Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What is Apparent Weight?
When a person of mass m stands on a floor, a downward force mg acts on the floor due to his weight. At the same time, the floor exerts an upward reaction force or normal force R on the man. As R acts on the person, the person feels his weight. From the third law of motion, R = mg. So, in this case, the actual weight mg and the normal reaction, R, is the same. If the normal force R is absent, the person feels weightless. For example, while jumping from a height, one feels weightless before touching the earth’s surface, because the normal force R is absent.
The normal force R varies inside a lift due to its vertical motion. The floor of the lift, in this case, provides the reaction. A person feels lighter or heavier than his actual weight depending on the reaction force R. This is called apparent weight.
Let the mass of a man in a lift be m [Fig.].
Downward attractive force acting on him = real weight = mg Upward normal force of the floor = apparent weight = R
∴ The net upward force F = R – mg.
We can determine the apparent weight R using F = ma for different types of motion of the lift.
i) The lift is moving up with an acceleration a (or, moving down with a retardation a) [Fig.]:
Here the upward acceleration is a.
∴ R – mg = ma
or, R = mg+ ma = m(g+ a) …….. (1)
∴ As R > mg, the man feels heavier.
ii) The lift is moving down with an acceleration a (or, moving up with a retardation a) [Fig.]:
Here the downward acceleration = a or upward acceler-ation = -a (here a < g).
∴ R – mg = -ma
or, R = m(g – a) …….. (2)
He feels lighter as m(g – a) < mg.
iii) The cable of the lift snaps and the lift with the man, falls freely due to gravitational pull:
Here, the acceleration, a = g. Hence from (2), R = m(g – g) = 0 ……… (3)
As no normal force acts on the man, he feels weightless. This apparent weightlessness is true for all freely falling bodies.
iv) The lift is at rest or in uniform motion: The value of a is zero and so the reaction force, R = mg. Hence, the apparent weight is the same as the real or true weight.
v) The lift falls with a downward acceleration greater than g [Fig.]:
Here, a > g. From equation (2),
R = m(g – a) or, R = -m(a – g)
The negative sign indicates that the apparent weight is negative, i.e., it is directed upwards. The man thus loses contact with the floor and hits the ceiling of the lift. Any item on the floor of the lift will hit the roof when exposed to similar conditions. This is termed as super-weightlessness.
Numerical Examples
Example 1.
A body of mass 1 kg is suspended from a spring bal-ance calibrated for acceleration due to gravity of 10 m ᐧ s-2. What is the reading on the spring balance when the system
(i) is ascending with an acceleration of 5 m ᐧ s-2 and
(ii) is descending with the same acceleration? [g = 10 m ᐧ s-2]
Solution:
Reaction force on the balance when it ascends,
R = m(g + a) = 1(10 + 5) = 15 N.
∴ In this case, the reading of the spring balance
= \(\frac{15}{10}\) = 1.5 kg.
ii) Reaction force when the system descends,
R = m(g – a) = 1(10 – 5) = 5N.
∴ In this case, the reading of the spring balance
= \(\frac{5}{10}\) = 0.5 kg.
Example 2.
A man of 50 kg is standing on a weighing machine in a lift. As the lift moves with a constant acceleration, the weighing machine registers the man’s weight as 45 kg. State whether the lift is ascending or descending. Give reasons for your answer. What is the acceleration of the lift? [g = 9.8 m ᐧ s-2]
Solution:
The weighing machine shows a reading lower than the real weight of the man. So, the lift is descending with an acceleration, because in such cases, apparent weight R = m(g – a) < mg.
The downward acceleration,
a = g – \(\frac{R}{m}\) = 9.8 – \(\frac{45 \times 9.8}{50}\) = 0.98 m ᐧ s-2.
Example 3.
A man of mass 70 kg is sitting in a motor car. The car is moving with an acceleration of 5 m ᐧ s-2. What is the gravitational force on the man? [g = 9.8 in ᐧ s-2]
Solution:
Let the horizontal acceleration of the car be \(\vec{a}\).
The apparent acceleration due to gravity with respect to the car,
\(\overrightarrow{g^{\prime}}\) = \(\vec{g}\) – \(\vec{a}\) = [\(\vec{g}\) + (-\(\vec{a}\))]
which is the resultant of the vectors \(\vec{g}\) and –\(\vec{a}\).
From Fig.,
g’ = \(\sqrt{g^2+a^2}\) = \(\sqrt{(9.8)^2+5^2}\)
Force of gravitation on the man = 70 × 11 = 770 N, which acts in the direction of g’.
[The apparent weight of the man is, W’ = \(\frac{770}{9.8}\) = 78.6 kg ]
The force of gravity, acting on the man, remains constant but, because of acceleration, the man feels a greater pull of gravity.
Example 4.
A man of mass 60 kg is standing in a lift at rest. What will be the reaction force on the man when the lift is
(i) stationary,
(ii) moving up with an acceleration of 4.9 m ᐧ s-2,
(iii) moving up at a constant speed, and
(iv) moving up with a retardation of 4.9 m ᐧ s-2? [g = 9.8 m ᐧ s-2]
Solution:
i) Reaction of the lift floor when the lift is stationary, R = mg = 60 × 9.8 N = 588 N.
ii) The lift is moving up with an acceleration, a = 4.9 m ᐧ s-2.
Reaction, R = m(g + a) = 60(9.8 + 4.9) N = 882 N.
iii) When the lift is moving up at a constant speed, a = 0
∴ R = mg = 60 × 9.8 N = 588 N
iv) When the lift moves up with a retardation of 4.9 m ᐧ s-2,
R = m[g + (-a)]
= m[g – a] = 60(9.8 – 4.9) N = 294 N.
Example 5.
A man of mass 98 kg, is standing on a weighing machine in a lift. What will be the readings of the weighing machine in the following cases:
(i) The lift ascends at 100 cm per second,
(ii) The lift descends with an acceleration of 30 cm ᐧ s-2 ? [g = 980 cm ᐧ s-2]
Solution:
i) When the lift ascends with uniform velocity (acceleration a = 0) then, R = mg = 98 × 9.8 N.
Reading of the weighing machine
= \(\frac{98 \times 9.8}{9.8}\)kg = 98 kg
ii) When the lift descends,
R = m(g- a)
= 98(9.8 – 0.3) [30 cm ᐧ s-2 = 0.3 m ᐧ s-2]
= 98 × 9.5 N
∴ Reading of the weighing machine
= \(\frac{98 \times 9.5}{9.8}\)kg = 95 kg.
Example 6.
A man weighing 60 kg is in a lift that descends with an acceleration of 4 cm ᐧ s-2. What force will the man exert on the floor of the lift? If the lift begins to ascend with the same acceleration, what reaction force will act on the man? For what acceleration of the lift, while descending, will the man experience weightlessness? [g = 980 cm ᐧ s-2] [HS ’01]
Solution:
If the lift descends with an acceleration a, and if the reaction force exerted by the lift on the man in the upward direction is R, then
mg – R = ma ….. (1)
Force exerted by the man on the lift’s floor,
R = mg – ma [4 cm ᐧ s-2 = 0.04 m ᐧ s-2]
= 60(9.8 – 0.04) = 585.6 N
If the lift ascends with the same acceleration, then reaction force on the man is,
R’ = mg + ma [4 cm ᐧ s-2 = 0.04 m ᐧ s-2]
= 60(9.8 + 0.04) = 590.4 N
When the acceleration of the lift, while descending, is a = g, then from equation (1), we get, R – mg – mg = 0. For this zero reaction force, the man feels weightless. So, for a lift falling freely with an acceleration g, the man inside it will feel no weight.
Example 7.
A lift of mass 200 kg is moving up with an acceleration of 4 m ᐧ s-2. What is the tension in the lift cable? If the lift moves down with the same acceleration, what will be the tension in that case? [g = 9.8 m ᐧ s-2] [HS ’04]
Solution:
Let the tension in the cable be T when the lift is moving with an acceleration a [Fig.]. The equation of motion for the lift is,
T – mg = ma
or, T = m(g + a) …….. (1)
Substituting the values for m, g and a in equation (1), for upward motion,
T = 200(9.8 + 4)
= 200 × 13.8 = 2760 N
For the lift moving downwards, a = -4 m ᐧ s-2,
T = 200(9.8 – 4) = 200 × 5.8 = 1160 N.
Example 8.
A lift of mass 2000 kg is supported by thick steel ropes. If maximum upward acceleration of the lift be 1.2 m/s2, and the breaking stress for the ropes be 2.8 × 108 N/m2, what should be the minimum diameter of the rope?
Solution:
Here, m = 2000 kg, a = 1.2 m/s2
Breaking stress = 2.8 × 108 N/m2
Let the diameter of the rope be D.
When the lift moves upwards, the tension in the rope is T = m(g + a) = 2000(9.8 + 1.2) = 22000 N.
Now, breaking stress \(\frac{\text { force }}{\text { area }}\) = \(\frac{T}{\pi D^2 / 4}\) = \(\frac{4 T}{\pi D^2}\)
or, 2.8 × 108 = \(\frac{4 \times 22000 \times 7}{22 \times D^2}\)
or, D2 = \(\frac{4 \times 22000 \times 7}{22 \times 2.8 \times 10^8}\) = 10-4 or, D = 1 cm