Which Elevator Movement Requires the Least Force to Move a Crate?

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Moving a crate in an elevator requires different forces depending on the elevator's movement. The least force is needed when the elevator is accelerating downwards, while the most force is required when it is accelerating upwards. When the elevator travels at a constant speed, the force needed falls between these two extremes. The difficulty in moving the crate is primarily due to static friction and the weight of the object. Understanding how static friction interacts with the crate's weight is crucial for analyzing the force required.
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Move objects inside elevator...

Homework Statement


You are moving a large crate from the back of the freight of the elevator to the front as the elevator is moving to the next floor.In which situation is the force you must apply to move the crate the smallest and in which is it the largest
1> When the elevator is accelerating upwards
2> When the elevator is accelerating downwards.
3> When it is traveling a constant speed.

Homework Equations





The Attempt at a Solution


The less force would be needed when the quantity m * ax would be less.
So when the elevator is accelerating downwards the force required would be least
and when the object is accelerating upwards the force required would be maximum .
Is my analogy true ?
 
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When you're moving the crate in the way the problem describes, what makes it difficult?
 


Its really hard for me to analyze this can some one just elaborate the answer for me ??
 


Imagine yourself trying to move a large crate. What makes it difficult to move the crate?

You can do this. You must have moved large objects before.
 


Its because of static friction and the weight of the object ...
 


rulzwrld said:
Its because of static friction and the weight of the object ...

OK, and what does the static friction depend on?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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