Friction force acting on object is given as function of angle '$\theta$' $f_r=\frac{\mu mg}{\sin \theta +\mu \cos \theta }$ then find $\theta$ where $f_r$ will be minimum.

Consider the system shown below.

A horizontal force $F$ is applied to a block $X$ of mass $8 \mathrm{kg}$ such that the block $Y$ of mass $2 \mathrm{kg}$ adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block $X$. The coefficient of friction between the surfaces of the blocks $X$ and $Y$ is $0.5$. Take acceleration due to gravity to be $10 \mathrm{m} {\mathrm{s}}^{-2}$. The minimum value of $F$ is

A block of mass $5 \mathrm{k}\mathrm{g}$ is (i) pushed in case $\left(\mathrm{A}\right)$ and (ii) pulled in case $\left(\mathrm{B}\right),$ by a force $\mathrm{F}=20\mathrm{ }\mathrm{N},$ making an angle of ${30}^o$ with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is $\mu =0.2.$ The difference between the accelerations of the block, in case $\left(\mathrm{B}\right)$ and case $\left(\mathrm{A}\right)$ will be:
$\left(\mathrm{g}=10\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2}\right)$