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)

Question

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)  (A) 2.0 kg (B) 4.0 kg (C) 0.2 kg (D) 0.4 kg

Tardigrade Admin · Accepted Answer

Free body diagram of two masses is

We get equations T+m a=fμ or T=μ NA( for a=0) and T=m a+m g or T=mB g (for a=0 ) ∴ μ NA=mB g ⇒ mB=μ mA =0.2 × 2=0.4 kg

Q. The coefficient of static friction, μs​, between block A of mass 2kg 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=10m/s2)

2.0 kg

4.0 kg

0.2 kg

0.4 kg

Free body diagram of two masses is We get equations T+ma=fμ​ or T=μNA​( for a=0) and T=ma+mg or T=mB​g (for a=0 ) ∴μNA​=mB​g ⇒mB​=μmA​ =0.2×2=0.4kg

Free body diagram of two masses is

We get equations
$T + ma = f_{μ}$
or $T = μ N_{A} ($ for $a = 0)$
and $T = ma + m g$
or $T = m_{B} g$ (for $a = 0$ )
$∴ μ N_{A} = m_{B} g$
$\Rightarrow m_{B} = μ m_{A}$
$= 0.2 \times 2 = 0.4 k g$