Chapter 1: Q8E (page 124)

You walk into an elevator, step onto a scale, and push the “up” button. You recall that your normal weight is 625 N. Draw a free-body diagram.
(a) When the elevator has an upward acceleration of magnitude $2.5 m / S^{2}$ , what does the scale read?
(b) If you hold a 3.85 kg package by a light vertical string, what will be the tension in this string when the elevator accelerates as in part (a)?

Short Answer

Expert verified

The free body diagram is as follows:

(a) When the elevator has an upward acceleration of magnitude $2.5 m / s^{2}$ , the scale reads $80.05 k g$ .

(b) If a3.85 kg package is held by a light vertical string inside the accelerating elevator, the tension in this string will be $47.36 N$ .

Step by step solution

Given data

The normal weight of the person stepping in the elevator is

$W = 625 N$

The acceleration of the elevator is

$a = 2.5 {m/s}^{2}$

The mass of the package is

$m = 3.85 kg$

Net weight inside an elevator

The weight of a body of mass inside an elevator moving up with an acceleration a is

$W = m (g + a) . . . . . (1)$

Here, g is the acceleration due to gravity.

Weight of the person

The mass of the person is

$\begin{array}{rcl} M & = & \frac{W}{g} \\ = & \frac{625 N}{9.8 {m/s}^{2}} \\ = & 63.78 kg \end{array}$

From equation (1), the weight inside the elevator is

$\begin{array}{rcl} W' & = & M (g + a) \\ = & 63.78 kg \times (9.8 {m/s}^{2} + 2.5 {m/s}^{2}) \\ = & 784.5 N \end{array}$

The scale will read

$\begin{array}{rcl} M' & = & \frac{W'}{g} \\ = & \frac{784.5 N}{9.8 {m/s}^{2}} \\ = & 80.05 kg \end{array}$

The scale reads is 80.05 Kg

Tension in the string

Tension in the string is equal to the weight of the object.

From equation (1), the weight of the object inside the elevator is

$W = m (g + a) = 3.85 kg \times (9.8 {m/s}^{2} + 2.5 {m/s}^{2}) = 47.36 N$

Thus, the tension is 47.36N

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Short Answer

Step by step solution

Given data

Net weight inside an elevator

Weight of the person

Tension in the string

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