Question

An equilateral triangular wedge of mass m is kept on frictionless surface. A horizontal force F is applied at top end of wedge. Coefficient of friction is $\mu =0.8$.

• A. As F is increased slowly wedge will slide before toppling.
• B. As F is increased slowly wedge will topple before sliding.
• C. Minimum force required to topple wedge is $\frac{mg}{\sqrt{3}}$.
• D. For $F=\frac{mg}{2}$, wedge will both topple and slide.

Answer 1

Answer: (B), (C)

The correct options are
B As F is increased slowly wedge will topple before sliding.
C Minimum force required to topple wedge is $\frac{mg}{\sqrt{3}}$.


To slide wedge $F=f\Rightarrow F=\mu mg=0.8mg$

By torque equilibria about P,

$F\times \frac{\sqrt{3}}{2}L=mg\frac{L}{2}\Rightarrow F=\frac{mg}{\sqrt{3}}$

$\frac{mg}{\sqrt{3}}<0.8mg$,
So block will not slide but topple.