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My question is regarding work by a force and it is one which is giving me many issues in understanding when work done by force is 0.

My question can be described as a multiple choice question.

Q) No work is done by a force on an object if -

a) the force is always perpendicular to velocity

b) the force is always perpendicular to acceleration

c) the object is stationary but point of application of force moves on the object

d) the object moves in such a way that the point of application of force remains fixed

The answers are (a), (c) and (d)

Now obviously I know (a) is true. Can somebody please explain :

  • What the OPTIONS (c) and (d) mean. I'm unable to understand the options clearly. What does THIS POINT OF APPLICATION OF FORCE REFER TO? Is it the force or the point on which force is acting?

  • Relate (c) and (d) to a sphere pure rolling. I know in pure rolling velocity of point of contact is 0 and hence point of contact doesn't move and hence work done by friction is 0. Now how to analyse using (c) or (d)? Which option does sphere rolling satisfy? Is it (c) or (d)?

  • Relate (c) and (d) with a pulley of mass M and moment of Inertia I about axis of rotation assuming string doesn't slip on pulley. I'm unable to analyse this situation in any way. How is work done by both tensions on a ROTATING PULLEY with mass 0? (The string doesn't slip on pulley and pulley has mass M and moment of inertia I about axis).Which option does pulley rotating satisfy? Is it (c) or (d)?

  • Examples for (c) and (d). I want to first understand (c) and (d) and then relate it with many other examples

I'd really like a SIMPLE explanation highlighting the above issues in order to analyse WHEN WORK DONE BY A FORCE IS 0 without any problems.

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  • $\begingroup$ there are a lot of questions in this, i'm voting to close as being too broad for now $\endgroup$ – Alex Robinson Mar 30 '18 at 13:58
  • $\begingroup$ Its only one question. I'd like to know the reason to the answer to the mcq question and relate it with a pulley rotating. Forget everything else and explain what (c) and (d) mean by defining POINT OF APPLICATION OF FORCE and relate the appropriate option to a pulley rotating to explain why work done by tension on a pulley is 0. $\endgroup$ – Hola Mar 30 '18 at 14:08
  • $\begingroup$ Ola, you state it is one question, but every bullet point has a question at the start of it, before you explain a bit about your thinking $\endgroup$ – Alex Robinson Mar 30 '18 at 14:12
  • $\begingroup$ See my comment, that is all I require. I dont think asking for examples makes a question long now does it? $\endgroup$ – Hola Mar 30 '18 at 14:20
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    $\begingroup$ @Ola, I count 8 question marks in your "one question", which also happens to look very much like a homework problem (which is also, okay). However, if it is a homework problem, I suggest that you read the answer to this question: meta.stackoverflow.com/questions/334822/…. Then, consider reformulating your question using the suggested guidelines. I hope that this suggestion helps, and I've also provided a general answer to the question in your title myself. $\endgroup$ – Charles Mar 31 '18 at 5:48
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Option c is correct as, even though there are many forces the displacement due to every force is 0,therefore, summation(f.ds)=0 .

U cannot relate it to rolling case as there are many forces acting but on rolling there is friction only.

For opt d, u can see it as the body divided into 2 small pieces,namely

1 The point of application piece

2 the rest of the body piece.

So since work is a scalar quantity we can add the work done on the two pieces after calculating them separately. Note: force acts only on piece 1 and not on piece 2.

Work no

1.On point of application piece:F*(displacement) =F*0=0

  1. On other pieces : F*(d) =0*(d)=0.

Work total=W1+W2=0+0=0.

Hence ur answer.

U can relate it to the rolling problem, the point of applications of force of friction is the same i. e. the lower most point, Hence the work done by friction is zero.

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  • $\begingroup$ How is work done by both tensions on a rotating pulley(having mass) zero ? $\endgroup$ – Hola Mar 30 '18 at 14:33
  • $\begingroup$ That is because even though the force is applied the Centre of mass never moves the pulley is fixed, hence the work that is force times displacement is zero as displacement is zero $\endgroup$ – Pkman Mar 30 '18 at 14:38
  • $\begingroup$ I hope this has cleared ur doubts? $\endgroup$ – Pkman Mar 30 '18 at 14:51
  • $\begingroup$ Yes, but work done by tension in a pulley rotating with angular acceleration a/R and translating down with acceleration a is also 0. Here R is radius of pulley. Why? I was actually referring to a pulley rolling and translating down (and the rope doesn't slip on pulley) $\endgroup$ – Hola Mar 30 '18 at 15:30
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The net work done by a force is equal to zero... when the net work is zero, of course! Let's first define what the net work is, then try to answer this question. Work is defined as the inner product of force and distance:

$\mathbf{W} = \mathbf{F} \bullet \mathbf{d}$

Note that these are vector quantities, so the line of force and line of direction must match for any work to occur. This means that we have a zero net force when one of two things happen:

1) There is net zero force along the direction $\mathbf{d}$.

or (that is an inclusive or, meaning one or both of these may be true)

2) There is net zero distance along which the force ($\mathbf{F}$) is applied.

How do we know what the force is? Newton's law of motion:

$\mathbf{F} = m\mathbf{a}$.

I'll give two examples of when zero net work may occur:

1) A circular orbiting satellite. A circular orbiting satellite, moving in a circle, is always accelerating towards the center of the orbit (let's say Earth). The direction of acceleration, and force (see Newton's equation), happen to be perpendicular to the velocity (i.e. motion, or distance traveled). Therefore, the gravitational force (which is the force acting on the satellite) is doing zero net work on the satellite because the satellite is not moving along the direction of the force (they are perpendicular).

2) A stationary building in strong winds. Strong winds acting on a stationary building perform zero net work on the building, no matter where the wind blows on the building (the force application point) nor which direction the wind is blowing, because the building is stationary ($\mathbf{d} = \mathbf{0}$).

You can similarly construct more examples based on my approach. I hope that helps.

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  • $\begingroup$ Look, forget my question. I understand it might be many questions so answer me one thing- All I want is this (as I stated on my comment) - Work done by both tensions in a pulley with moment of inertia I rotating with angular acceleration a/R and translating down with acceleration a is also 0. Here R is radius of pulley and the string doesnt slip on the pulley. Why? Why is the work done by both tensions 0? $\endgroup$ – Hola Mar 31 '18 at 6:53
  • $\begingroup$ I'm explaining the situation (not asking another question).Consider a massless string attached to fixed support on one side and the string joins a massless SPRING fixed to the SAME support on the other side passing through a pulley along the way, why is work done by both tensions 0?Obviously here, the pulley falls under gravity , if it falls by distance x , elongation in spring is 2x as string length is constant. Assume it has linear acceleration a and the tensions (T1 and T2) provide for angular acceleration a/R as the string doesnt slip on pulley. How is work done by T1 and T2 zero?? $\endgroup$ – Hola Mar 31 '18 at 6:55
  • $\begingroup$ And the reason I'm saying the work done by both tensions is 0 is because the situation I explained is an SHM problem as thr centre of mass of pulley executes SHM and to prove that we write mechanical energy and derivate wrt TIME and EQUATE TO 0 and obtain acceleration and we obtain it as -(constant) x which proves it executes shm. Now this derivative will be 0 only if WORK done by TENSIONS T1 and T2 =0 (as system must be conservative). Hence my question, why is work done by T1 and T2 0? (Please note, in MEAN position T1 = T2 as pulley is in rotational and translational equilibrium) $\endgroup$ – Hola Mar 31 '18 at 7:01
  • $\begingroup$ And actually the pulley relation in my original question was of the same type(although i exactly didnt convey it in the question) I wasnt refering to a pulley only in rotational motion and fixed. I was refrrring to a situation where a pulley having mass rotates and translates down and the strings dont slip on pulley, why is the work done by tensions =0? Answer me this ONE QUESTION. Thats ALL I need. $\endgroup$ – Hola Mar 31 '18 at 7:06
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    $\begingroup$ @Ola, I suggest you either start a new question, or edit this question. And before simply copying and pasting your comments, please research how to ask effective questions on Stack Exchange. Then, continue with writing your question and include diagrams to help convey your problem, show supporting equations that are relevant to the problem, and show that you’ve tried to answer the problem yourself. I guarantee that more people will join in the effort to help. I’m going to sleep, I’ll take a look again tomorrow sometime. Good luck. $\endgroup$ – Charles Mar 31 '18 at 7:24

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