Answer
The magnitude of the force on the spheres from the left side of the container is $~~mg$
Work Step by Step
The force from the bottom of the container must be equal in magnitude to the total weight of the spheres:
$F = 2mg$
The vertical component of the force on the bottom sphere due to the top sphere must be equal to $~~mg$
We can find the force on the bottom sphere due to the top sphere:
$F~sin~\theta = mg$
$F = \frac{mg}{sin~\theta}$
$F = \frac{mg}{sin~45^{\circ}}$
$F = \frac{mg}{1/\sqrt{2}}$
$F = \sqrt{2}~mg$
The force due to the left side of the container must be equal in magnitude to the horizontal component of the force on the bottom sphere due to the top sphere. We can find the force due to the left side of the container:
$F = \sqrt{2}~mg~cos~\theta$
$F = \sqrt{2}~mg~cos~45^{\circ}$
$F = \sqrt{2}~mg~(\frac{1}{\sqrt{2}})$
$F = mg$
The magnitude of the force on the spheres from the left side of the container is $~~mg$
