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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

In the figure $A=10 kg,B=5 kg$ and coefficient of friction between A and table is $ \mu=0.2$ . find the minimum value of mass C which may stop A from slipping

a) 5 kg b) 10 kg c) 20 kg d) 15 kg

1 Answer

Since A and B are stationary the net force on each is zero.
Let m be mass of C and $m_1$ mass of A
At $A$: $N=(m+m_1)g$
$f=\mu N$
$\quad= \mu _s(m_1+m)g$-----(1)
At $B$ :$ \;T=m_2 g$-----(2)
from (1) and (2)
$m_2 g=\mu_s (m_1+m)g$
$m=\Large\frac{m_2-\mu _s m_1}{\mu _s}$
$\quad=\Large\frac{m_2}{\mu_s}$$-m_1$
$\quad=\large\frac{5}{0.2}$$-10$
$\quad=15 \;kg$
Hence d is the correct answer
answered Jul 10, 2013 by meena.p
edited May 27, 2014 by lmohan717
 

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