Apparent Weight of an Object in a Lift

  1. When a man standing inside an elevator, there are two forces acting on him.
    1. His weight, (W) which acting downward.
    2. Normal reaction (R), acting in the opposite direction of weight.
  2. The reading of the balance is equal to the normal reaction (R).
  3. Figure below shows the formula to calculate the reading of the balance at different situation.


Example 1:
Subra is standing on a balance inside an elevator. If Subra’s mass is 63kg, find the reading of the balance when the lift,
  1. stationary
  2. moving upward with a constant velocity, 15 ms-1.,
  3. moving upward with a constant acceleration, 1 ms-2.
  4. moving downward with a constant acceleration, 2 ms-2.

Answer:
a.
W = mg
W = (63)(10) = 630N

b.
W = mg
W = (63)(10) = 630N

c.
R=mg+ma R=(63)(10)+(63)(1) R=693N
d.
R=mg−ma R=(63)(10)−(63)(2) R=504N

Example 2:
A 54kg boy is standing in an elevator. Find the force on the boy's feet when the elevator
  1. stands still
  2. moves downward at a constant velocity of 3 m/s
  3. decelerates downward with at 4.0 m/s2, 
  4. decelerates upward withat 2.0 m/s2.

Answer:
a.
W = mg
W = (54)(10) = 540N

b.
W = mg
W = (54)(10) = 540N

c.
R=mg+ma R=(54)(10)+(54)(4) R=756N
d.
R=mg−ma R=(54)(10)−(54)(2) R=432N