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Virtual Work Method

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

Note by Swapnil Das
5 months, 1 week ago

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The crux of the Virtual Work Method is that the sum of work done by all internal forces of a system is 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 the total work done by all internal forces is 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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@Miraj Shah This one helped me ;) Swapnil Das · 4 months, 2 weeks ago

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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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@Aryan Goyat it sounds too boring but its going to be interesting i found it to be useful. try this

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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@Aryan Goyat

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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@Harsh Shrivastava It is the point about which the body appers to be in purely rotational motion (no translatory motion). Deeparaj Bhat · 5 months, 1 week ago

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@Deeparaj Bhat How to apply this concept in problem solving? Harsh Shrivastava · 5 months, 1 week ago

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@Harsh Shrivastava 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). Deeparaj Bhat · 5 months, 1 week ago

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

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@Swapnil Das 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. Deeparaj Bhat · 5 months, 1 week ago

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@Deeparaj Bhat Could you add one example? Thanks. Swapnil Das · 5 months, 1 week ago

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