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 m s-1, (b) downwards with a uniform acceleration of 5 m s-2, (c) upwards with a uniform acceleration of 5 m 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)
Given: The mass of the man is 70 kg and the lift moves upward with the speed of 10 m / s .
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
According to Newton’s second law of motion,
N − m g = m a N = m ( g + a ) (1)
Where, m is the mass of the man, a is the acceleration of the lift and N is the reaction force applied by the floor on the man.
The lift is moving with uniform speed. Therefore, the acceleration of the lift is zero.
By substituting the given values in the equation (1), we get
N = ( 70 kg ) ( 10 + 0 ) = 700 N
The reading shown by the scale is,
m = N g
By substituting the given values in the above expression, we get
m = 700 10 = 70 kg
Thus, the reading is 70 kg.
(b)
Given: The mass of the man is 70 kg and the lift moves downwards with the uniform acceleration of 5 m / s 2 .
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
Newton’s second law of motion is given as,
m g − N = m a N = m ( g − a ) (1)
By substituting the given values in the equation (1), we get
N = ( 70 ) ( 10 − 5 ) = 350 N
The reading shown by the scale is,
m = N g
By substituting the given values in the above expression, we get
m = 350 10 = 35 kg
Thus, the reading is 35 kg.
(c)
Given: The mass of the man is 70 kg and the lift moves upwards with the uniform acceleration of 5 m / s 2 .
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
Newton’s second law of motion is given as,
N − m g = m a N = m ( g + a ) (1)
By substituting the given values in the equation (1), we get
N = ( 70 ) ( 10 + 5 ) = 1050 N
The reading shown by the scale is,
m = N g
By substituting the given values in the above expression, we get
m = 1050 10 = 105 kg
Thus, the reading is 105 kg.
(d)
Given: The mass of the man is 70 kg and the lift is falling freely under the gravity.
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
Newton’s second law of motion is given as,
m g − N = m a N = m ( g − a ) (1)
By substituting the given values in the equation (1), we get
N = ( 70 ) ( 10 − 10 ) = 0 N
The reading shown by the scale is,
m = N g
By substituting the given values in the above expression, we get
m = 0 10 = 0 kg
Thus, the reading is 0 kg.
(a)
Given: The mass of the man is
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
According to Newton’s second law of motion,
Where,
The lift is moving with uniform speed. Therefore, the acceleration of the lift is zero.
By substituting the given values in the equation (1), we get
The reading shown by the scale is,
By substituting the given values in the above expression, we get
Thus, the reading is 70 kg.
(b)
Given: The mass of the man is
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
Newton’s second law of motion is given as,
By substituting the given values in the equation (1), we get
The reading shown by the scale is,
By substituting the given values in the above expression, we get
Thus, the reading is 35 kg.
(c)
Given: The mass of the man is
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
Newton’s second law of motion is given as,
By substituting the given values in the equation (1), we get
The reading shown by the scale is,
By substituting the given values in the above expression, we get
Thus, the reading is 105 kg.
(d)
Given: The mass of the man is
The reading on the scale is a measure of force on the floor by the man. According to Newton’s third law, this force is equal and opposite of the normal force on the man by the floor.
Newton’s second law of motion is given as,
By substituting the given values in the equation (1), we get
The reading shown by the scale is,
By substituting the given values in the above expression, we get
Thus, the reading is 0 kg.
