Question

A particle of weight $W$ rests on a rough inclined plane which makes an angle $\alpha$ with the horizontal. If the coefficient of static friction $\mu =2\tan \alpha$, find the horizontal force $H$ acting transverse to the slope of the plane when the particle is about to slip.

• A. $W\sqrt{3}\sin \alpha$
• B. $\frac{\sqrt{3}}{2}W\sin \alpha$
• C. $W\sin \alpha$
• D. $2W\sin \alpha$

Answer 1

The correct option is A $W\sqrt{3}\sin \alpha$

A component of weight will act along the plane downwards which is $W\sin \alpha$.
And the other component of weight will act perpendicular to plane and will balance normal force on the particle as shown in figure.
Net force, $F=\sqrt{(W\sin \alpha {)}^2+H^2}$
For particle about to get slip, $F=\mu N$
$\sqrt{(W\sin \alpha {)}^2+H^2)}=2\tan \alpha \left(W\cos \alpha \right)$
$=2W\sin \alpha$
Squaring both sides
$(W\sin \alpha {)}^2+H^2)=4W^2{\sin }^2\alpha$
$H=\sqrt{3}W\sin \alpha$