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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A system consists of three blocks A,B and C arranged as shown. The system is pushed by a force as shown in figure. All surfaces are smooth except between B and C,where coefficient of friction is $\mu$. Minimum value of F to prevent block B from slipping downward is

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Horizontal acceleration of the system is $a=\large\frac{F}{m+2m+2m}$
$\qquad=\large\frac{F}{5m}$
Normal force between B and C is N
is given by $N=2m \times a$
$\qquad=2m \times \large\frac{F}{5m}$
$\qquad=\large\frac{2}{5}$$F$
B will not slide down if frictional force is more that the weight mg of block B
$\mu \geq mg$
$\mu \large\frac{2}{5} $$F \geq mg$
$F \geq \large\frac{5}{2 \mu}$$mg$
Hence b is the correct answer.

 

answered Jul 10, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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