Hello Physicists!

I am studying Newton's Laws of Motion, and am working out some problems on Pulleys and Constrained motion. I have heard of a technique called the **Virtual Work Method**, also called the *T dot Trick* or the *Tension Trick*. I would like to learn about it, to make my problem solving effective. So please write about it in this note and attach an example problem. I'll be highly grateful.

Thanks.

*Swapnil*

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TopNewestThe crux of the Virtual Work Method is that the sum of work done by all

internal forces of a systemis zero!. So to explain this in the context of pulley block system I'll consider bodies \(B_1,B_2,B_3\dots B_N\) which, due to the strings they are attached to, are experiencing a tension of \(\vec{T_1},\vec{T_2}\dots \vec{T_N}\). Also, consider that in a certain time \(t\) these blocks get displaced by \(\vec{S_1},\vec{S_2}\dots \vec{S_N}\). Since thetotal work done by all internal forcesis zero, we can write:\(\boxed{\displaystyle \sum_{i=0}^{N} \vec{T_{i}}.\vec{S_{i}} = 0}\)...\((1)\)

Differentiating the above equation twice we get:

\(\boxed{\displaystyle \sum_{i=0}^{N} \vec{T_{i}}.\vec{a_{i}} = 0}\) (Here \(\vec{a_i}\) denotes the accelleration of the body \(B_i\))

This method is basically used to find the relationship between the accelerations of the various moving bodies of the system.

Hope this will help you! – Miraj Shah · 4 months, 2 weeks ago

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– Swapnil Das · 4 months, 2 weeks ago

This one helped me ;)Log in to reply

virtual work is a really a god trick when you get fed up with constrains writing static equilibrium condition . virtual work principle state that when a body in static equilibrium is given a virtual displacement dx then net work done is 0. – Aryan Goyat · 5 months, 1 week ago

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a mass tied with a string in vertical plane with horizontal force acting find it if it is at angle theta with the vertical using virtual work method. (it s too easy) – Aryan Goyat · 5 months, 1 week ago

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just give a displacement d(theta) and let net work done =0

mg(-lsin(theta)d(theta)+F(lcos(theta)d(theta)=0

F=mgtan(theta)

{note-this may appear long this time but its time saving trick } – Aryan Goyat · 5 months, 1 week ago

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Also can anybody please explain in brief about instantaneous axis of rotation. – Harsh Shrivastava · 5 months, 1 week ago

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– Deeparaj Bhat · 5 months, 1 week ago

It is the point about which the body appers to be in purely rotational motion (no translatory motion).Log in to reply

– Harsh Shrivastava · 5 months, 1 week ago

How to apply this concept in problem solving?Log in to reply

– Deeparaj Bhat · 5 months, 1 week ago

You can figure out the ratios of veloties of certain points if you know their distance from the ICR and the angular velocity. Plus, you can calculate kinetic energy by knowing the moment of intertia about that point (you don't need to calculate translational kinetic energy).Log in to reply

Please reply and enlighten me. – Swapnil Das · 5 months, 1 week ago

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– Deeparaj Bhat · 5 months, 1 week ago

It basically says \(\displaystyle \sum \vec{T} \cdot \vec{a} = 0\) where \(T\) is the tension of the string and \(a\) is the acceleration of bodies in contact with it.Log in to reply

– Swapnil Das · 5 months, 1 week ago

Could you add one example? Thanks.Log in to reply