Will a block slide in a moving lift with equilibrium?

In summary, the conversation discusses the concept of frictional force and tension in relation to the acceleration of gravity. The equations provided are correct, but may feel counterintuitive. The important lesson is that the maximum force of static friction and the tension both scale with the acceleration of gravity, meaning that the inequality is preserved regardless of the lift's acceleration.
  • #1
ProfManas
ATTACH]

Homework Statement



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Homework Equations



The Attempt at a Solution


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

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  • #2
Hi ProfManas and welcome to PF.

So what is your answer? Your equations appear to be correct.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 

FAQ: Will a block slide in a moving lift with equilibrium?

What is meant by "moving lift equilibrium"?

"Moving lift equilibrium" refers to the balance of forces that allows a lift (elevator) to move up and down smoothly and safely.

What factors affect the equilibrium of a moving lift?

The equilibrium of a moving lift is affected by the weight of the lift and its contents, the tension in the cables, and the force of gravity pulling the lift down.

How is equilibrium maintained in a moving lift?

Equilibrium is maintained in a moving lift through the use of a counterweight system. This system involves a counterweight on the opposite side of the lift, which helps balance the weight of the lift and keep it moving smoothly.

Can the equilibrium of a moving lift be affected by external factors?

Yes, external factors such as temperature, humidity, and wind can affect the equilibrium of a moving lift. These factors can cause the cables to expand or contract, which can impact the tension and balance of the lift.

What happens if the equilibrium of a moving lift is disrupted?

If the equilibrium of a moving lift is disrupted, it can lead to jerky movements, uneven weight distribution, and potential safety hazards. It is important for regular maintenance and inspections to be conducted to ensure the equilibrium of a moving lift is maintained.

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