Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

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
Can you answer this question?

1 Answer

0 votes
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)
$m_2 g=\mu_s (m_1+m)g$
$m=\Large\frac{m_2-\mu _s m_1}{\mu _s}$
$\quad=15 \;kg$
Hence d is the correct answer
answered Jul 10, 2013 by meena.p
edited May 27, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App