I am stuck in this question. Please help!!

Three identical cylinders are arranged in a triangle**(one lying on top of the other two and all 3 are in contact)**, with the bottom two lying on the ground.The ground and the cylinders are frictionless. You apply a constant horizontal force (directed to the right) on the left cylinder. Let **a** be the acceleration you give to the system. For what range of **a** will all three cylinders remain in contact with each other ?

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TopNewestwere you doing david-morin at that time ? – Shubham Dhull · 5 months, 4 weeks ago

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At the time when the upper cylinder just leaves the contact with the bottom cylinder.

N Sin60 = mg;

N cos60 = ma;

a = g cot60 = g \(\frac{1}{\sqrt3}\) – Rohit Gupta · 2 years ago

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@satvik pandey @Kushal Patankar please help!! – Ch Nikhil · 2 years ago

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img

It's is obvious that the cylinder at the top will first loose contact with cylinder at right. The moment it looses contact with that the normal force acting between them would be zero. If we consider a frame of reference which is moving with the CoM of the sytem then pseudo force on top cylinder is \(ma\).

So

\(N_{1}cos(30)=mg\)....(1)

and \(N_{1}sin(30)=ma\)......(2)

Solving this we get a value of ,say \(a_{0}\). So the cylinders would loose contact for all accelerations greater than \(a_{0}\). – Satvik Pandey · 2 years ago

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@satvik pandey !! – Ch Nikhil · 2 years ago

Thanks a lotLog in to reply

– Satvik Pandey · 2 years ago

You are welcome!Log in to reply

– Rohit Gupta · 2 years ago

Hi Satvik, Can you tell me how you draw the diagram and upload it..??Log in to reply

First upload the diagram on imgur. Then type:

! [ anything] ( url of the image). Without spaces. :) – Satvik Pandey · 2 years ago

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– Rohit Gupta · 2 years ago

Thanks.... ;)Log in to reply

– Satvik Pandey · 2 years ago

You are welcome, Sir! :)Log in to reply