Question

The coefficient of static friction between the block of 2 kg and the table shown in figure is $\mu$s = 0.2. what should be the maximum value of m so that the blocks do not move? Take g = 10 $m/s^2$. The string and the pulley are light and smooth.

• A. 2 kg
• B. 1 kg
• C. 0.4 kg
• D. 0.5 kg

Answer 1

The correct option is C

0.4 kg

Consider the equilibrium of the block of mass m. The forces on this block are

(a) mg downward by the earth and (b) T upward by the string.

Hence, T-mg=0 or, T=mg. ...............(i)

Now consider the equillibrium of the 2kg block.This forces on this block are

(a) T towards right by 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 = 20 N. . . . . . . (II)

As m is the largest mass which can be used without moving the system, the friction is limiting.

Thus, f = ${\mu }_s$ N. . . . . . . (III)

For horizontal equilibrium of the 2 kg block,

f = T . . . . . . (IV)

Using equations (I), (III) and (IV)

$\mu$s N = mg

Or, 0.2 $\times$ 20 N = mg

Or, m=$\frac{0.2\times 20}{10}kg=0.4kg.$