Question 246: Laws of Motion

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 m/s
2
)

Accepted Answer

Free body diagram of two masses is
We get equations
$$T + ma = {f_\mu }$$ or $$T = \mu {N_A}(for\,a = 0)$$
and T = ma + mg or T = m
B
g (for a = 0)
$$ 	herefore $$ $$\mu {N_A} = {m_B}g$$
$$ \Rightarrow {m_B} = \mu {m_A} = 0.2 	imes 2 = 0.4\,kg$$

Answer

2.0 kg

Answer

4.0 kg

Answer

0.2 kg

Answer

0.4 kg

Laws of Motion

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 m/s 2 )

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