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

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TopNewestWe 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

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 · 2 years, 5 months ago

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– Yogesh Ghadge · 2 years, 5 months ago

I solve as you told and I got the answer as 0.4 static friction can you tell this ans is right or wrongLog in to reply

– Satvik Pandey · 2 years, 5 months ago

Could you please show your steps. I am getting its value slightly differently.Log in to reply

– Yogesh Ghadge · 2 years, 5 months ago

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= sfLog in to reply

fig

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

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

Find

Nfrom eq(1) and put its value in eq(2). Then find \( \mu \) – Satvik Pandey · 2 years, 5 months agoLog in to reply

– Yogesh Ghadge · 2 years, 5 months ago

thank you for clearing the doubtLog in to reply

– Satvik Pandey · 2 years, 5 months ago

You are welcome. :DLog in to reply

– Yogesh Ghadge · 2 years, 5 months ago

Thanks you r the bestLog in to reply

– Satvik Pandey · 2 years, 5 months ago

Thanks!Log in to reply