If the normal force is doubled,the coefficient of friction is

$A$ body of mass $5 \,kg$ is placed on a rough horizontal surface of coefficient of static friction $\frac{1}{3}$. The least pulling force to be applied on the body at an angle $45^{\circ}$ with the horizontal to slide it,is . . . . . . $\left(g=10 \,ms^{-2}\right)$

As shown in the figure,a block of mass $\sqrt{3} \text{ kg}$ is kept on a horizontal rough surface with a coefficient of friction $\mu = \frac{1}{3 \sqrt{3}}$. $A$ force $F$ is applied on the vertical face of the block at an angle of $60^{\circ}$ with the horizontal. The minimum force $F$ required to just move the block is $3x$. Find the value of $3x$.
$\left[ g = 10 \text{ m/s}^2; \sin 60^{\circ} = \frac{\sqrt{3}}{2}; \cos 60^{\circ} = \frac{1}{2} \right]$

The coefficient of static friction,$\mu_s$,between block $A$ of mass $2\, kg$ and the table as shown in the figure is $0.2$. Find the maximum mass value of block $B$ in $kg$ so that the two blocks do not move. The string and the pulley are assumed to be smooth and massless. $(g = 10\, m/s^2)$

$A$ body of mass $m$ rests on a horizontal surface. The coefficient of friction between the body and the surface is $\mu$. If the mass is pulled by a force $P$ at an angle of $30^{\circ}$ with the horizontal as shown in the figure,the limiting friction between the body and the surface will be:

$A$ block of mass $m$ lying on a rough horizontal plane is acted upon by a horizontal force $P$ and another force $Q$ inclined at an angle $\theta$ to the vertical. The block will remain in equilibrium,if the coefficient of friction between it and the surface is