In the arrangement shown, the relation between acceleration of wedge $1$ and wedge $2$ is

In the situation given, all surfaces are frictionless, pulley is ideal and string is light, $F=\frac{mg}{2},$ find the acceleration of block 2 .

In the arrangement shown, all surfaces are frictionless. The rod $R$ is constrained to move vertically. The vertical acceleration of $R$ is $a_1$ and the horizontal acceleration of the $W$ is $a_2.$ The ratio $\frac{a_1}{a_2}$ is equal to

A block of mass $2 \mathrm{kg}$ is placed on the floor $(\mu =0.4).$ A horizontal force of $7 \mathrm{N}$ is applied on the block. The force of friction between the block and floor is

A bus starts from rest and travels a distance $s$ along a straight horizontal road. The coefficients of static and kinetic friction between the road and the tyres are ${\mu }_s$ and ${\mu }_k$ respectively. The minimum time of travel is proportional to

Find the contact force on the block having mass $1 \mathrm{kg}.$ (Assume wedge to be fixed)

The coefficient of friction of a surface is $\frac{1}{\sqrt{3}}.$ What should be the angle of inclination so that a body placed on the surface just begins to slide down?

The coefficient of static friction, µ_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 do not move ? The string and the pulley are assumed to be smooth and massless. (g = 10 m/s²)