The wedge is fixed and the mass of the block is $m$. The minimum value of $F$ so that the block is in equilibrium is (neglect the dimension of the block).

$A$ 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 $(g = 10 \, m/s^2)$.

$A$ box when dropped from a certain height reaches the ground with a speed $v$. When it slides from rest from the same height down a rough inclined plane inclined at an angle $45^{\circ}$ to the horizontal,it reaches the ground with a speed $v/3$. The coefficient of sliding friction between the box and the plane is:

Two blocks $A$ and $B$ of equal masses are sliding down along straight parallel lines on an inclined plane of $45^{\circ}$. Their coefficients of kinetic friction are $\mu_A = 0.2$ and $\mu_B = 0.3$ respectively. At $t = 0$,both the blocks are at rest and block $A$ is $\sqrt{2} \text{ m}$ behind block $B$. Calculate the time and distance from the initial position where the front faces of the blocks come in line on the inclined plane as shown in the figure. (Use $g = 10 \text{ m/s}^2$)

$Assertion$ : Angle of repose is equal to the angle of limiting friction.
$Reason$ : When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

$A$ block is placed on a rough inclined plane of inclination $\theta = 30^o$. If the force required to drag it up along the plane is to be smaller than the force required to lift it vertically,the coefficient of friction $\mu$ should be less than: