Question

Two forces with magnitudes of 150 and 100 pounds act on an object at angles of 30° and 120°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Answer 1

Answer:

Magnitude:

$|R|=180.27756$

Direction:

$\theta=63.69^{\circ \:}$

Step-by-step explanation:

Calculation of first force:

we are given

$|F_1|=150$

$\theta=30$

now, we can find components

$x=|F_1|cos(\theta)$

$x=150cos(30)$

$x=75\sqrt{3}$

$y=|F_1|sin(\theta)$

$y=150sin(30)$

$y=75$

now, we can find force

$F_1=(75\sqrt{3},75)$

Calculation of Second force:

we are given

$|F_2|=100$

$\theta=120$

now, we can find components

$x=|F_2|cos(\theta)$

$x=100cos(120)$

$x=-50$

$y=|F_2|sin(\theta)$

$y=100sin(120)$

$y=50\sqrt{3}$

now, we can find force

$F_2=(-50,50\sqrt{3})$

now, we can find resultant force

$R=F_1+F_2$

$R=(75\sqrt{3},75)+(-50,50\sqrt{3})$

$R=(75\sqrt{3}-50,75+50\sqrt{3})$

$R=(79.90381 ,161.60254)$

Magnitude of Resultant:

$|R|=\sqrt{79.90381^2+161.60254^2}$

$|R|=180.27756$

Direction of Resultant:

$\theta=tan^{-1}(\frac{161.60254}{79.90381})$

$\theta=63.69^{\circ \:}$

Answer 2

Answer:

$Magnitude=180.27 \ lbf \\ \\ Direction=63.69 \ degrees$

Step-by-step explanation:

We have two forces as follows:

First force:

Magnitude: 150 pounds

Angle: 30°

First force:

Magnitude: 100 pounds

Angle: 120°

So the components can be found as follows:

$F_{1x}=150cos(30)=129.90 \ lbf\\F_{1y}=150sin(30) = 75 \ lbf \\ \\ F_{2x}=100cos(120)=-50 \ lbf\\F_{2y}=100sin(120) = 86.60 \ lbf$

So the components of the resultant force can be found by adding each component of the individual forces as follows:

$R_{x}=\Sigma F_{x} \\ R_{y}=\Sigma F_{y} \\ \\ R_{x}=129.90-50=79.90 \ lbf \\ R_{y}=75+86.60=161.6 \ lbf$

Finally, the magnitude and direction of the resultant force is:

$Magnitude \rightarrow R=\sqrt{R_{x}^2+R_{y}^2}=\sqrt{79.90^2+161.6^2}=180.27 \ lbf \\ \\ Direction \rightarrow \theta=tan^{-1}(\frac{R_{y}}{R_{x}})=tan^{-1}(\frac{161.6}{79.90})=63.69 \ degrees$