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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 $k < \frac{3}{2}$.

How do you deduce the k in this question?

MCCCS
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Laggger246
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    Please type out your question. It makes the question easier to read and allows for search engines to read it, thus helping the community much more. – 5xum Jun 29 '17 at 14:08
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    Welcome to the site. You should share your thoughts on the problem to get a favourable response. And rotate the image at least if you do not use $\LaTeX$. – StubbornAtom Jun 29 '17 at 14:09

1 Answers1

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Sum of the forces equals to mass times acceleration. In the first case, the normal force on the object is $N_1$ acting upwards. You get $$ma=N-W=kW-W=(k-1)W$$ You therefore get $$a=\frac{(k-1)W}{m}$$ When the platform is moving downward, the weight is greater than the normal force $$m\cdot 2a=W-N_2\\N_2=W-m\cdot 2a=W-m\frac{2(k-1)W}{m}=(3-2k)W$$ Note that $N_2$ has to be positive, so this will get you the condition for $k$

Andrei
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