Chapter 9, Problem 73GP

To determine

To Find: The magnitude of the force acting on the sphere exerted by each plane.

Answer to Problem 73GP

The magnitude of the force which makes $30°$ with the horizontal surface is $\text{187}\text{.3 N}$ .

The magnitude of the force which makes $70°$ with the horizontal surface is $\text{99}\text{.7 N}$ .

Explanation of Solution

Given:

Mass of sphere, $\text{m = 20 kg}$

Reaction force F makes and angle of $\text{60°}$ at the centre of gravity, ${\text{θ}}_1\text{ = 60°}$

Reaction force R makes $\text{20°}$ at the centre of gravity, ${\text{θ}}_2\text{=20°}$

Formula used:

Write the expression for the force balance in the horizontal direction.

  $\text{F}\times {\text{cos θ}}_{\text{1}}\text{ = R}\times {\text{cosθ}}_{\text{2}}$

Write the expression for the force balance in the vertical direction.

  $\text{F}\times {\text{sin θ}}_{\text{1}}\text{ + R}\times {\text{sinθ}}_{\text{2}}\text{ = mg}$

Calculation:

Substitute the given values,

  $\begin{array}{l}\text{F}\times \text{cos 60}°\text{ = R}\times \text{cos 20}°\\ \text{F=1}\text{.879}\times \text{R}\text{-(1)}\end{array}$

  $\begin{array}{l}\text{F}\times \text{sin 60}°\text{ + R}\times \text{sin 20}°\text{ = 20}\times \text{9}\text{.8 = 196}\text{.2}\\ 0.866\times \text{F}+0.342\times \text{R}=196.2\text{-(2)}\end{array}$

Substitute the value of F in equation 2 from equation 1.

  $\begin{array}{l}0.866\times 1.879\times \text{R}+0.342\times \text{R}=196.2\\ \text{R=99}\text{.7}\text{N}\end{array}$

  $\begin{array}{c}\text{F = 1}\text{.879}\times \text{99}\text{.7 N}\\ \text{=187}\text{.3}\text{N}\end{array}$

Conclusion:

The magnitude of the force which makes $30\text{°}$ with the horizontal surface is $\text{F = 187}\text{.3 N}$ .

The magnitude of the force which makes ${70}^0$ with the horizontal surface is $\text{R = 99}\text{.7 N}$ .