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Give step-by-step solution with explanation and final answer:v PartA In (Figure 1), 8= 44° Determine the coordinate direction angles of the force. Express your answers in degrees to tree significant figures and separated by commas. se om a B= | B Figure Tort z [ | Ferm

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Give step-by-step solution with explanation and final answer:Uploaded Imagev PartA In (Figure 1), 8= 44° Determine the coordinate direction angles of the force. Express your answers in degrees to tree significant figures and separated by commas. se om a B= | B Figure Tort z [ | Ferm

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Asked by: Rahul
Created at: 2026-02-18 04:18:47

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Let's break down the problem step by step: ## **Step 1: Write the Force Components** Given: - Magnitude of force \( F = 75 \) lb - \( \theta = 44^\circ \) (angle with the \( z \)-axis) - The projection of \( F \) on the \( xy \)-plane makes \( 45^\circ \) with the \( x \)-axis. Let the coordinate direction angles be \( \alpha \) (with \( x \)), \( \beta \) (with \( y \)), and \( \gamma \) (with \( z \)). ### **Find the components:** - The component along \( z \): \( F_z = F \cos \theta \) - The projection on \( xy \)-plane: \( F_{xy} = F \sin \theta \) - This projection makes \( 45^\circ \) with the \( x \)-axis: - \( F_x = F_{xy} \cos 45^\circ \) - \( F_y = F_{xy} \sin 45^\circ \) Plug in the values: - \( \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \) - \( F_{xy} = 75 \sin 44^\circ \) So: \[ F_x = 75 \sin 44^\circ \cos 45^\circ \] \[ F_y = 75 \sin 44^\circ \sin 45^\circ \] \[ F_z = 75 \cos 44^\circ \] ## **Step 2: Calculate the Components** \[ \sin 44^\circ \approx .6947 \] \[ \cos 44^\circ \approx .7193 \] \[ \cos 45^\circ = \sin 45^\circ = .7071 \] \[ F_{xy} = 75 \times .6947 = 52.1025 \] \[ F_x = 52.1025 \times .7071 \approx 36.836 \] \[ F_y = 52.1025 \times .7071 \approx 36.836 \] \[ F_z = 75 \times .7193 \approx 53.4475 \] ## **Step 3: Find Direction Cosines** Direction cosines: \[ \cos \alpha = \frac{F_x}{F} \] \[ \cos \beta = \frac{F_y}{F} \] \[ \cos \gamma = \frac{F_z}{F} \] \[ \cos \alpha = \frac{36.836}{75} = .4911 \] \[ \cos \beta = \frac{36.836}{75} = .4911 \] \[ \cos \gamma = \frac{53.4475}{75} = .7126 \] ## **Step 4: Find the Angles** \[ \alpha = \cos^{-1}(.4911) \approx 60.6^\circ \] \[ \beta = \cos^{-1}(.4911) \approx 60.6^\circ \] \[ \gamma = \cos^{-1}(.7126) \approx 44.6^\circ \] ## **Step 5: Final Answers (to three significant figures)** \[ \boxed{60.6^\circ,\ 60.6^\circ,\ 44.6^\circ} \] --- ### **Summary Table** | Angle | Value (degrees) | |--------|----------------| | \( \alpha \) | 60.6 | | \( \beta \) | 60.6 | | \( \gamma \) | 44.6 | --- **Final Answer:** \[ \boxed{60.6^\circ,\ 60.6^\circ,\ 44.6^\circ} \]

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