Apparent Weight of a Person in a Lift
The word Apparent means 'it seems to be" or "feel like that". Like the same way, the apparent weight in the lift means that, "the weight that a person feel when he move in the lift." There are totally 5 situation under which a person feels his weight differently(in each case)
Let us consider a person of mass 'm', has a weight 'mg'(W = mg) standing in the weighing machine inside the lift(force acts downward); moving in the lift that has the acceleration 'a'. The weighing machine offers a reaction force 'R' against the weight, which is force (mg). This R is also known as the apparent weight.
case (i) a = 0
When acceleration of the lift is 0 (When lift moves up or down)(case I)
First, we have to calculate the net force.
To calculate the net force sum up the force acting on the same direction and subtract the force acting on opposite direction[subtract the smaller from greater].
F = R - mg (since, force acting on opposite direction) ------------ 1
According to Newtons second law of motion,
F = ma
substitute in 1
ma = R - mg ----------- 2
we know that a = 0 in this case, so substitute a = 0 in eq 2
m(0) = R - mg
0 = R - mg
R = mg
When acceleration is 0, we feel our original weight
case (ii)
When lift moves up, acceleration is not equal to zero.(case II)
Calculate the Net force,
F = R - mg (∵ lift moves upward and reaction force 'R' is also upward force, a is not equal to 0) ------------ 3
According to Newtons second law of motion,
F = ma
substitute in eq 3
ma = R - mg ----------- 4
R = mg + ma
Here apparent weight = real weight + ma.
The apparent weight > real weight (∵ 'ma' is added to our real weight)
case (iii)
When lift moves down, acceleration is not equal to zero. (case III)
Calculate the Net force,
F = mg - R (∵ lift moves downward and weight 'mg' is also downward force, a is not equal to 0) --------5
According to Newtons second law of motion,
F = ma
substitute in eq 5
ma = mg - R ----------- 6
R = mg - ma
Here apparent weight = real weight - ma.
The apparent weight < real weight (∵ 'ma' is deducted to our real weight)
case (iv)
When lift moves up, acceleration is equal to gravity 'g'. (case IV)
Calculate the Net force,
F = R - mg (∵ lift moves upward and reaction force 'R' is also upward force, a is equal to g) ------------ 7
According to Newtons second law of motion,
F = ma
Here in this case, a = g
substitute F =ma and a = g in eq 7
mg = R - mg ----------- 8
R = mg + mg
R = 2mg
Here apparent weight = real weight + real weight
The apparent weight > real weight (∵ apparent weight is 2 times of real weight)
case (v)
When lift moves down, acceleration is equal to gravity 'g'. (case v)
Calculate the Net force,
F = mg - R (∵ lift moves down and 'mg' is also downward force, a is equal to g) ------------ 9
According to Newtons second law of motion,
F = ma
Here in this case, a = g
substitute F =ma and a = g in eq 9
mg = mg - R ----------- 10
R = mg - mg
R = 0
Here apparent weight = real weight - real weight
= 0
To calculate the net force sum up the force acting on the same direction and subtract the force acting on opposite direction[subtract the smaller from greater].
F = R - mg (since, force acting on opposite direction) ------------ 1
According to Newtons second law of motion,
F = ma
substitute in 1
ma = R - mg ----------- 2
we know that a = 0 in this case, so substitute a = 0 in eq 2
m(0) = R - mg
0 = R - mg
R = mg
When acceleration is 0, we feel our original weight
case (ii)
When lift moves up, acceleration is not equal to zero.(case II)
Calculate the Net force,
F = R - mg (∵ lift moves upward and reaction force 'R' is also upward force, a is not equal to 0) ------------ 3
According to Newtons second law of motion,
F = ma
substitute in eq 3
ma = R - mg ----------- 4
R = mg + ma
Here apparent weight = real weight + ma.
The apparent weight > real weight (∵ 'ma' is added to our real weight)
case (iii)
When lift moves down, acceleration is not equal to zero. (case III)
Calculate the Net force,
F = mg - R (∵ lift moves downward and weight 'mg' is also downward force, a is not equal to 0) --------5
According to Newtons second law of motion,
F = ma
substitute in eq 5
ma = mg - R ----------- 6
R = mg - ma
Here apparent weight = real weight - ma.
The apparent weight < real weight (∵ 'ma' is deducted to our real weight)
When lift moves up, acceleration is equal to gravity 'g'. (case IV)
Calculate the Net force,
F = R - mg (∵ lift moves upward and reaction force 'R' is also upward force, a is equal to g) ------------ 7
According to Newtons second law of motion,
F = ma
Here in this case, a = g
substitute F =ma and a = g in eq 7
mg = R - mg ----------- 8
R = mg + mg
R = 2mg
Here apparent weight = real weight + real weight
The apparent weight > real weight (∵ apparent weight is 2 times of real weight)
case (v)
When lift moves down, acceleration is equal to gravity 'g'. (case v)
Calculate the Net force,
F = mg - R (∵ lift moves down and 'mg' is also downward force, a is equal to g) ------------ 9
According to Newtons second law of motion,
F = ma
Here in this case, a = g
substitute F =ma and a = g in eq 9
mg = mg - R ----------- 10
R = mg - mg
R = 0
Here apparent weight = real weight - real weight
= 0
The apparent weight = 0
∴, we feel weightless.
