Two forces

Two forces with magnitudes of 25 and 30 pounds act on an object at 10° and 100° angles. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and your final answer.

Correct answer:

Φ =  60.19 °
F =  39.05

Step-by-step explanation:

$$\begin{gathered} F_1=25\text{lb} \\ {F}_2=30\text{lb} \\ \alpha =10°\to \text{rad}=10°\cdot \frac{\pi }{180}=10°\cdot \frac{3.1415926}{180}=0.17453=\pi /18 \\ \beta =100°\to \text{rad}=100°\cdot \frac{\pi }{180}=100°\cdot \frac{3.1415926}{180}=1.74533=5\pi /9 \\ \gamma =\beta -\alpha =\pi /2=90° \\ {F}_1\perp F_2 \\ F=F_1+F_2=(x;y) \\ x=F_1\cdot \cos \alpha +F_2\cdot \cos \beta =25\cdot \cos 0.1745+30\cdot \cos 1.7453\doteq 19.4107 \\ y=F_1\cdot \sin \alpha +F_2\cdot \sin \beta =25\cdot \sin 0.1745+30\cdot \sin 1.7453\doteq 33.8854 \\ {\Phi }_1=\arctan (\frac{y}{x})=\arctan (\frac{33.8854}{19.4107})\doteq 1.0506\text{rad} \\ \Phi ={\Phi }_1\to \text{°}={\Phi }_1\cdot \frac{180}{\pi }\text{°}=1.0506\cdot \frac{180}{\pi }\text{°}=60.19\text{°}=60.19^{\circ }=60°11^{\prime }40" \end{gathered}$$

$$\begin{gathered} F^2=x^2+y^2 \\ F=\sqrt{x^2+y^2}=\sqrt{19.4107^2+33.8854^2}=5\sqrt{61}\doteq 39.0512 \\ \text{ Verifying Solution: } \\ {F}_2=\sqrt{F_1^2+F_2^2}=\sqrt{25^2+39.0512^2}=5\sqrt{61}\doteq 39.0512 \end{gathered}$$

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