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

$A$ block of mass $2 \, kg$ 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 $0.7$. The frictional force on the block is ....... $N$.

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

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:

$A$ body of mass $2 \ kg$ slides down with an acceleration of $4 \ m/s^2$ on an inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be (Acceleration due to gravity $= 10 \ m/s^2$) (in $N$)

$A$ uniform chain of length $L$ hangs partly from a table and is kept in equilibrium by friction. If the maximum length that can hang without slipping is $\ell$,then the coefficient of friction between the table and the chain is: