar 31 '17 at 19:37
  • $\begingroup$ I cannot find where it says in the Wikipedia article Work(Physics) where it is claimed that work is only done in the case when $\theta=0$. $\endgroup$ – Farcher Mar 31 '17 at 19:48
  • $\begingroup$ @Farcher: Okay, I may be having problem with understanding the meaning of the statements. I haven't yet understood on where I am misunderstanding. Doesn't the above quoted text from wiki (posted in the question), a definition (?) put at the start of the main article of wiki, imply on work to be done if the force is in the direction of displacement? Thank you for the time. This may be trivial, but will help in identifying the fragile understanding of me. $\endgroup$ – Immortal Player Apr 1 '17 at 12:32
  • $\begingroup$ If the angle is 60 degrees then there is a component of the force in the direction of the displacement or a component of the displacement in the direction of the force. $\endgroup$ – Farcher Apr 1 '17 at 13:02
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You have to consider that the Work done by a constant force F by displacing a body along a directed distance r is W=FrcosΘ (Θ is the angle between the directions of the force vector and the displacement vector). In other words, the Work done is the inner vector product of the force vector and the displacement vector. That also justifies the case when the directions of the vectors of F and r are perpendicular to each other since then the angle Θ =90 degrees and thus cosΘ=0 which gives the result W=0 for that case.

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  • $\begingroup$ Thank you for the answer. I seem to have been not clear in the question. I agree with you. If I am understanding you, force does work (except the case of 90 degree) even if displacement is not in the direction of force; but wikipedia is saying force to do work if, when acting, there is a displacement of the point of application in the direction of the force. It seems that according to Wikipedia, work is not done by force, if the direction of force and displacement are not the same. $\endgroup$ – Immortal Player Mar 31 '17 at 19:27
  • $\begingroup$ @ImmortalPlayer, it is not that far complicated. Simply, when the net applied constant force F moves a body by a distance, the only case no work is produced or consumed is when the directions of the displacement of the body and the direction of the force are perpendicular to each other. I hope you agree to that and you might not need to do any further perplexing considerations. $\endgroup$ – SteveTheGrk Mar 31 '17 at 19:42
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if [...] there is a displacement [...] in the direction of the force.

Let's say that you have a rope around a large stone. You can't lift it. But you can slide it sideways.

Your force has an angle so it has both a horizontal and a vertical component. The sideways displacement that you cause is only horizontal.

So, only the horizontal component of your force made the stone move. Only the horizontal force component did work.

The horizontal component is a part of the total force you applied. So your total force did this work.

This way of splitting into components is what they refer to on the wiki quote. Because you could equivalently look at it from the opposite perspective and say that the stone does want to move in the same direction as your pulling force, but that gravity then diverts the motion, pulling it more downwards. The result is a horizontal motion.

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We take displacement in the direction of force while calculating the work done by the force. A little bit of history about work tells us that work done was invented just to simplify the concept of energy transfer. They have just given some mathematical expression for energy and said that they can be transferred - but how? That is a different topic, but if you somehow get convinced that energy is transferred, then work done is defined as that amount of energy transfer.

But which type of energy transfer? The answer is kinetic energy. If a body having constant mass gains kinetic energy, it means that the energy is being transferred to the body, and the amount of energy transfer is work done by the force.

Why is work done by force? It is because of Newton's second law of motion, which tells us that force has the physical effects of altering the velocity in the direction of force, and for constant mass, if velocity increases, kinetic energy also increases. It later turns out that the change in kinetic energy was found to be force×displacement in the direction of force. You can prove it easily using $F=ma$ and $a=(v-u)/t$. Hence we take displacement in the direction of force, and since work=change in kinetic energy, therefore, work=-change in potential energy. The expression $F×d$ has no meaning and was invented just to simplify calculation because every time you calculate the change in kinetic energy is very annoying, but the concept of energy transfer is far more convincing.

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  • $\begingroup$ Welcome to PSE! Please consider rewording the phrase: "every time you calculate the change in kinetic energy is very annoying, but the concept of energy transfer is far more convincing." Using feeling/emotion-words like "annoying" in a physics argument is usually inappropriate. $\endgroup$ – Thomas Lee Abshier ND Oct 20 at 6:00
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You can think $W=FSCosx$, either force is along the direction of displacement, or you can think work will only by the force cos component. On wiki, there is no such article which claims that work will be done only if $x=0$, let say you have this example. enter image description here

As they angle $x$ changes towards zero the force is approaching towards maximum, hence the work done will be maximum at this position.

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