@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 ago

Log in to reply

@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 ago

Log in to reply

@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 ago

## Comments

Sort by:

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 ago

Log in to reply

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 ago

Log in to reply

– Aishwary Omkar · 2 years 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 ago

Log in to reply

– Aishwary Omkar · 2 years 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 ago

Log in to reply

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

Log in to reply

\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 ago

Log in to reply

– Aishwary Omkar · 2 years ago

thanksLog in to reply