Apparent weight of a man in a lift

Apparent weight of a man in a lift

In this topic we will discuss about , a man of mass m standing in a lift . What will be its apparent weight of a man in a lift when lift has some motion with uniform acceleration or velocity?

But before to know about this topic students must know Newton’s third law of motion. To get the notes on Newton’s third law of motion click here-

Apparent weight of a man in a lift  (elevator)  –

Suppose a man of mass ‘m’ is standing on a weighing machine placed inside a lift . initially when the lift is in rest weight of the man acts vertically downward direction which acts on the   weighing machine and machine offers a resistance ‘R’ against its weight shows the actual weight which is mg .

Case i – When the lift is in rest or moving with uniform velocity (fig-c)- In that case there is no acceleration in the lift and man , and hence reaction force is equal to the weight of the man.

In that case , R  = mg

i.e  Apparent weight = Actual weight

Case ii – When lift moves upward with acceleration ‘a’ fig (a).

Since there is upward motion then  ;

So  R-mg =ma

Or , R = mg+ma = m(g+a)

i.e. R ( apparent weight) > mg(actual weight) ;

i.e  Weighing machine shows more weight the actual .

Case iii – When lift moves downward with acceleration ‘a’ fig (b).

Since there is downward motion then  ;

So  mg-R =ma

Or , R = mg-ma = m(g-a)

i.e. R ( Apparent weight) < mg(Actual weight) ;

i.e  Weighing machine shows less weight the actual .

case iv– When the lift falls freely –  In that case lift accelerates downward with acceleration ‘a=g’

So we can write , R= mg-ma = mg-mg ( since a=g) = o

Show apparent weight of the man will be zero . So person feel’s weightlessness.

 

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