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Please someone tell me how to get constraint equations especially in movable pulleys and movable wedges

Please Help Me!!

Note by Aishwary Omkar
2 years, 3 months ago

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Suppose you are asked to find relation between acceleration of \(m_{1}\) and \(m_{2}\).

Figure

Figure

I like to do this with the help of calculus.Let \(l\) be tha length of the string.

Then

\(x+y+2z=l\)

On differentiating this wrt time (two times) we will get

\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +2\frac { { d }^{ 2 }z }{ { d{ t }^{ 2 } } } =0\)

Let the acceleration of \(m_{1}\) and \(m_{2}\) be \(a_{1}\) and \(a_{2}\) respectively

then \({ a }_{ 1 }=-2{ a }_{ 2 }\)

Here (-)sign with \(a_{2}\) just denotes direction.

If we want to find relation between magnitudes of accelerations then the relation will be

\({ a }_{ 1 }=2{ a }_{ 2 }\)

Search for 'string constraint' on utube. You will get some similar types of example there. Satvik Pandey · 2 years, 3 months ago

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@Satvik Pandey Well I always Prefer Virtual work Method :

That is Net work done By Internal Forces ( Like here it is Tension in Strings) on a system will be zero

\(\sum { \xrightarrow { T } \xrightarrow { x } } \quad =\quad 0\\ \\ or\\ \\ \sum { \xrightarrow { T } \xrightarrow { v } } \quad =\quad 0\\ \\ or\\ \\ \sum { \xrightarrow { T } \xrightarrow { a } } \quad =\quad 0\\ \). Deepanshu Gupta · 2 years, 3 months ago

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@Deepanshu Gupta that is some thing new thanks Aishwary Omkar · 2 years, 3 months ago

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@Deepanshu Gupta Nice method.

Image

Image

\(T_{1}a_{1}=0\) and \(T_{2}a_{2}\)

But \(2T_{1}=T_{2}\)

So \(T_{1}a_{1}=2T_{1}a_{2}\)

So \(a_{1}=2a_{2}\)

By using calculus or virtual work method ultimately we will reach the same answer.

Thank you Deepanshu for sharing this method. :D

This method is very good for finding relation between accelerations of different masses. Satvik Pandey · 2 years, 3 months ago

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@Satvik Pandey please explain this virtual work method once more Aishwary Omkar · 2 years, 3 months ago

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@Aishwary Omkar In a system work done by internal force is zero. In this case internal force is tension.

So \( \vec { T } \cdot \vec { x }=0\)

If you differentiate this equation w.r.t time (two times) you will get

\( \vec { T } \cdot \vec { a }=0\).

By using this equation you can easily find relation between accelerations of masses. Satvik Pandey · 2 years, 3 months ago

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@Deepanshu Gupta Sorry But This Latex is Poor.... Here Arrow means vector and Yes it is Dot product. I wonder How This arrow is shifted down side while In Latex Coding it seems to be correct.

will Any Body Tell what is Exact Latex Code For representing an vector dot product ?? Deepanshu Gupta · 2 years, 3 months ago

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@Deepanshu Gupta If you type \vec { A } \cdot \vec { B } and enclosed it within brackets then you will see it's output as --

\(\vec { A } \cdot \vec { B } \) Satvik Pandey · 2 years, 3 months ago

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@Satvik Pandey thanks Aishwary Omkar · 2 years, 3 months ago

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