Question

A horizontal force $F$ is applied at the top of an equilateral triangular block having mass $m$. The minimum coefficient of friction required to topple the block before translation will be :

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

The attached picture shows the forces acting and the center of mass.

To avoid sliding $\mu mg=F$

In order to topple the clockwise torque across C produced by F should be more than anti clockwise torque by mg

As a result of the torque the triangle will start to rotate about point C (a,0)

The normal reaction will act on point C as it is the only point in contact with the ground when toppling will start. So, the Torque of friction and Normal about C will be zero.

${\tau }_c<0$

$-Facos\left(30\right)+mga/2<0$

$\mu mgacos\left(30\right)>mga/2$

$\mu >\frac{1}{2cos\left(30\right)}$

$\mu >\frac{1}{\sqrt{3}}$