In the given arrangement the maximum value of $F$ for which there is no relative motion between the blocks
$\mu m_1 g$
$\mu\left(m_1+m_2\right) g$
$\mu m_1 g\left(\frac{m_1}{m_2}+1\right)$
Zero
Explain -“Static friction force opposes impending motion”.
A pen of mass $m$ is lying on a piece of paper of mass $M$ placed on a rough table. If the coefficients of friction between the pen and paper and the paper and the table are $\mu_1$ and $\mu_2$, respectively. Then, the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by
A rope of length $L$ and mass $M$ is being pulled on a rough horizontal floor by a constant horizontal force $F$ = $Mg$ . The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is $1/2$ . Then, the tension at the midpoint of the rope is
Determine the maximum acceleration in $m/s^2$ of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is $0.15.$
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 ....... $kg$ $(g = 10\,m/s^2)$