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static friction

This is a problem which I am confused and able to solve So please tell me how to solve

A block of mass market is kept on an incline plane of a lift moving down with acceleration of 2m/s^-2 . What should be the coefficient of friction for the block to move down with constant velocity relative to lift

Note by Yogesh Ghadge
3 years, 3 months ago

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We have to use concept of pseudo force here.

First we shall consider a reference frame which is moving along with lift. As it is non-inertial frame of reference so we have to include a pseudo force here.

fig

fig

Resolve the vectors in direction parallel to the plane and perpendicular to the plane. Then use newtons's second law to find equations.

If the block has to move with constant velocity then \(F_{net}=0\).

Can you solve it from here?

Satvik Pandey - 3 years, 2 months ago

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I solve as you told and I got the answer as 0.4 static friction can you tell this ans is right or wrong

Yogesh Ghadge - 3 years, 2 months ago

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Could you please show your steps. I am getting its value slightly differently.

Satvik Pandey - 3 years, 2 months ago

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@Satvik Pandey Incline is at 30 ° so a= g sin 30°=5 So normal force = mg = 5m So the pseudo is acting as friction force here Then friction force (f)=static friction × normal force F=sf×5m 2m=sf×5m 2m/5m=sf 0.4= sf

Yogesh Ghadge - 3 years, 2 months ago

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@Yogesh Ghadge

fig

fig

From the diagram

\(N+2mcos\theta=mgcos\theta\)............(1)

and \(mgsin\theta=\mu N+2msin\theta\) ...................(2)

Find N from eq(1) and put its value in eq(2). Then find \( \mu \)

Satvik Pandey - 3 years, 2 months ago

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@Satvik Pandey thank you for clearing the doubt

Yogesh Ghadge - 3 years, 2 months ago

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@Yogesh Ghadge You are welcome. :D

Satvik Pandey - 3 years, 2 months ago

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Thanks you r the best

Yogesh Ghadge - 3 years, 2 months ago

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Thanks!

Satvik Pandey - 3 years, 2 months ago

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