Question

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

Answer 1

Here, $\mu =0.8,\frac{b}{h}=\frac{1}{3}=0.33;\tan \theta =\frac{3}{4}=0.75$
Since $\mu >b/h;\mu >\tan \theta$, therefore the block of translation tendency to topple over. From the condition of translation equilibrium, we have
$N=Mg\cos {37}^o=\left(50\right)\left(10\right)\left(\frac{4}{5}\right)orN=400N$
$f=Mg\cos {37}^o=\left(50\right)\left(10\right)\left(\frac{3}{5}\right)=300N<f_{max}=\mu N=0.8\left(400\right)=320N$
for rotational equilibrium $\sum {\tau }_C=0$
$N_x=f\frac{h}{2}$
thus, $x=\frac{f}{N}\frac{h}{2}=\left(\frac{300}{400}\right)\left(\frac{3}{2}\right)=\frac{9}{8}>\frac{b}{2}$
Since the value of $x$ is more than $b/2$ which not possible therefore the block topples over