The coefficient of static friction between the block of 2 kg and the table shown in figurwe is `mu_s=0.2` What should be the maximum value of m so tht the blocks do not move? Take `g=10 m/s^2.` The string and the pulley are light and smooth.

Explanation

Consider the equilibrium of the block of mass m. The forces on this block are
a. mg downward by the earth and
T upward by the string.
hence `T-mg=0 or, T=mg`…….i
Now consider the equilibrium of the 2 kg block. The forces on this block are
a. T towards rightby the string,
b. f towards left (friction)by the table,
c. 20 N downward (weight) by the earth and
d. N upward (normal force) by the table
For vertical equilibrium of this block,
N=20N...........ii
A m is the largest mass which can be used without moving the system, the friction is limiting
Thus, `f=mu_sN` ...........iii
For horizontal equilibrium of the 2 kg block
f=T
Using equations i, iii and iv
`mu_sN=mg`
or `02xx20N=mg`
or, `m=(02xx20)/10kg=04kg`