Question

1. a 980 n man stands on a scale in a motionless elevator. when the elevator begins to move, the scale reads 1200 n. what is the magnitude and direction of the elevator’s acceleration?
2. an 784 n woman stands on a scale in a motionless elevator. when the elevator begins to move, the scale reads 600 n. what is the magnitude and direction of the elevator’s acceleration?

Explanation:

Problem 19 (A 980 N man in elevator)

Step1: Find mass of the man

Weight \( W = mg \), so mass \( m=\frac{W}{g} \). Given \( W = 980\ N \), \( g = 9.8\ m/s^2 \), so \( m=\frac{980}{9.8}=100\ kg \).

Step2: Analyze forces

The normal force \( N = 1200\ N \) (scale reading), weight \( W = 980\ N \). Using Newton's second law \( \sum F = ma \), where \( \sum F=N - W \).

Step3: Calculate acceleration

\( N - W=ma \), so \( a=\frac{N - W}{m}=\frac{1200 - 980}{100}=\frac{220}{100}=2.2\ m/s^2 \). Since \( N>W \), the net force is upward, so acceleration is upward.

Step1: Find mass of the woman

Using \( W = mg \), \( m=\frac{W}{g} \). Given \( W = 784\ N \), \( g = 9.8\ m/s^2 \), so \( m=\frac{784}{9.8}=80\ kg \).

Step2: Analyze forces

Normal force \( N = 600\ N \), weight \( W = 784\ N \). Newton's second law: \( \sum F=ma=W - N \) (since \( W>N \), net force is downward).

Step3: Calculate acceleration

\( a=\frac{W - N}{m}=\frac{784 - 600}{80}=\frac{184}{80}=2.3\ m/s^2 \). Direction is downward as \( W>N \).

Answer:

Magnitude: \( 2.2\ m/s^2 \), Direction: Upward

Problem 20 (A 784 N woman in elevator)