Two identical uniform and frictionless spheres each of mass

Two identical, uniform and frictionless spheres, each of mass m, rest in a rigid rectangular container as shown in the figure. A line connecting their centers is at 45° to the horizontal. Find, in terms of a ratio to the weight (mg),
a.) the magnitude of the forces on the spheres from the left side of the container?

b.)And what is the magnitude of the forces on the spheres from the right side of the container?

c.)And what is the magnitude of the forces on the spheres from the bottom of the container?

d.) And what is the magnitude of the forces on the spheres from each other?

e.)If the angle is increased to 90.° think about what happens to the forces. Now consider another angle, 34°. What is now the magnitude of the forces on the spheres from the left side of the container?

f.)And what is now the magnitude of the forces on the spheres from the right side of the container?

g.) And what is now the magnitude of the forces on the spheres from the bottom of the container?

h.) And what is now the magnitude of the forces on the spheres from each other?

Please explain in detail! Thank you!

Solution

As you can see the question is very long, So I will answer the initial four part, you have to ask rest of the four part in a new question.

A.

Since all the surface are frictionless.

The contact force F exerted by lower sphere on the upper sphere is in 45 deg of horizontal.

Forces exerted by walls and floors are in the normal direction of them.

Equilibrium on the top force gives two equations

Fwr = F*cos 45 deg

F*sin 45 deg = m*g

equilibrium of forces on bottom sphere

Fwl = F*cos 45 deg

Ff = F*sin 45 deg + mg

Solving the above four equation gives

force by the bottom of the container

Ff = 2*m*g

force by left side of container

Fwl = m*g

force by right side of container

Fwr = m*g

force by the contact of esch other

F = mg/sin 45 deg

F = [sqrt (2)]*m*g