Chapter 2, Problem 27P

To determine

Rank the objects according to the magnitude of net force from smallest to largest value.

Answer to Problem 27P

The objects ranked according to the magnitude of net force from smallest to largest value is $F_{\text{net,C}}<F_{\text{net,A}}=F_{\text{net,B}}<F_{\text{net,E}}<F_{\text{net,D}}$.

Explanation of Solution

The figure 1 shows the resultant magnitude of the net force.

Write the expression for the resultant force by using Pythagoras theorem.

  $F_{\text{net,A}}=\sqrt{F_1^2+F_2^2}$        (I)

Here, $F_{\text{net}}$ is the magnitude of net force, $F_1$ and $F_2$ is the force of the objects.

The figure 2 shows the resultant magnitude of the net force.

Write the expression for the resultant force.

  $F_{\text{net,B}}=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$        (II)

The figure 3 shows the resultant force.

Write the expression for the resultant force.

  $F_{\text{net,C}}=F_2-F_1$        (III)

The figure 4 shows the resultant net force.

Write the expression for the resultant force.

  $F_{\text{net,D}}=F_1+F_2$        (IV)

The figure 5 shows the resultant force.

Write the expression for the force along x axis.

  $F_x=F_1+F_2\cos \theta$        (V)

Write the expression for the force along y axis.

  $F_y=F_2\sin \theta$        (VI)

Write the expression for net magnitude of the force.

  $F_{\text{net,E}}=\sqrt{F_x^2+F_y^2}$        (VII)

Conclusion:

Substitute $2000\text{N}$ for $F_1$ and $3000\text{N}$ for $F_2$ in equation (I) to find $F_{\text{net,A}}$.

  $\begin{array}{c}{F}_{\text{net,A}}=\sqrt{{\left(2000\text{N}\right)}^2+{\left(3000\text{N}\right)}^2}\\ =3605\text{N}\end{array}$

Substitute $2000\text{N}$ for $F_1$, $90°$ for $\theta$, and $3000\text{N}$ for $F_2$ in equation (II) to find $F_{\text{net,B}}$.

  $\begin{array}{c}{F}_{\text{net,B}}=\sqrt{{\left(2000\text{N}\right)}^2+{\left(3000\text{N}\right)}^2+2\left(2000\text{N}\right)\left(3000\text{N}\right)\cos \left(90°\right)}\\ =3605\text{N}\end{array}$

Substitute $2000\text{N}$ for $F_1$ and $3000\text{N}$ for $F_2$ in equation (III) to find $F_{\text{net,C}}$.

  $\begin{array}{c}{F}_{\text{net,C}}=3000\text{N}-2000\text{N}\\ \text{=1000}\text{N}\end{array}$

Substitute $2000\text{N}$ for $F_1$ and $3000\text{N}$ for $F_2$ in equation (IV) to find $F_{\text{net,D}}$.

  $\begin{array}{c}{F}_{\text{net,D}}=2000\text{N}+3000\text{N}\\ \text{=5000}\text{N}\end{array}$

Substitute $2000\text{N}$ for $F_1$, $45°$ for $\theta$, and $3000\text{N}$ for $F_2$ in equation (V) to find $F_x$.

  $\begin{array}{c}{F}_x=2000\text{N}+\left(3000\text{N}\right)\cos \left(45°\right)\\ =4121\text{N}\end{array}$

Substitute $2000\text{N}$ for $F_1$, $45°$ for $\theta$, and $3000\text{N}$ for $F_2$ in equation (V) to find $F_x$.

  $\begin{array}{c}{F}_y=\left(3000\text{N}\right)\sin \left(45°\right)\\ =2121\text{N}\end{array}$

Substitute $2121\text{N}$ for $F_y$ and $4121\text{N}$ for $F_x$  in equation (VII) to find $F_{\text{net}}$.

  $\begin{array}{c}{F}_{\text{net,E}}=\sqrt{{\left(2121\text{N}\right)}^2+{\left(4121\text{N}\right)}^2}\\ =4635\text{N}\end{array}$

Therefore, the objects ranked according to the magnitude of net force from smallest to largest value is $F_{\text{net,C}}<F_{\text{net,A}}=F_{\text{net,B}}<F_{\text{net,E}}<F_{\text{net,D}}$.