A heavy block of length `b` and height `h` is placed at rest on a rough inclined plane of inclination `theta` with the horizontal, as shown in figure.

Explanation

a. The free body diagram of the block is shown in figure.

Note that the point of application of the normal reaction `N` is displaced through `x` in the downward direction. It has to porduce a clockwise torque about the centre of mass that may balance the anticlockwise torque produced by the friction force.
There are two tendencies of the block: to slide down
to rotate (or topple) about the point `A`
For translation equation `sumvecF=0`
`f=Mgsintheta`.......i
`N=mgcostheta`...........ii
For rotational equilibrium `sum vectau_(C)=0`
`Nxx x=fxxh/2impliesMgcosthetaxx x=(fh)/2`
`implies f=(2Mgcosthetaxx x)/h`
If the block topples over about `A, x=b/2`
Hence, `(bMgcostheta)/h`......iii
From eqn if block do not slide down `fltmuN`
`Mgsintheta,muyxxMgcosthetaimpliestanthetalemu`
or we can say if the block `multtantheta`
from eqn iii if the block topples before sliding,
`flemu(Mgcostheta)`
`(bMgcostheta)/hlemuMgcosthetaimpliesb/hlemu`
Hence if the block topples before sliding `mugeb/h`