Answer:
a. Bearing: 343.4°
b. 5.09 N
Step-by-step explanation:
You want the bearing of a 3.5 N force and the magnitude of the resultant when the 3.5 N force is added to a 2.0 N force on a bearing of 30° to produce a resultant acting northerly.
Law of sines
Referring to the attachment, we can draw triangle OAB with OA = 2.0, AB = 3.5, and angle O = 30°. Angle B can be found from the law of sines as ...
[tex]\dfrac{\sin(B)}{OA}=\dfrac{\sin(O)}{AB}\\\\\\B=\sin^{-1}\left(\dfrac{OA}{AB}\sin(O)\right)=\sin^{-1}\left(\dfrac{2}{3.5}\sin(30^\circ)\right)\approx16.6^\circ[/tex]
The direction of the 3.5 N force applied to the object is 360° less this angle:
θ = 360° -16.6° = 343.4° . . . . . the bearing of the 3.5 N force
The magnitude of the resultant can also be found using the law of sines.
[tex]\dfrac{OB}{\sin(A)}=\dfrac{AB}{\sin(O)}\\\\\\OB=AB\cdot\dfrac{\sin(A)}{\sin(O)}=3.5\cdot\dfrac{\sin(180^\circ-30^\circ-16.6^\circ)}{\sin(30^\circ)}\approx 5.09[/tex]
The magnitude of the resultant force is about 5.09 N.
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Additional comment
Since sin(θ) = sin(180°-θ), we could use sin(30°+16.6°) in the numerator of that last fraction.
Finding angle B is essentially balancing easterly components of the vectors so they total 0. The magnitude of OB is essentially OB = OAcos(30°) +ABcos(16.6°), the sum of the northerly components of the vectors.
