Answer:
The magnitude of the resultant is 30.4 N.
The resultant angle direction is 25.3°.
Explanation:
To find the resultant of the magnitude and direction for given forces “P” and “Q” are 20 N and 15 N respectively, the angle (θ) between them is 60°.
We know that from triangle law of forces,
[tex]R=\sqrt{P^{2}+2 P Q \cos \theta+Q^{2}}[/tex]
Substitute the given values in the above formula,
[tex]R=\sqrt{20^{2}+2 (20)(15) Q \cos 60+15^{2}}[/tex]
[tex]R=\sqrt{400+600(0.5) + 225}[/tex]
[tex]R=\sqrt{400+300 + 225}[/tex]
[tex]R=\sqrt{925}[/tex]
R = 30.4 N
The magnitude of the resultant is 30.4 N.
To find the direction of the resultant we know that [tex]\text {Resultant angle}=\tan ^{-1} \frac{Q \sin \theta}{P+Q \cos \theta}[/tex]
Substitute the given values in the above formula,
[tex]\text {Resultant angle}=\tan ^{-1} \frac{15 \sin 60}{20+15 \cos 60}[/tex]
[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{20+7.5}[/tex]
[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{27.5}[/tex]
[tex]\text { Resultant angle }=\tan ^{-1} 0.472[/tex]
Resultant angle=25.3°
The resultant angle direction is 25.3°.
