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Two blocks are in contact on a frictionless table. One has mass m and another mass '2m'. A force F is applied on '2m' as shown. Now the force F is applied on 'm' from right. In the two cases respectively the ratio of force of contact between the blocks is

a) same b) 2 :1 c) 1:2 d) 1:3

case I
$a=\large\frac{F}{3a}$
$N_1=ma$
$\qquad=m \times \large\frac{F}{3a}=\large\frac{F}{3}$
case II
$a=\large\frac{F}{3a}$
$N_2=2m \times a$
$\qquad=2m \times \large\frac{F}{3a}=\large\frac{2}{3}$$F$
$\large\frac{N_1}{N_2}=\frac{1}{2}$
Hence c is the correct answer.

edited Feb 10, 2014 by meena.p