If for an inclined plane the coefficient of static friction is $\mu_s = \frac{3}{4}$,then the angle of repose for the inclined plane will be ........ $^o$.

$A$ block of mass $15 \, kg$ is resting on a rough inclined plane as shown in the figure. The block is tied by a horizontal string which has a tension of $50 \, N$. The minimum coefficient of friction between the surfaces of contact is $(g = 10 \, m/s^2)$.

$A$ block of mass $m$ rests on a rough inclined plane. The coefficient of friction between the surface and the block is $\mu$. At what angle of inclination $\theta$ of the plane to the horizontal will the block just start to slide down the plane?

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is:

$A$ block of mass $5 \ kg$ is kept against an accelerating wedge with a wedge angle of $45^{\circ}$ to the horizontal. The coefficient of friction between the block and the wedge is $\mu = 0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? Assume $g = 10 \ m \ s^{-2}$.

When a body slides down an inclined plane with a coefficient of friction $\mu$,its acceleration will be: