Question

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.)

Short answer

The magnitude of forces on the sphere

a) from the bottom of the container,${F^{\text{'}}}_{floor}^{}=2mg$ .

b) from the left side of the container,${F^{\text{'}}}_{wall}^{}=mg$ .

c) from the right side of the container, ${\text{F}}_{\text{wall}}=\text{mg}$.

d) on each other, $\text{F}=\sqrt{2}\text{mg}$.

Step-by-step solution

Understanding the given information

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.

Concept and formula used in the given question

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.$\mathit{F}\mathbf{=}\mathit{F}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{+}\mathit{F}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$

(a) Calculation for the magnitudes of the forces on the spheres from the bottom of the container

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

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

(b) Calculation for themagnitudes of the forces on the spheres from the left side of the container

The left side of the container:

As

$sin45°=cos45°$

Then you can write

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

(c) Calculation for themagnitudes of the forces on the spheres from the left side of the container to the right side of the container

The right side of the container

As

$sin45°=cos45°$

So,you can write

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

(d) Calculation for the magnitudes of the forces on the spheres from the each other

You know

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