Question

A man of mass m kg is standing on a weighing machine which is kept in an elevator. Find the reading of the spring balance if the elevator is going up with an acceleration $g/10$.

Answer 1

Hint: This problem has to be solved in the frame of the elevator and not the inertial frame. After applying the pseudo force, we will have to calculate the normal force exerted by the weighing machine on the man, which will be the reading of the spring balance.

Formula used:
$\overrightarrow{F}=m\overrightarrow{a}$ ---(Newton’s second law of motion)
where $\overrightarrow{F}$ is the force vector applied on or by a body, m is the mass of the body and $\overrightarrow{a}$ is the acceleration of the body due to that force.

Complete step-by-step answer:

This problem will become very easy if we solve it from the frame of reference of the elevator and not the inertial frame.
Thus, we assume a frame of reference that is moving along with the elevator, that is, with an upward acceleration of $g/10$. Now, since our frame of reference is accelerating upward, we will have to apply a pseudo force with the same acceleration $g/10$ on all bodies, but in the downward direction.
Hence, now let us analyse the problem.
The mass of the man is $m$ kg.
The magnitude of normal force exerted by the man on the weighing machine will be the reading on the weighing machine.
Therefore first, let us draw the free body diagram of the man and put all the proper pseudo forces.

Where $N$ is the normal force on the man, $mg$ is its weight and $mg/10$ is the pseudo force on the body.
Since, in this frame of reference, the man is in static equilibrium, the sum of the forces on him should be zero.
$\sum{\overrightarrow{F}}=0$ --(for static equilibrium)--(1)
$\overrightarrow{F}=m\overrightarrow{a}$ ---(Newton’s second law of motion)--(2)

where $\overrightarrow{F}$ is the force vector applied on or by a body, m is the mass of the body and $\overrightarrow{a}$ is the acceleration of the body due to that force.
Therefore, putting all the proper forces and using equation (1) and (2),
$N-mg-\dfrac{mg}{10}=0$
where forces in the upward direction are considered positive.
$\therefore N=mg+\dfrac{mg}{10}=\dfrac{11mg}{10}=1.1mg$
Hence, the magnitude of normal force is $1.1mg$.
Hence, the reading of the spring balance is $1.1mg$.

Note: Many problems such as these can be easily solved by assuming a proper frame of reference. Always solving in the inertial frame of reference can bring in unnecessary variables and complications in the calculations of some problems. However, while assuming a non-inertial frame of reference, one must always first determine the acceleration of the frame, if any and assign proper pseudo forces opposite to the acceleration of the frame on all bodies in the frame.
However, not all moving non-inertial frames have to be assigned pseudo forces. Only the frames which are accelerating have to be assigned pseudo forces, those with constant velocity do not.