Question

A tall block of mass $M=50\text{kg}$ and base width $b=1\text{m}$ and height $h=3\text{m}$ is kept on a rough inclined surface with coefficient of friction $\mu =0.8$ as shown in figure. The angle of inclination with the horizontal is ${37}^{\circ }.$ Determine whether the block slides down or topple over. Take $g=10{\text{m/s}}^2$.

• A. Block will topple
• B. Block will slide
• C. Block will slide, then topple
• D. Block will neither slide nor topple.

Answer 1

The correct option is A Block will topple

Given, $\mu =0.8,b=1\text{m},h=3\text{m}$
$\tan {37}^{\circ }=\frac{3}{4}$

Let $F=F_{max}$ when block is just about to slide
Then, $F_{max}=f_{limit}$ for just sliding
where, $f_{limit}=\mu N$
$=0.8\times 50\times 10\times \frac{4}{5}=320\text{N}$



Now, let $F=F_{max}$ when block is just about to topple (Normal will shift at point $\text{O}$)
Applying torque balance about point $\text{O}$,
$F_{max}\times \frac{h}{2}=mg\cos {37}^{\circ }\times \frac{b}{2}$
$F_{max}\times \frac{3}{2}=50\times 10\times \frac{4}{5}\times \frac{1}{2}$
$F_{max}=\frac{400}{3}\text{N}=133.3\text{N}$ for just toppling
As we can see,
$F_{max}$ for just toppling $<F_{max}$ for just sliding

and the component of force of gravity acting down the incline
$=mg\sin {37}^{\circ }=50\times 10\times \frac{3}{5}=300\text{N}$
So, $mg\sin {37}^{\circ }>$ $F_{max}$ for just toppling.
$\Rightarrow$ Block will topple.