Will a block slide in a moving lift with equilibrium?

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ProfManas
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Homework Statement



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Homework Equations



The Attempt at a Solution


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Effective gravity = g-a
So
Frictional force = tension
Tension = m(g-a)
u 2m(g-a) =m(g-a)
 

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kuruman said:
Hi ProfManas and welcome to PF.

So what is your answer? Your equations appear to be correct.
Answer through equations is g=a which feels wrong.
I think maybe lift can descend with whatever deceleration less than g and the equilibrium will still hold? I m stumped.
 
ProfManas said:
Answer through equations is g=a which feels wrong.
It feels wrong but it isn't. There is an important lesson to be learned here.

If the lift were at rest or moving at constant velocity, the block would not slide because μs(2mg) > mg. Would it slide on the Moon where the acceleration of gravity is about 1/6 of 9.8 m/s2? No. That's because both the tension and the maximum force of static friction scale as whatever value the acceleration of gravity has. If the acceleration of gravity is reduced by a factor of 1/6, so is the maximum force of static friction and the inequality is preserved. Conversely, if the block does slide when the acceleration of the lift is zero, then it will also slide when the lift is accelerating up or down, it doesn't matter.
 
kuruman said:
It feels wrong but it isn't. There is an important lesson to be learned here.

If the lift were at rest or moving at constant velocity, the block would not slide because μs(2mg) > mg. Would it slide on the Moon where the acceleration of gravity is about 1/6 of 9.8 m/s2? No. That's because both the tension and the maximum force of static friction scale as whatever value the acceleration of gravity has. If the acceleration of gravity is reduced by a factor of 1/6, so is the maximum force of static friction and the inequality is preserved. Conversely, if the block does slide when the acceleration of the lift is zero, then it will also slide when the lift is accelerating up or down, it doesn't matter.
That does make sense, thanks.