Chapter SH, Problem 3.2APP

To determine

To Calculate: The force applied by the weight lifter’s friend.

Answer to Problem 3.2APP

  $300\text{N}$

Explanation of Solution

Given:

Weight, $w=1500\text{N}$

Force exerted by weight lifter, $F_1=1200\text{N}$

Velocity is constant, so acceleration $a=0$

Formula Used:

From Newton’s second law of motion, the net force would be:

  ${\displaystyle \sum F=ma}$

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

Calculations:

Let the force exerted by weight lifter’s friend be $F_2$ .

Substitute the values in the formula and solve:

  $\begin{array}{l}w+F_1+F_2=0\\ 1500\text{N}-1200\text{N}+F_2=0\\ F_2=-300\text{N}\end{array}$

Here, negative force indicates force is being applied in the opposite direction of the weight.

Conclusion:

Thus, the force applied by the weight lifter’s friend is 300 N.