In Fig. 12-72, two identical, uniform, and frictionless spheres, each of mass m, rest in a rigid rectangular container. A line connecting their centers is at$\mathbf{45}\mathbf{°}$to the horizontal. Find the magnitudes of the forces on the spheres from (a) the bottom of the container, (b) the left side of the container, (c) the right side of the container, and (d) each other. (Hint:The force of one sphere on the other is directed along the center–center line.)

The mass of spheres and the angle of force between them.

Hint: The force of one sphere on the other is directed along the center–center line.

The force of one sphere on the other is directed along the centercenterline. As seen from the figure, the force from the sphere would be along $\mathbf{45}\mathbf{°}$and forces from the wall on the balls would be perpendicular. Therefore, you can resolve the forces and write the equations for the vertical and horizontal directions of the forces. Solving this, you would get the forces in terms of weight.$\mathit{F}\mathbf{=}\mathit{F}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{+}\mathit{F}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$

First, you have to resolve all the forces along the X and Y-axis.

Then you get the forces on the upper sphere. As shown in the figure:

$\begin{array}{l}{F}_{wall}=Fcos45°\\ Fsin45°=mg\end{array}$

As well as forces on the bottom sphere.

By solving the above equations,you get

$\begin{array}{c}{F^{\text{'}}}_{floor}^{}=Fsin45°+mg\\ {F^{\text{'}}}_{floor}^{}=mg+mg\\ {F^{\text{'}}}_{floor}^{}=2mg\end{array}$

The left side of the container:

As

$sin45°=cos45°$

Then you can write

$\begin{array}{c}{F^{\text{'}}}_{wall}^{}=Fcos45°\\ =Fsin45°\\ =mg\end{array}$

The right side of the container

As

$sin45°=cos45°$

So,you can write

$\begin{array}{c}{F}_{wall}=Fcos45°\\ =Fsin45°\\ =mg\end{array}$

You know

$\begin{array}{l}Fsin45°=mg\\ F=\frac{mg}{sin45°}\\ F=\sqrt{2}mg\end{array}$