Question

Find the friction force between the blocks in the figure:-

• A. $6N$
• B. $17.6N$
• C. $5.2N$
• D. $12N$

Answer 1

The correct option is B $17.6N$

$\sum F_y=m{a}_y$

The block does not move in the $y-$ direction$,$ therefore $a_y=0$

$\sum F_y=0$

$F_N-19.6=0$

$0r{F}_N=19.6N$

$\sum F_x=m{a}_x$

$\mu F_N=m{a}_x$

$0.3\left(19.6N\right)=2kg\left(a_z\right)$

$a_z=2.94\frac{m}{s^2}$

The upper block will slip when the acceleration exceeds $2.94\frac{m}{s^2}$

Now consider both blocks connected together$.$

$\sum F_x=\left(m_{total}\right)a_x$

$F=\left(2kg+4kg\right)2.94\frac{m}{s^2}=17.6N$

Hence,

option $\left(B\right)$ is correct answer.