Chapter 10.2, Problem 31SSC

To determine

To Calculate: The maximum weight of the object which could be lifted.

Answer to Problem 31SSC

  $\text{5250}\text{N}$

Explanation of Solution

Given:

Largest force that worker can exert, $F_e=875\text{}\text{N}$

Amount of rope moved, $l_e=1.5\text{m}$

Object moves, $l_o=0.25\text{m}$

Formula Used:

  $MA=\frac{l_e}{l_o}=\frac{F_o}{F_e}$

  $F_e\times l_e=F_o\times l_o$

Here, $F_o$ is the weight of the object that can be lifted.

Calculations:

Substitute the values and solve:

  $\begin{array}{l}{F}_e\times l_e=F_o\times l_o\\ F_o=\frac{1.5}{0.25}\times 875\text{}\text{N}\\ F_o\text{=5250}\text{N}\end{array}$

Conclusion:

Thus, the maximum weight of the object which could be lifted is $\text{5250}\text{N}$ .