Chapter 3, Problem 42SP

Two forces act on a point object as follows: 100 N at $170.0°$ and 100 N at $50.0°$.

Find their resultant.

To determine

The resultant force due to the two forcesacting on an object, which are as follows: $100\text{ N}$ at $170°$ and $100\text{ N}$ at $50°$.

Answer to Problem 42SP

Solution:

$100\text{ N}$, $110°$

Explanation of Solution

Given data:

Two forcesacting on a point are as follows:

$100\text{ N}$ at $170°$

$100\text{ N}$ at $50°$

Formula used:

The resultant $\overrightarrow{R}$ has two components, which are $F_x$ and $F_y$, and the expression of the magnitude of the resultant is

$\overrightarrow{R}=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2}$

Here, $F_x$ the force in the x-directionand $F_y$ is the force in the y-direction.

The expression for the direction of the resultant force is

$\begin{array}{c}\tan \theta =\frac{F_y}{F_x}\\ \theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)\end{array}$

Here, $\theta$ is the direction of the resultant forces.

Explanation:

Draw the schematic diagram of the problem:

Find the x- and y-components of each force. These components are as follow:

Force	x-component	y-component
$100\text{ N}$	$\left(100\text{ N}\right)\cos 50°=64.27\text{ N}$	$\left(100\text{ N}\right)\sin 50°=76.60\text{ N}$
$100\text{ N}$	$\left(100\text{ N}\right)\cos 170°=-98.48\text{ N}$	$\left(100\text{ N}\right)\sin 170°=17.36\text{ N}$
Sum	${\displaystyle \sum F_x}=-34.21\text{ N}$	${\displaystyle \sum F_y}=93.96\text{ N}$

Recall the expression of the magnitude of the resultant:

$R=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2}$

Substitute $\text{300 N}$ for $F_x$ and $\text{400 N}$ for $F_y$

$\begin{array}{c}R=\sqrt{{\left(-34.21\text{ N}\right)}^2+{\left(93.96\text{ N}\right)}^2}\\ =\sqrt{\left(1170.32\text{ N}\right)+\left(8828.48\text{ N}\right)}\\ \approx 100\text{ N}\end{array}$

Draw a schematic diagram for the resultant of both forces.

In the above diagram, $\theta$ is the direction of the resultant force from the negative $x$-axis.

Recall the expression of the angle:

$\text{tan}\theta =\frac{F_y}{F_x}$

Rearrange for $\theta$

$\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$

Substitute $93.96\text{ N}$ for $F_y$ and $34.35\text{ N}$ for $F_x$

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{93.96\text{ N}}{34.21\text{ N}}\right)\\ ={\tan }^{-1}\left(2.736\right)\\ \approx 70°\end{array}$

This is the angle of the resultant with the negative x-axis. Therefore, the angle of the resultant with the positive x-axis should be ${180}^{\circ }-{70}^{\circ }={110}^{\circ }$.

Conclusion:

The magnitude of the resultant is $100\text{ N}$ and the angle is $110°$.