91Ó°ÊÓ

The given data can be listed below as:

• The mass of the packing case is m = 40.0kg.
• The mass of the pickup truck is M = 1500kg.
• The static friction coefficient is ${\mathrm{μ}}_s=0.30$.
• The kinetic friction coefficient is ${\mathrm{μ}}_k=0.20$.
• The acceleration of the truck in the northward direction is $a_1=2.20{\text{m/s}}^{\text{2}}$.
• The acceleration of the truck in the southward direction is $a_2=3.40{\text{m/s}}^{\text{2}}$.

The friction force mainly resists the motion of two surfaces when they come in contact with each other. The frictional force is described as the product between normal force and the coefficient of friction.

The free body diagram of the truck is drawn below:

In the above diagram, the normal force N is acting in the upward direction and the weight of the packing case that is the product of the mass of the case and acceleration due to gravity mg is acting in the downwards direction. The frictional force$f_k$is acting in the north direction and the reaction force$F_{\text{reac}}$is acting in the opposite direction.

The summation of the forces in the x and the y direction is zero according to the free body diagram.

The equation of the summation of the forces in the x direction is expressed as:

$$\begin{gathered} N-mg=0 \\ N=mg \end{gathered}$$

Here, N is the normal force, m is the mass of the packing case and g is the acceleration due to gravity.

Here, the static frictional force has been applied as the packing box is at rest as the reaction force is higher than the frictional force.

The equation of the friction force is expressed as:

$f_k={\mathrm{μ}}_{s}N$

Here,$f_k$is the friction force and${\mathrm{μ}}_s$is the static friction.

Substitute mg for N in the above equation.

$f_k={\mathrm{μ}}_{s}mg$

Substitute the values in the above equation.

$\begin{array}{rcl}{f}_k& =& 0.3\left(40\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ & =& \left(12\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ & =& 117.6\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 117.6\text{N}\end{array}$

The equation of the force exerted on the truck is expressed as:

$F=m{a}_1$

Here, F is the force exerted on the truck, m is the mass of the packing case and$a_1$ is the acceleration of the truck in the northward direction.

Substitute the values in the above equation.

$\begin{array}{rcl}F& =& 40.0\text{kg}×2.20{\text{m/s}}^{\text{2}}\\ & =& 88\text{kg}·{\text{m/s}}^{\text{2}}\\ & =& 88\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 88\text{N}\end{array}$

The force exerted on the truck is equal to the frictional force of the truck.

Thus, the friction force acting on the case is 88N and it is directed in the northward direction.

The free body diagram of the truck is drawn below:

In the above diagram, the normal force N is acting in the upward direction and the weight of the packing case that is the product of the mass of the case and acceleration due to gravity mg is acting in the downwards direction. The frictional force $f_k$is acting in the south direction and the reaction force $F_{\text{reac}}$is acting in the opposite direction.

The summation of the forces in the x and the y direction is zero according to the free body diagram.

The equation of the summation of the forces in the x direction is expressed as:

$$\begin{gathered} N-mg=0 \\ N=mg \end{gathered}$$

Here, N is the normal force, m is the mass of the packing case and g is the acceleration due to gravity.

Here, the kinetic frictional force has been applied as the packing box is at motion as the reaction force is lesser than the frictional force.

The equation of the friction force is expressed as:

$f_k={\mathrm{μ}}_{k}N$

Here,$f_k$ is the friction force and${\mathrm{μ}}_k$ is the kinetic friction.

Substitute mg for N in the above equation.

$f_k={\mathrm{μ}}_{k}mg$

Substitute the values in the above equation.

$\begin{array}{rcl}{f}_k& =& 0.2\left(40\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ & =& \left(8\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ & =& 78.4\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 78.4\text{N}\end{array}$

Thus, the friction force acting on the case is 78.4N and it is directed in the southward direction.