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

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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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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– Aishwary Omkar · 2 years, 3 months ago

that is some thing new thanksLog in to reply

Image

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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– Aishwary Omkar · 2 years, 3 months ago

please explain this virtual work method once moreLog in to reply

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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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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\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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– Aishwary Omkar · 2 years, 3 months ago

thanksLog in to reply