Question

A block of mass $M=50\text{kg}$, base $b=1\text{m}$ and height $h=3\text{m}$ is kept on a rough inclined surface with coefficient of static friction $\mu =0.8$ as shown in figure. The angle of inclination with horizontal is ${37}^{\circ }$. Determine whether the block slides down or topples over.

• A. slides down
• B. topples over
• C. remains in complete equilibrium
• D. cannot be determined.

Answer 1

The correct option is B topples over

Coefficient of static friction, $\mu =0.8$
Here, $\frac{b}{h}=\frac{1}{3}=0.33$
where $b=$ base and $h=$ height of block.
$\tan \theta =\tan {37}^{\circ }=\frac{3}{4}=0.75$
Condition for no slipping is,
$\mu \ge \tan \theta ...\left(i\right)$
Condition for toppling is,
$\mu \ge \frac{b}{h}...\left(ii\right)$

From the Eq. $\left(i\right)$ and $\left(ii\right)$, we find that as per given data, $\mu >\tan \theta \&\mu >\frac{b}{h}$, hence block will have no sliding tendency, rather it will tend to topple over.

Method II:
The normal reaction $N$ will shift towards left to counter the torque of friction, i.e at distance $x$ from centre.
For translational equilibrium:
$N=Mg\cos {37}^{\circ }$
$\Rightarrow N=50\times 10\times \frac{4}{5}=400\text{N}$
$f=Mg\sin {37}^{\circ }$
$\Rightarrow f=50\times 10\times \frac{3}{5}=300\text{N}$
$f_{\text{max}}=\mu N=0.8\times 400=320\text{N}$

Clearly $f\le f_{\text{max}}$,supports no sliding of block.

For rotational equilibrium, net torque about centre of block should be zero.
$\sum {\tau }_C=0...\left(iii\right)$
$\because \tau =F\times r_{\perp }$, considering anticlockwise sense of rotation for torque as $+ve$
${\tau }_N=-(N\times x),{\tau }_f=+f\left(\frac{h}{2}\right)$
& ${\tau }_{Mg}=0$ as $Mg$ passes through centre $C$.

Substituting in Eq $\left(iii\right)$,
$\Rightarrow {\tau }_{Mg}+{\tau }_N+{\tau }_f=0$
$\Rightarrow -Nx+(f\times \frac{h}{2})=0$
$\Rightarrow x=\frac{fh}{2N}=\frac{300\times 3}{2\times 400}$
$\therefore x=\frac{9}{8}\text{m}$
The value of $x$ cannot be more than the value of the base of the block $b=1\text{m}$
$\therefore$ value of $x$, justifies that block will topple over.