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

Please Help Me!!

#Mechanics

Note by Aishwary Omkar
6 years, 8 months ago

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Comments

Suppose you are asked to find relation between acceleration of m1m_{1} and m2m_{2}. Figure Figure

I like to do this with the help of calculus.Let ll be tha length of the string.

Then

x+y+2z=lx+y+2z=l

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

d2xdt2+2d2zdt2=0\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +2\frac { { d }^{ 2 }z }{ { d{ t }^{ 2 } } } =0

Let the acceleration of m1m_{1} and m2m_{2} be a1a_{1} and a2a_{2} respectively

then a1=2a2{ a }_{ 1 }=-2{ a }_{ 2 }

Here (-)sign with a2a_{2} just denotes direction.

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

a1=2a2{ a }_{ 1 }=2{ a }_{ 2 }

Search for 'string constraint' on utube. You will get some similar types of example there.

satvik pandey - 6 years, 7 months ago

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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

Tx=0orTv=0orTa=0\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 - 6 years, 7 months ago

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Nice method.

Image Image T1a1=0T_{1}a_{1}=0 and T2a2T_{2}a_{2}

But 2T1=T22T_{1}=T_{2}

So T1a1=2T1a2T_{1}a_{1}=2T_{1}a_{2}

So a1=2a2a_{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 - 6 years, 7 months ago

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@Satvik Pandey please explain this virtual work method once more

Aishwary Omkar - 6 years, 7 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 Tx=0 \vec { T } \cdot \vec { x }=0

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

Ta=0 \vec { T } \cdot \vec { a }=0.

By using this equation you can easily find relation between accelerations of masses.

satvik pandey - 6 years, 7 months ago

that is some thing new thanks

Aishwary Omkar - 6 years, 7 months ago

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 - 6 years, 7 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 --

AB\vec { A } \cdot \vec { B }

satvik pandey - 6 years, 7 months ago

thanks

Aishwary Omkar - 6 years, 7 months ago
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