The coefficient of static friction, $\mu_s$, between block A of mass 2 kg and the table as shown in the figure

Question

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 do not move? The string and the pulley are assumed to be smooth and massless. ($g = 10 	ext{ m/s}^2$)

Accepted Answer

1. **System Analysis:** Block A is on a horizontal surface, held in place by the static friction. Block B is hanging and provides the tension in the string trying to pull Block A. For the blocks not to move, the system must be in equilibrium.
2. **Forces on Block B:** The tension $T$ in the string supports the weight of Block B. 
   $$ T = m_B g $$
3. **Forces on Block A:** The tension $T$ pulls Block A horizontally. This is opposed by the static friction force $f_s$. The normal reaction on A is $N = m_A g$. 
   For equilibrium: $T = f_s$
4. **Limiting Condition:** The maximum mass of B corresponds to the maximum tension Block A can withstand without sliding. This occurs when static friction reaches its limiting value:
   $$ (f_s)_{max} = \mu_s N = \mu_s m_A g $$
   [Source 56, 63].
5. **Calculation:** Equating tension to limiting friction:
   $$ m_B g = \mu_s m_A g $$
   $$ m_B = \mu_s m_A $$
   $$ m_B = 0.2 	imes 2 	ext{ kg} = 0.4 	ext{ kg} $$

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