laggger246
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A box of weight $W$ rests on platform of a lift. When the platform is moving upwards with acceleration $a$, the normal contact force of the platform on the box has magnitute $kW$. When the platform is moving downwards with acceleration $2a$, the box remains in contact with it. Find the normal contact force in terms of $k$ and $W$, and deduce that $k < \frac{3}{2}$.

It is about the mechanic in lift
How do you deduce the k in this question?
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Nulliverse
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What is it that you're finding hard? Remember you're given weight and not mass so make sure you know the difference between the 2 and know how to interchange as you're using (F=ma) for this question. Alrighty, now that we got that sorted, first isolate it so that it's going up with acceleration with a, then this means that the force going upwards is equal to a force that is able to lift the weight of the box as well as accelerate it with speed a, thus the force going up is kW=W+ma. W=mg, means m=(W/g), now get an equation. For the 2nd part, "the box remains in contact with it" what does this mean? Can you form an inequality using this information?
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Physics Enemy
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(Original post by laggger246)
A box of weight $W$ rests on platform of a lift. When the platform is moving upwards with acceleration $a$, the normal contact force of the platform on the box has magnitute $kW$. When the platform is moving downwards with acceleration $2a$, the box remains in contact with it. Find the normal contact force in terms of $k$ and $W$, and deduce that $k < \frac{3}{2}$.

It is about the mechanic in lift
How do you deduce the k in this question?
Part 1: The normal contact force is a result of the box's weight and also the accelerating force by the lift. So: kW = W + ma

Part 2: Now the accelerating force by the lift subtracts from the contact force. So: F = W - 2ma = W - 2(kW - W) = 3W - 2kW = W(3 - 2K)

We require the box to remain in contact with the lift i.e) F > 0, so 3 - 2K > 0, so K < 3/2.
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