Apparent Weight of a Body in a Lift
Category : JEE Main & Advanced
When a body of mass m is placed on a weighing machine which is placed in a lift, then actual weight of the body is mg.
This acts on a weighing machine which offers a reaction R given by the reading of weighing machine. This reaction exerted by the surface of contact on the body is the apparent weight of the body.
Apparent weight in a lift
| Condition | Figure | Velocity | Acceleration | Reaction | Conclusion |
| Lift is at rest |  |
\[\upsilon =0\] | a = 0 |
R - mg = 0 \[\therefore \] R = mg |
Apparent weight = Actual weight |
| Lift moving upward or downward with constant velocity |  |
\[\upsilon \] = constant |
a = 0 |
R - mg = 0 \[\therefore \] R = mg |
Apparent weight = Actual weight |
| Lift accelerating upward at the rate of 'a' |  |
\[\upsilon \] = variable |
a < g |
R - mg = ma \[\therefore \] R = m(g + a) |
Apparent weight > Actual weight |
| Lift accelerating upward at the rate of 'g' |  |
\[\upsilon \] = variable |
a = g |
R - mg = mg R = 2mg |
Apparent weight = 2 Actual weight |
| Lift accelerating downward at the rate of 'a' |  |
\[\upsilon \] = variable |
a < g |
mg - R = ma \[\therefore \] R = m(g - a) |
Apparent weight < Actual weight |
| Lift accelerating downward at the rate of 'g' |  |
\[\upsilon \] = variable |
a = g |
mg - R = mg R = 0 |
Apparent weight = Zero (weightlessness) |
| Lift accelerating downward at the rate of a(>g) |  |
\[\upsilon \] = variable |
a > g |
mg - R = ma R = mg - ma R = \[-\,\upsilon e\] |
Apparent weight negative means the body will rise from the floor of the lift and stick to the ceiling of the lift. |
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