Question

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration 9m/s², what would be the reading of the weighing scale? (take g= 10m/s²)

Answer 1

Answer:

• Reading will be (W) = 50 Newton's.

Given:

1. Mass of the Body (M) = 50 Kg.
2. Acceleration due to gravity (g) = 10 m/s².
3. Acceleration of the Lift (a) = 9 m/s².

Explanation:

$\rule{300}{1.5}$

From F.B.D of the Lift.

#refer the attachment for figure.

As the Lift is Moving Down there is an Pseudo Force(F) acting on it in the opposite direction I.e. In Upward direction. (F = Ma).

Now, From Equilibrium of Body.

$\large\star \: {\boxed{\bold{F_{net} = 0}}}$

$\large{\tt \hookrightarrow F_{net} = 0}$

Therefore,

$\large{\tt \hookrightarrow N + ma = mg}$

$\bold{Here}\begin{cases}\text{N Denotes Normal Reaction} \\ \text{ma = F Denotes Pseudo Force} \\ \text{mg = W Denotes Weight of Man}\end{cases}$

Simplifying,

$\large{\tt \hookrightarrow N + ma = mg}$

$\large{\tt \hookrightarrow N = mg - ma}$

$\large{\tt \hookrightarrow N = m(g - a)}$

• Here "N" is the Reading of the Weighing Scale.

Substituting the values,

$\large{\tt \hookrightarrow N = 50 \: Kg \times(10 - 9) \: m/s^2}$

$\large{\tt \hookrightarrow N = 50 \times(10 - 9)}$

$\large{\tt \hookrightarrow N = 50 \times 1}$

$\huge{\boxed{\boxed{\tt N = 50 \: N}}}$

∴ Reading of Weighing scale will be 50 Newton's.

$\rule{300}{1.5}$