Question: Construct the parallelogram law to show $F_R = F_1 + F_2$. Then establish the triangle rule, where $F_R = F_1 + F_2$. Label all known and unknown sides and internal angles. $F_1 = 450$ N $20^\circ$ $F_2 = 100$ N $15^\circ$

1. **State the problem:** We need to construct the parallelogram law to show that the resultant force $F_R$ is the vector sum of $F_1$ and $F_2$, and then establish the triangle rule for the same. 2. **Parallelogram Law:** The parallelogram law states that if two vectors $\vec{F_1}$ and $\vec{F_2}$ are represented by two adjacent sides of a parallelogram, then their resultant $\vec{F_R}$ is represented by the diagonal of the parallelogram passing through the common point. 3. **Triangle Rule:** The triangle rule states that if two vectors are represented by two sides of a triangle taken in order, then their resultant is represented by the third side of the triangle taken in the opposite order. 4. **Given:** - $F_1 = 450$ N at $20^\circ$ above the horizontal to the right. - $F_2 = 100$ N at $15^\circ$ from the vertical to the left (equivalently $75^\circ$ from horizontal to the left). 5. **Calculate the components:** - $F_1$ components: $$F_{1x} = 450 \cos 20^\circ$$ $$F_{1y} = 450 \sin 20^\circ$$ - $F_2$ components: $$F_{2x} = -100 \sin 15^\circ$$ (negative because to the left) $$F_{2y} = 100 \cos 15^\circ$$ 6. **Sum components to find $F_R$ components:** $$F_{Rx} = F_{1x} + F_{2x} = 450 \cos 20^\circ - 100 \sin 15^\circ$$ $$F_{Ry} = F_{1y} + F_{2y} = 450 \sin 20^\circ + 100 \cos 15^\circ$$ 7. **Calculate $F_R$ magnitude:** $$F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2}$$ 8. **Calculate $F_R$ direction angle $\theta$ relative to horizontal:** $$\theta = \tan^{-1} \left( \frac{F_{Ry}}{F_{Rx}} \right)$$ 9. **Interpretation:** The parallelogram formed by $F_1$ and $F_2$ has diagonal $F_R$ which equals the vector sum $F_1 + F_2$. The triangle formed by $F_1$, $F_2$, and $F_R$ also shows the triangle rule. **Final result:** - $F_R = \sqrt{(450 \cos 20^\circ - 100 \sin 15^\circ)^2 + (450 \sin 20^\circ + 100 \cos 15^\circ)^2}$ N - $\theta = \tan^{-1} \left( \frac{450 \sin 20^\circ + 100 \cos 15^\circ}{450 \cos 20^\circ - 100 \sin 15^\circ} \right)$ degrees This confirms $F_R = F_1 + F_2$ by both parallelogram and triangle rules.