Find the minimum value of mass $m$ that should be placed on the $20 \, kg$ block so that the system remains at rest. (Given coefficient of friction $\mu = 0.25$) (in $, kg$)

$A$ block of mass $50 \, kg$ can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is $0.6$. The least force of pull acting at an angle of $30^\circ$ to the upward drawn vertical which causes the block to just slide is ........ $N$.

Why is the coefficient of static friction considered a property of the surfaces in contact?

$A$ uniform chain of $6 \, m$ length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is $0.5$. The maximum length of the chain hanging from the table is.......$m.$

$A$ block of mass $1 \, kg$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.5$. The magnitude of the force acting upwards at an angle of $60^{\circ}$ from the horizontal that will just start the block moving is

Block $B$ lying on a table weighs $W$. The coefficient of static friction between the block and the table is $\mu$. Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is