Question

The resultant of two forces of 628 grams and 532 grams is a force of 718 grams. Find the angle which the resultant makes with the forces.

Answer 1

Answer on Question #63908, Physics / Mechanics | Relativity

**Problem:** The resultant of two forces of 628 grams and 532 grams is a force of 718 grams. Find the angle which the resultant makes with the forces.

**Solution:** Let us denote

$$\left| \overrightarrow {F _ {1}} \right| = 5 3 2 g, \left| \overrightarrow {F _ {2}} \right| = 6 2 8 g, \left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| = 7 1 8 g$$

Angle between $\overrightarrow{F_2}$ and $\overrightarrow{F_1} + \overrightarrow{F_2}$ $\alpha$, angle between $\overrightarrow{F_1}$ and $\overrightarrow{F_1} + \overrightarrow{F_2}$ $\beta$.

Using cosine rule for triangles $\left(\overrightarrow{F_1}; \overrightarrow{F_2}; \overrightarrow{F_1} + \overrightarrow{F_2}\right)$ and $\left(\overrightarrow{F_2}; \overrightarrow{F_2} + \overrightarrow{F_1}; \overrightarrow{F_1}\right)$

$$\cos \alpha = \frac {\left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| ^ {2} + \left| \overrightarrow {F _ {2}} \right| ^ {2} - \left| \overrightarrow {F _ {1}} \right| ^ {2}}{2 \cdot \left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| \cdot \left| \overrightarrow {F _ {2}} \right|}$$ $$\cos \beta = \frac {\left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| ^ {2} + \left| \overrightarrow {F _ {1}} \right| ^ {2} - \left| \overrightarrow {F _ {2}} \right| ^ {2}}{2 \cdot \left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| \cdot \left| \overrightarrow {F _ {1}} \right|}$$

Derive final formula for angles:

$$\alpha = \arccos \left(\frac {\left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| ^ {2} + \left| \overrightarrow {F _ {2}} \right| ^ {2} - \left| \overrightarrow {F _ {1}} \right| ^ {2}}{2 \cdot \left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| \cdot \left| \overrightarrow {F _ {2}} \right|}\right) = \arccos \left(\frac {7 1 8 ^ {2} + 6 2 8 ^ {2} - 5 3 2 ^ {2}}{2 \cdot 7 1 8 \cdot 6 2 8}\right) = \arccos 0. 6 9 5 1 = 4 6 {}^ {\circ}$$ $$\beta = \arccos \left(\frac {\left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| ^ {2} + \left| \overrightarrow {F _ {1}} \right| ^ {2} - \left| \overrightarrow {F _ {2}} \right| ^ {2}}{2 \cdot \left| \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}} \right| \cdot \left| \overrightarrow {F _ {1}} \right|}\right) = \arccos \left(\frac {7 1 8 ^ {2} + 5 3 2 ^ {2} - 6 2 8 ^ {2}}{2 \cdot 7 1 8 \cdot 5 3 2}\right) = \arccos 0. 5 2 9 = 5 8 {}^ {\circ}$$

**Answer:** Angle between $\overrightarrow{F_2}$ and $\overrightarrow{F_1} + \overrightarrow{F_2}$ $\alpha = 46{}^{\circ}$, angle between $\overrightarrow{F_1}$ and $\overrightarrow{F_1} + \overrightarrow{F_2}$ $\beta = 58{}^{\circ}$

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