Will a block slide in a moving lift with equilibrium?

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The discussion centers on whether a block will slide in a moving lift under equilibrium conditions. It is established that the effective gravity in the lift is influenced by its acceleration, leading to the conclusion that if the lift accelerates downward with a deceleration less than gravity, the block may remain in place. The equations suggest that the block will not slide if the lift is at rest or moving at constant velocity, as the static friction force exceeds the gravitational force acting on the block. The conversation also highlights that this principle holds true regardless of the gravitational context, such as on the Moon, where both tension and static friction scale down proportionately. Ultimately, the key takeaway is that the block's behavior is consistent across different acceleration scenarios, reinforcing the understanding of static friction and equilibrium.
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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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Hi ProfManas and welcome to PF.

So what is your answer? Your equations appear to be correct.
 
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.
 

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