Chapter 4, Problem 13P

To determine

The strategy used by a thief to climb down through a rope which can’t support the mass of the thief.

Answer to Problem 13P

Solution:

Thief climbs down with the acceleration of $2.22{\text{ m/s}}^{\text{2}}$

Explanation of Solution

Given:

Mass of petty thief, $m_1=75\text{ kg}$

Mass which generates maximum tension T in the rope. $m_2=5\text{8}\text{kg}$

Formula Used:

According to Newton’s second law of motion:

  ${\displaystyle \sum F} =ma$

Where, m is the mass and a is the acceleration.

Calculation:

Maximum tension in the rope would be equal to the weight it can support i.e. $T=m_{2}g$

Weight of thief acts downwards i.e. $m_{1}g$ .

Free body diagram for the given situation is:

  

Substitute the values and solve for acceleration ( a ).

  ${\displaystyle \sum F} =ma$

  $\begin{array}{l}T-m_{1}g=ma\\ m_{2}g-m_{1}g=ma\\ 58\left(9.8\right)–75\left(9.8\right)=ma \\ a=-2.22{\text{ m/s}}^{\text{2}}\end{array}$

Rope can’t support tension due to the weight of the thief. To reduce tension due to weight, thief climbs down with the acceleration $2.22{\text{ m/s}}^{\text{2}}$ .