A block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and $\tan \theta>\mu$. The block is held stationary by applying a force $\mathrm{P}$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $\mathrm{P}$ is varied from $\mathrm{P}_1=$ $m g(\sin \theta-\mu \cos \theta)$ to $P_2=m g(\sin \theta+\mu \cos \theta)$, the frictional force $f$ versus $P$ graph will look like

In the figure shown, horizontal force $F_1$ is applied on a block but the block does not slide. Then as the magnitude of vertical force $F_2$ is increased from zero the block begins to slide; the correct statement is

A uniform rod of length $L$ and mass $M$ has been placed on a rough horizontal surface. The horizontal force $F$ applied on the rod is such that the rod is just in the state of rest. If the coefficient of friction varies according to the relation $\mu = Kx$ where $K$ is a $+$ ve constant. Then the tension at mid point of rod is

In the given figure the acceleration of $M$ is $(g = 10 \,ms^{-2})$

A wooden block of mass $M$ resting on a rough horizontal surface is pulled with a force $F$ at an angle $\phi $ with the horizontal. If $\mu $ is the coefficient of kinetic friction between the block and the surface, then acceleration of the block is

A block pressed against the vertical wall is in equilibrium. The minimum coefficient of  friction is:-