91Ó°ÊÓ

The given data can be listed below as:

• The weight of the block A is .$W_1=1.90\text{ N}$
• The weight of the block B is .$W_2=4.20\text{ N}$
• The coefficient of the kinetic friction is .${\mathrm{μ}}_k=0.30$

The friction is described as the force which opposes the motion of a particular object. The friction is also described as the pair of the action and the reaction forces.

The free body diagram of the system has been drawn below:

For the block B, the equation of the horizontal force exerted by the block B is expressed as:

$F=f+T+f_c$ …(1)

Here,$F$is described as the horizontal force,$f$is the upper frictional force,$T$is the tension of the block B and $f_c$is the lower frictional force.

The equation of the upper frictional force is expressed as:

$f={\mathrm{μ}}_k{W}_1$

Here,${\mathrm{μ}}_k$is the coefficient of the kinetic friction and$W_1$is the weight of the block A.

According to the free body diagram of the block A, the tension exerted by the block is the upper frictional force.

The equation of the lower frictional force is expressed as:

$f_c={\mathrm{μ}}_k(W_1+W_2)$

Here,$f_c$is the lower frictional force, ${\mathrm{μ}}_k$is the coefficient of the kinetic friction and$W_2$is the weight of the block B.

Substitute ${\mathrm{μ}}_k(W_1+W_2)$for $f$and $T$and ${\mathrm{μ}}_k{W}_1$for$f_c$in the above equation.

$\begin{array}{c}F={\mathrm{μ}}_k(W_1+W_2)+{\mathrm{μ}}_k(W_1+W_2)+{\mathrm{μ}}_k{W}_1\\ =2{\mathrm{μ}}_k(W_1+W_2)+{\mathrm{μ}}_k{W}_1\\ ={\mathrm{μ}}_k(3W_1+W_2)\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}F=\left(0.30\right)(3\left(1.90\text{ N}\right)+\left(4.20\text{ N}\right))\\ =\left(0.30\right)(\left(5.7\text{ N}\right)+\left(4.20\text{ N}\right))\\ =\left(0.30\right)\left(9.9\text{ N}\right)\\ =2.97\text{ N}\\ ≈\text{3 N}\end{array}$

Thus, the magnitude of the horizontal force $\stackrel{→}{F}$is $3\text{ N}$.