Chapter 4, Problem 72A

To determine

The force exerted by the wall on the Pwhen she pushes off the wall.

Answer to Problem 72A

  $1.80\times {10}^2\text{N}$ .

Explanation of Solution

Given:

The weight of the P is $F_{\text{g}}=588\text{N}$ .

The vertical acceleration with which P pushes the wall is $a_{\text{P}}=3.00{\text{m/s}}^2$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Newton’s Third Law of Motion:

According to Newton’s Third Law, for every action, there is an equal and opposite reaction. When a body $A$ applies a force on the body $B$ then the body $B$ exerts an equal and opposite force on the body $A$ .

  $F_{\text{A on B}}=-F_{\text{B on A}}$

Calculation:

Consider the acceleration due to the gravity of Earth is $g=9.80{\text{m/s}}^2$ .

Consider the mass of P is denoted by $m_{\text{P}}$ .

Find the mass of P as follows:

Apply Newton’s Second Law of Motion,

  $\begin{array}{c}{m}_{\text{P}}=\frac{F_{\text{g}}}{g}\\ =\frac{588\text{N}\times \left(\frac{1\text{kg}\cdot {\text{m/s}}^2}{1\text{N}}\right)}{9.80{\text{kgm/s}}^2}\\ =60\text{kg}\end{array}$

Find the force exerted by the wall on P as follows:

  $\begin{array}{c}{F}_{\text{wall on P}}=-F_{\text{P on wall}}\\ =m_{\text{P}}{a}_{\text{P}}\\ =60\times 3\text{N}\\ =1.80\times {10}^2\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the wall on P is $1.80\times {10}^2\text{N}$ .