Question.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$,
(b) downwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$,
(c) upwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$,
(b) downwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$,
(c) upwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
solution:
(a) Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
Where, ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
$\therefore R=m g$
= 70 × 10 = 700 N
$\therefore$ Reading on the weighing scale $=\frac{700}{\mathrm{~g}}=\frac{700}{10}=70 \mathrm{~kg}$
(b) Mass of the man, $m=70 \mathrm{~kg}$
Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ downward
Using Newton's second law of motion, we can write the equation of motion as:
$R+m g=m a$
$R=m(q-a)$
$=70(10-5)=70 \times 5$
$=350 \mathrm{~N}$
$\therefore$ Reading on the weighing scale $=\frac{350}{\mathrm{~g}}=\frac{350}{10}=35 \mathrm{~kg}$
(c) Mass of the man, $m=70 \mathrm{~kg}$
Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ upward
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5) = 70 × 15
= 1050 N
$\therefore$ Reading on the weighing scale $=\frac{1050}{\mathrm{~g}}=\frac{1050}{10}=105 \mathrm{~kg}$
(d) When the lift moves freely under gravity, acceleration a = g
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= m(g – g) = 0
$\therefore$ Reading on the weighing scale $=\frac{0}{g}=0 \mathrm{~kg}$
The man will be in a state of weightlessness.
(a) Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
Where, ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
$\therefore R=m g$
= 70 × 10 = 700 N
$\therefore$ Reading on the weighing scale $=\frac{700}{\mathrm{~g}}=\frac{700}{10}=70 \mathrm{~kg}$
(b) Mass of the man, $m=70 \mathrm{~kg}$
Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ downward
Using Newton's second law of motion, we can write the equation of motion as:
$R+m g=m a$
$R=m(q-a)$
$=70(10-5)=70 \times 5$
$=350 \mathrm{~N}$
$\therefore$ Reading on the weighing scale $=\frac{350}{\mathrm{~g}}=\frac{350}{10}=35 \mathrm{~kg}$
(c) Mass of the man, $m=70 \mathrm{~kg}$
Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ upward
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5) = 70 × 15
= 1050 N
$\therefore$ Reading on the weighing scale $=\frac{1050}{\mathrm{~g}}=\frac{1050}{10}=105 \mathrm{~kg}$
(d) When the lift moves freely under gravity, acceleration a = g
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= m(g – g) = 0
$\therefore$ Reading on the weighing scale $=\frac{0}{g}=0 \mathrm{~kg}$
The man will be in a state of weightlessness.
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