$A$ weight $W$ can be just supported on a rough inclined plane by a force $F$ either acting along the plane or horizontally. If $\theta$ is the angle of friction,then $F / W$ 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 $g=10 \ m/s^2$)

When a cube of side length $a$ is in limiting equilibrium on an inclined plane and it is about to topple,what is the coefficient of friction between the cube and 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 $4 \ kg$ is at rest on a rough inclined plane making an angle of $\theta$ with the horizontal. The coefficient of static friction between the block and the plane is $0.5$ and the frictional force on the block is $14.14 \ N$. Find the value of $\theta$. (in $^{\circ}$)

$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.