Pushing a chest across the floor.?

Question

A man wishes to push a chest across the horizontal floor.
To do this, he applies a horizontal force of magnitude F0 to
the rear of the chest, as indicated in figure. The rear and
front legs of the chest are separated by a distance b. The
chest has a mass M and its center of mass C is located
midway between its legs at a height h above the floor. The
coefficient of kinetic friction between the floor and the
legs of the chest is μ. What is the maximum vertical height
y above the floor that the force can be applied so that the
rear legs of the moving chest do not leave the floor and so
that the chest does not begin to flip over?

Answer 1

To solve this, I assume the rate of motion will be constant speed so there is no horizontal acceleration of the chest.
Also, the reaction force between the rear legs and the floor will be zero since the balance point is just on the verge of tipping. So no friction develops here either. I will also solve this in two dimensions.

Summing torques at the front edge of the chest at a height h from the floor with rotational torque as clockwise being positive

Note that all of the weight of the chest is on the front legs, so friction is
M*g*μ
and the torque from friction is
M*g*μ*h
the torque from the center of mass is
=-M*g*b/2

The torque from the applied force (note it is applied a distance of y from the floor which has to be above h for this solution to be valid, or even be the right solution for that matter)
When a line is drawn from the point where I am summing torques horizontally through the center of mass, it forms the base of a triangle that has a vertical side of y-h and a hypotenuse of sqrt((y-h)^2+4*b^2)
The torque is
F0*(2h-y)

Summing all the torques and setting equal to zero

F0*(2h-y)=M*g*(b/2-μ*h)

y (max)=2*h-M*g*(b/2-μ*h)/F0
Y (max) is also less than the height of the chest.

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