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

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

A mass of $4\; kg$ rests on a horizontal plane. The plane is gradually inclined until at an angle $\theta= 15^o$ with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?

If a ladder weighing $250\,N$ is placed against a smooth vertical wall having coefficient of friction between it and floor is $0.3$, then what is the maximum force of friction available at the point of contact between the ladder and the floor ........ $N$

Among the forces in nature, friction can be classified into

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