The blocks are in equilibrium. The friction force acting on the $10 \, kg$ block 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 $g = 10 \ m \ s^{-2}$.

$A$ block of mass $5 \ kg$ is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and the inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is . . . . . . $N$.

Consider a small block sliding down an inclined plane of inclination $30^{\circ}$ with the horizontal. The coefficient of friction is $\mu = \frac{2}{3} x$, where $x$ is the distance (in meters) through which the mass slides down. The distance covered by the mass before it stops is

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If $\mu$ is the coefficient of friction,the inclination of the plane to the horizontal is:

$A$ uniform rod of mass $20 \ kg$ and length $5 \ m$ leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g=10 \ m/s^2$)