Answer
Coefficient of static friction $=0.65$,
Angle of inclined plane $=20^{\circ}$
The crater will slide if only $mgsin20^{\circ}>F_{s}$
Now, $F_{s}=0.65\times mg$
Again, $mgsin20^{\circ}=0.342 \times mg$
From above, it is obvious that $mgsin20^{\circ}$.
Work Step by Step
Coefficient of static friction $=0.65$,
Angle of inclined plane $=20^{\circ}$
The crater will slide if only $mgsin20^{\circ}>F_{s}$
Now, $F_{s}=0.65\times mg$
Again, $mgsin20^{\circ}=0.342 \times mg$
From above, it is obvious that $mgsin20^{\circ}$.
