Question

Two forces have magnitudes of 8N and 12 N and the angle between them is 130°. Find the magnitude of the resultant and the angle it makes with the larger of the two forces.

Answer 1

Answer on Question #51334 – Math – Vector Calculus

Two forces have magnitudes of 8N and 12 N and the angle between them is $130{}^{\circ}$. Find the magnitude of the resultant and the angle it makes with the larger of the two forces.

Figure

Solution

Step 1.

Let's use Parallelogram Law of Vector Addition: "If two vector quantities are represented by two adjacent sides of a parallelogram then the diagonal of the parallelogram will be equal to the resultant of these two vectors."

Let's denote $\angle BOA = 130{}^{\circ}$, $\left|\overrightarrow{OB}\right| = b = 12N$, $\left|\overrightarrow{OA}\right| = a = 8N$, $\left|\overrightarrow{OC}\right| = c$, $\angle BOC = \beta$.

Resulting force is $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{OA}$, OBCA is a parallelogram, $\angle OBC = \angle OAC = 180{}^{\circ} - 130{}^{\circ} = 50{}^{\circ}$, Let's consider triangle $OBC$. We find the magnitude of the resultant according to the Law of Cosines.

$$c^{2} = b^{2} + a^{2} - 2ab\cos\angle OBC = 12^{2} + 8^{2} - 2\cdot12\cdot8\cos 50{}^{\circ} = 208 - 192\cos \frac{50\pi}{180} = 208 - 192\cdot0.64 = 84.58$$ $$c = \sqrt{c^{2}} = \sqrt{84.58} = 9,2 \Rightarrow |\overrightarrow{OC}| = 9,2N$$

Step 2.

According to the Law of Sines, $\frac{BC}{\sin\angle BOC} = \frac{OC}{\sin\angle OBC}$, that is,

$$\frac{a}{\sin\beta} = \frac{c}{\sin 50{}^{\circ}} \Rightarrow \sin\beta = \frac{a}{c} \sin 50{}^{\circ} = \frac{8}{9,2} \sin 50{}^{\circ} = 0,67 \Rightarrow \beta = 41,8{}^{\circ}.$$

Answer: $|\overrightarrow{OC}| = 9,2N$, $\beta = 41,8{}^{\circ}$.

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