Different Lift Cases in NLM=> (Tutorial)

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\(\Huge\color{Blue}{Different~Lift~Cases} \)
in Newtons Laws of motion
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 |dw:1440296184019:dw| 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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Other answers:

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
Case 2> If a lift is moving downward with acceleration a. |dw:1440296743979:dw| Therefore downward force on man mg-N Therefore, mg-N=ma \[N=m(g-a) \] Apparent weight
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 .
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 |dw:1440297484077:dw| 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. |dw:1440298160222:dw| Now when head of mine (man) collided with ceiling of lift N+mg=ma \[N=ma-mg\]
Hope this helps. Happy learning.... :) Enyone can share their knowledge here....
  • rvc
good work
Thank you @rvc
Good job! Easily understandable :)
Thank you @shreehari499

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