1. **Problem Statement:** Find the magnitude and direction of the resultant force $\mathbf{R}$ from two forces $\mathbf{P}$ and $\mathbf{Q}$ given their magnitudes and angles. 2. **Given:** - $P = 450$ N, angles: $40^\circ$ with $y$-axis, $25^\circ$ with $z$-axis - $Q = 600$ N, angles: $55^\circ$ with $y$-axis, $30^\circ$ with $x$-axis 3. **Force Components:** - $\mathbf{P} = 450[\sin 40^\circ \sin 25^\circ \mathbf{i} + \cos 40^\circ \mathbf{j} + \sin 40^\circ \cos 25^\circ \mathbf{k}]$ - Calculate each component: $$P_x = 450 \times \sin 40^\circ \times \sin 25^\circ = 122.244$$ $$P_y = 450 \times \cos 40^\circ = 344.72$$ $$P_z = 450 \times \sin 40^\circ \times \cos 25^\circ = 262.154$$ - $\mathbf{Q} = 600[\cos 55^\circ \cos 30^\circ \mathbf{i} + \sin 55^\circ \mathbf{j} - \cos 55^\circ \sin 30^\circ \mathbf{k}]$ - Calculate each component: $$Q_x = 600 \times \cos 55^\circ \times \cos 30^\circ = 298.04$$ $$Q_y = 600 \times \sin 55^\circ = 491.49$$ $$Q_z = -600 \times \cos 55^\circ \times \sin 30^\circ = -172.073$$ 4. **Resultant Vector Components:** $$R_x = P_x + Q_x = 122.244 + 298.04 = 420.28$$ $$R_y = P_y + Q_y = 344.72 + 491.49 = 836.21$$ $$R_z = P_z + Q_z = 262.154 - 172.073 = 90.081$$ 5. **Magnitude of Resultant:** $$R = \sqrt{420.28^2 + 836.21^2 + 90.081^2} = 940.21$$ 6. **Direction Angles:** - Using $\cos \theta_x = \frac{R_x}{R}$: $$\theta_x = \cos^{-1} \left( \frac{420.28}{940.21} \right) = 63.4^\circ$$ - Using $\cos \theta_y = \frac{R_y}{R}$: $$\theta_y = \cos^{-1} \left( \frac{836.21}{940.21} \right) = 27.2^\circ$$ - Using $\cos \theta_z = \frac{R_z}{R}$: $$\theta_z = \cos^{-1} \left( \frac{90.081}{940.21} \right) = 84.5^\circ$$ **Final Answer:** - Magnitude of resultant force $R = 940$ N - Direction angles: $\theta_x = 63.4^\circ$, $\theta_y = 27.2^\circ$, $\theta_z = 84.5^\circ$