Question

A block with a square base measuring $a\times a$, and height $h$, is placed on an inclined plane. The coefficient of friction is $\mu$. The angle of inclination ($\theta$) of the plane is gradually increased. The block will :

• A. topple before sliding if $\mu >\frac{a}{h}$
• B. topple before sliding if $\mu <\frac{a}{h}$
• C. slide before toppling if $\mu >\frac{a}{h}$
• D. slide before toppling if $\mu <\frac{a}{h}$

Answer 1

Answer: (A), (D)

The correct options are
A topple before sliding if $\mu >\frac{a}{h}$
D slide before toppling if $\mu <\frac{a}{h}$

Using the free body diagram we get frictional force $f$ as
$f=Mgsin\theta$
and normal force as
$N=Mgcos\theta$
For rotational equilibrium we have the torque as
$N\times x=f\times \frac{h}{2}$
or
$Mgcos\theta \times x=\frac{fh}{2}$
or
$f=\frac{2Mgcos\theta \times x}{h}$
If the block topples about A, $s=\frac{a}{2}$
Thus
$f=\frac{aMgcos\theta }{h}$
If the block do not slide down $f<\mu N$
or
$Mgsin\theta <\mu \times Mgcos\theta$
or
$tan\theta \le \mu$
Also from the frictional force equation we have
$f\le \mu \left(Mgcos\theta \right)$
or
$\frac{aMgcos\theta }{h}\le \mu Mgcos\theta$
or
$\frac{a}{h}\le \mu$
i.e the block would slide for $\mu <\frac{a}{h}$