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.

A heavy block of length b and height h is placed at rest on a rough in

1.	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.
Answer» Solution :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`