# Accelerating Cylinders !!Someone pls 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 ? Note by Ch Nikhil
5 years, 6 months 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 $\frac{1}{\sqrt3}$

- 5 years, 6 months ago

- 5 years, 6 months ago

Hi Nikhil! 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 $a_{0}$.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

Hi Sir!! I draw these diagrams using paint.

First upload the diagram on imgur. Then type:

! [ anything] ( url of the image). Without spaces. :)

- 5 years, 6 months ago

Thanks.... ;)

- 5 years, 6 months ago

You are welcome, Sir! :)

- 5 years, 6 months ago

Thanks a lot @satvik pandey !!

- 5 years, 6 months ago

You are welcome!

- 5 years, 6 months ago