Question

Find the ratio of the force of interaction between the $4kg$ and $2kg$ blocks and $2kg$ and $3kg$ blocks. (Assume all surfaces to be smooth)

• A. $\frac{1}{3}$
• B. $\frac{5}{3}$
• C. $2$
• D. $\frac{7}{3}$

Answer 1

The correct option is B $\frac{5}{3}$

From Newton's second law,
$a=\frac{F}{m}=\frac{18}{4+2+3}=2m/s^2$

FBD of $4kg$ block,

we can write,
$18-N_1=ma=4\times 2⟹N_1=10N$
FBD of $3kg$ block,

$N_2=ma=3\times 2=6N$
$N_2$ is the force of interaction between $2kg$ and $3kg$ blocks.
$\therefore \frac{N_1}{N_2}=\frac{10}{6}=\frac{5}{3}$