A horizontal force `F` is applied to a homogeneous rectangular block of mass `m`, width `b` and height `H`. The block moves with constant velocity, the coefficient of friction is `mu_(k)`.
a. What is the greater height `h` at which the force `F` can be applied so that the block will slide without tipping over ?

b. Through which point on the bottom face of the block will the resultant of the friction and normal forces act if `h=H//2?`
c. If the block is at rest and coefficient of static friction is `mu_(s)` what are the various criteria for which sliding or tipping occurs?

Explanation

a. In the absence of any external force in horizontal direction the normal reaction `N` passes through the centre of mass of the block, when force `F` is applied, normal reaction shifts in the direction of applied force `F`.

Since right part of body is having tendency to lift from surface, at the instant of tipping over about the edge the normal reaction passes through edge. from the conditionns of equilibrium.
In horizontal directionn `f=F=mu_(k)mg`
In vertical direction `N=mg`
Balancing torque about edge `Fh-mgb/2`
or `h=(mgb)/(2F)=(mgb)/(2mu_(k)mg)=b/(2mu_(k))`
We can solve the problems by another approach also as the resultant of friction force `(mu_(k)N)` and the normal reactiong must pass through the same point through which `F` passes, since three coplanar forces keeping a body inequilibrium pass through a common point i.e. they should be concurrent.
From Lami's theorem.
`F'/(sin90^@)=F/(sin(180^@+theta))=(mg)/(sin(90^@+theta))`
`implies F/(sintheta)=(mg)/(costheta)` or `tan theta=F/(mg)=(mu_(k)mg)/(mg)=mu_(k)`
From figure it is clear `tan theta=(mu_(k)N)/N=(b/2)/h`
or `(b/2)/h=mu_(k)impliesh=b/(2mu_(k))`
b. In this case we cannot take normal reaction at the edge.

Let normal reaction acts at a distance `x` from the line of action of `mg`.
Torque of all the forces about `C` must be zero.
`(mu_(k)NxxH/2)=N_(x)`
`x=(mu_(k)H)/2`
Alternately we can proceed as in part a. Resultant of `mu_(k)N,N` passes through `C` the interaction of `F` and `mg`.
Therefore from geometry of the figure.
`x/(H/2)=tantheta=mu_(k),x=(mu_(k)H)/2`
c. As the point of application of force is raised higher, the location of the line of action of the normal reaction `N` moves to the let. In the limiting case (when the block is about to tip over ), `x=b//2`. the normal reaction passes through edge as shown in figure.

Initial toppling torque about edge:
`Fh_(max)=mgb/2` or `h_(max)=(mgb)/(2F)`........i
For initial sliding in horizontal direction `F=muN` In vertical direction `N=mg`
Hence, `F=mumg`......ii
If sliding and tipping are equally likely to occur, we can elimiate `F` from eqn i and ii to get `h_(max)=b/(2mu)`
which is independent of weight `mg` height `H` of the body and applied force `F`.