Different Lift Cases in NLM= Tutorial
Creator of Tutorial: arindameducationusc

Question

Different Lift Cases in NLM=> Tutorial
Creator of Tutorial: arindameducationusc

Tutorials · Answer

Case 1=> Consider a lift which is moving upward with acceleration 'a' and a person is standing on the floor of lift.

The person standing on the floor of lift is under the influence of two forces
1)Normal reaction due to flor of lift 
2)Weight(mg) due to Earth
<img src="http://image.prntscr.com/image/9aea108a3cc646d6857baa0e77824fef.png" alt="9aea108a3cc646d6857baa0e77824fef.png" class="bbcode_img" />

Net upward force on man= N-mg
Now, N-mg=ma         
Because $$F _{net}=ma$$
so,
N=mg+ma

$$N=m(g+a)$$

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If a platform balance is placed on the floor of lift and man is allowed to stand over it then the force exerted by man on balance will be mg+ma   (in case 1),
which is greater than the true weight of the man,i.e, mg

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Case 2> If a lift is moving downward with acceleration a

Therefore downward force on man mg-N Therefore, mg-N=ma $$N=m(g-a)$$ Apparent weight

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Case 3=> If cable of lift breaks

then a=g
putting in above equation N=m(g-a)
we get N=0
In case of free falling of lift, man is in the condition of weightlessness .

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Case 4=> If a>g in downward direction In this case man is also in free falling,i.e, lift and man both are coming downward wrt observer standing on the ground

Now if a>g then lift is moving faster with acceleration 'a' but you are in free fall with acceleration 'g'. This makes the lift go faster downward, eventually I collide with the top of the lift.

Now when head of mine (man) collided with ceiling of lift N+mg=ma N = ma - mg

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If you are having a hard time viewing the images provided above, take a look at these screenshots: Case 1 http://prnt.sc/df4ui4 Case 2 http://prnt.sc/df4uwt Case 4 http://prnt.sc/df4vaw http://prnt.sc/df4vkb