$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$).

The minimum value of $F$ so that the block is in equilibrium is

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction, when the angle of inclination of the plane is $60^{\circ}$ is

$A$ block kept on a rough inclined plane,as shown in the figure,remains at rest up to a maximum force $2 \ N$ down the inclined plane. The maximum external force up the inclined plane that does not move the block is $10 \ N$. The coefficient of static friction between the block and the plane is: [Take $g = 10 \ m/s^2$]

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

$A$ block of mass $m$ is stationary on a rough plane of mass $M$ inclined at an angle $\theta$ to the horizontal,while the whole setup is accelerating upwards at an acceleration $a$. If the coefficient of friction between the block and the plane is $\mu$,then the force that the plane exerts on the block is