- When a man standing inside an elevator, there are two forces acting on him.
- His weight, (W) which acting downward.
- Normal reaction (R), acting in the opposite direction of weight.
- The reading of the balance is equal to the normal reaction (R).
- Figure below shows the formula to calculate the reading of the balance at different situation.

**Example 1**:

Subra is standing on a balance inside an elevator. If Subra’s mass is 63kg, find the reading of the balance when the lift,

- stationary
- moving upward with a constant velocity, 15 ms
^{-1}., - moving upward with a constant acceleration, 1 ms
^{-2}. - moving downward with a constant acceleration, 2 ms
^{-2}.

**Answer**:

a.

W = mg

W = (63)(10) = 630N

b.

W = mg

W = (63)(10) = 630N

c.

$$\begin{array}{l}R=mg+ma\\ R=(63)(10)+(63)(1)\\ R=693N\end{array}$$

d.

$$\begin{array}{l}R=mg-ma\\ R=(63)(10)-(63)(2)\\ R=504N\end{array}$$

**Example 2**:

A 54kg boy is standing in an elevator. Find the force on the boy's feet when the elevator

- stands still
- moves downward at a constant velocity of 3 m/s
- decelerates downward with at 4.0 m/s
^{2}, - decelerates upward withat 2.0 m/s
^{2}.

**Answer**:

a.

W = mg

W = (54)(10) = 540N

b.

W = mg

W = (54)(10) = 540N

c.

$$\begin{array}{l}R=mg+ma\\ R=(54)(10)+(54)(4)\\ R=756N\end{array}$$

d.

$$\begin{array}{l}R=mg-ma\\ R=(54)(10)-(54)(2)\\ R=432N\end{array}$$