ProfManas
Will a block slide in a moving lift with equilibrium?
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Hi ProfManas and welcome to PF.
So what is your answer? Your equations appear to be correct.
So what is your answer? Your equations appear to be correct.
ProfManas
Answer through equations is g=a which feels wrong.kuruman said:Hi ProfManas and welcome to PF.
So what is your answer? Your equations appear to be correct.
I think maybe lift can descend with whatever deceleration less than g and the equilibrium will still hold? I m stumped.
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It feels wrong but it isn't. There is an important lesson to be learned here.ProfManas said:Answer through equations is g=a which feels wrong.
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.
ProfManas
That does make sense, thanks.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.
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