Question

Two forces have magnitudes of 10N and 15N and the angle between them is 45°. Find the magnitude of the resultant and the angle it makes with the smaller of the two forces.

Answer 1

Answer on Question #51335 – Math – Vector Calculus

Given:

$$\angle (F_1, F_2) = 45{}^\circ$$ $$\bar{F}_1 = 10\,N$$ $$\bar{F}_2 = 15\,N$$

Find:

$$\bar{F}_{12}, \alpha$$

Solution:

1) $\bar{F}_{12} = \bar{F}_1 + \bar{F}_2$

Using the law of cosines, we obtain

$$F_{12}^2 = F_1^2 + F_2^2 - 2F_1F_2\cos(135{}^\circ) = 100 + 225 - 300\cos(90{}^\circ + 45{}^\circ) = 325 + 300\sin(45{}^\circ) = 325 + 300 \cdot \frac{\sqrt{2}}{2} \approx 537$$ $$F_{12} = \sqrt{537} \approx 23.18\,N$$

2) $\cos\alpha = \frac{F_1^2 + F_{12}^2 - F_2^2}{2F_1F_{12}} = \frac{100 + 537 - 225}{2 \cdot 10 \cdot 23.18} = \frac{412}{463.6} \approx 0.889$

$$\alpha = \arccos\left(\frac{412}{463.6}\right) \approx \arccos(0.889) \approx 27{}^\circ,$$

where $\arccos(t)$ is the inverse of cosine function $\cos(t)$.

Answer: $F_{12} = 23.18\,N$

$$\alpha = 27{}^\circ$$

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