Suppose you pull a suitcase with a strap that makes a \( 45^{\circ} \) angle with the horizontal. The magnitude of the force you exert on the suitcase is 50 lb . a. Find the horizontal and vertical components of the force. b. Is the horizontal component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \) ? c. Is the vertical component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \) ? A. \( \mathbf{F}_{x}=|\mathbf{F}| \cos \theta \) and \( \mathbf{F}_{y}=|\mathbf{F}| \sin \theta \) B. \( \mathbf{F}_{x}=|\mathbf{F}| \sin \theta \) and \( \mathbf{F}_{y}=|\mathbf{F}| \cos \theta \) C. \( \mathbf{F}_{x}=|\mathbf{F}| \tan \theta \) and \( \mathbf{F}_{y}=|\mathbf{F}| \cot \theta \) D. \( \mathbf{F}_{x}=|\mathbf{F}| \cot \theta \) and \( \mathbf{F}_{y}=|\mathbf{F}| \tan \theta \) The horizontal and vertical components of the force are \( \langle 25 \sqrt{2}, 25 \sqrt{2}\rangle \). (Type exact answers.) b. Is the horizontal component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \) ? Yes N No

Let's work through each part of your problem step by step to ensure your understanding and verify your answers. --- ### **Problem Statement:** Suppose you pull a suitcase with a strap that makes a \( 45^{\circ} \) angle with the horizontal. The magnitude of the force you exert on the suitcase is 50 lb. **a.** Find the horizontal and vertical components of the force. **b.** Is the horizontal component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \)? **c.** Is the vertical component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \)? --- ### **Part A: Finding the Horizontal and Vertical Components** To resolve the force into its horizontal (\( F_x \)) and vertical (\( F_y \)) components, we use trigonometric functions based on the angle \( \theta \) the force makes with the horizontal. The correct formulas are: \[ \mathbf{F}_{x} = |\mathbf{F}| \cos \theta \] \[ \mathbf{F}_{y} = |\mathbf{F}| \sin \theta \] **Explanation of Options:** - **Option A:** \( \mathbf{F}_{x} = |\mathbf{F}| \cos \theta \) and \( \mathbf{F}_{y} = |\mathbf{F}| \sin \theta \) *This is correct.* - **Option B, C, D:** These options incorrectly assign trigonometric functions to the components. **Calculations:** Given: - \( |\mathbf{F}| = 50 \) lb - \( \theta = 45^{\circ} \) \[ F_x = 50 \cos 45^{\circ} = 50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \approx 35.36 \text{ lb} \] \[ F_y = 50 \sin 45^{\circ} = 50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \approx 35.36 \text{ lb} \] **Answer:** \( \mathbf{F}_{x} = 25\sqrt{2} \) lb and \( \mathbf{F}_{y} = 25\sqrt{2} \) lb --- ### **Part B: Comparing Horizontal Components at Different Angles** **Question:** Is the horizontal component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \)? **Your Answer:** *Yes* **Analysis:** Let's calculate the horizontal component for both angles to compare. 1. **At \( 45^{\circ} \):** \[ F_{x_{45}} = 50 \cos 45^{\circ} = 25\sqrt{2} \approx 35.36 \text{ lb} \] 2. **At \( 60^{\circ} \):** \[ F_{x_{60}} = 50 \cos 60^{\circ} = 50 \times 0.5 = 25 \text{ lb} \] **Comparison:** \[ F_{x_{60}} = 25 \text{ lb} \quad \text{vs.} \quad F_{x_{45}} \approx 35.36 \text{ lb} \] **Conclusion:** The horizontal component **decreases** when the angle increases from \( 45^{\circ} \) to \( 60^{\circ} \). **Correct Answer:** **No**, the horizontal component of the force is **not** greater; it actually decreases. --- ### **Part C: Comparing Vertical Components at Different Angles** **Question:** Is the vertical component of the force greater if the angle of the strap is \( 60^{\circ} \) instead of \( 45^{\circ} \)? **Analysis:** Let's calculate the vertical component for both angles. 1. **At \( 45^{\circ} \):** \[ F_{y_{45}} = 50 \sin 45^{\circ} = 25\sqrt{2} \approx 35.36 \text{ lb} \] 2. **At \( 60^{\circ} \):** \[ F_{y_{60}} = 50 \sin 60^{\circ} = 50 \times \frac{\sqrt{3}}{2} \approx 50 \times 0.8660 \approx 43.30 \text{ lb} \] **Comparison:** \[ F_{y_{60}} \approx 43.30 \text{ lb} \quad \text{vs.} \quad F_{y_{45}} \approx 35.36 \text{ lb} \] **Conclusion:** The vertical component **increases** when the angle increases from \( 45^{\circ} \) to \( 60^{\circ} \). **Answer:** **Yes**, the vertical component of the force is greater. --- ### **Summary of Answers:** - **Part A:** \( \mathbf{F}_{x} = 25\sqrt{2} \) lb and \( \mathbf{F}_{y} = 25\sqrt{2} \) lb - **Part B:** **No**, the horizontal component is not greater at \( 60^{\circ} \); it decreases. - **Part C:** **Yes**, the vertical component is greater at \( 60^{\circ} \). --- **Additional Insight:** This analysis highlights how changing the angle of the applied force affects its horizontal and vertical components: - **Increasing the angle** (\( \theta \)) **towards 90°**: - **Reduces** the horizontal component (\( F_x = F \cos \theta \)) - **Increases** the vertical component (\( F_y = F \sin \theta \)) - **Decreasing the angle** (\( \theta \)) **towards 0°**: - **Increases** the horizontal component - **Reduces** the vertical component Understanding this relationship is crucial in scenarios where you need to optimize either the horizontal or vertical force component, such as in ergonomics, engineering applications, and everyday tasks like pulling a suitcase.