Let's break down the problem step by step. --- ## **Problem Recap** Four forces are applied to the eyebolt as shown. If the resultant force on the bolt is directed out of the page (along the *z*-direction) and has a magnitude of **670 lb**, determine the values of **F₃** and **α**. Given: - \( F_1 = 400 \) lb - \( F_2 = 325 \) lb - \( F_4 = 350 \) lb - \( F_3 \) and \( \alpha \) are unknown - Resultant force \( R = 670 \) lb (in the *z*-direction) --- ## **Step 1: Resolve Each Force into Components** Let’s analyze the force directions from the diagram: - **\( F_1 \):** Along the *y*-axis (upward) - **\( F_2 \):** Along the *x*-axis (right) - **\( F_3 \):** At angle \( \alpha \) from the *x*-axis, in the *xy*-plane - **\( F_4 \):** At 45° to both axes, negative *x* and negative *y* direction Let’s resolve each force into x and y components: ### **\( F_1 \)** - \( F_{1x} = 0 \) - \( F_{1y} = 400 \) lb ### **\( F_2 \)** - \( F_{2x} = 325 \) lb - \( F_{2y} = 0 \) ### **\( F_3 \)** - \( F_{3x} = F_3 \cos \alpha \) - \( F_{3y} = F_3 \sin \alpha \) ### **\( F_4 \)** - \( F_{4x} = -350 \cos 45^\circ = -350 \frac{\sqrt{2}}{2} = -247.5 \) lb - \( F_{4y} = -350 \sin 45^\circ = -350 \frac{\sqrt{2}}{2} = -247.5 \) lb --- ## **Step 2: Resultant Force Condition** The resultant is **directed out of the page**, which means the **x and y components must sum to zero**. ### **Set up Equations** #### **Sum of x-components:** \[ F_{2x} + F_{3x} + F_{4x} = 0 \] \[ 325 + F_3 \cos \alpha - 247.5 = 0 \] \[ F_3 \cos \alpha = -325 + 247.5 = -77.5 \] #### **Sum of y-components:** \[ F_{1y} + F_{3y} + F_{4y} = 0 \] \[ 400 + F_3 \sin \alpha - 247.5 = 0 \] \[ F_3 \sin \alpha = -400 + 247.5 = -152.5 \] --- ## **Step 3: Resultant Magnitude (z-component)** The magnitude of the resultant is the sum of the **z-components** of all forces (assuming each force acts in the xy-plane except the resultant). \[ R = \sqrt{(F_{Rx})^2 + (F_{Ry})^2} \] But since \( F_{Rx} = 0 \) and \( F_{Ry} = 0 \), all the force is in the z-direction. Thus, the magnitude of the resultant is: \[ F_3 = \sqrt{(F_3 \cos \alpha)^2 + (F_3 \sin \alpha)^2} \] But we already have the components that sum to zero, so let's solve for \( F_3 \) and \( \alpha \). --- ## **Step 4: Solve the System** We have: - \( F_3 \cos \alpha = -77.5 \) - \( F_3 \sin \alpha = -152.5 \) Calculate \( F_3 \): \[ F_3 = \sqrt{(-77.5)^2 + (-152.5)^2} \] \[ F_3 = \sqrt{6006.25 + 23256.25} \] \[ F_3 = \sqrt{29262.5} \] \[ F_3 \approx 171 \text{ lb} \] Now, solve for \( \alpha \): \[ \tan \alpha = \frac{-152.5}{-77.5} \] \[ \tan \alpha = 1.9677 \] \[ \alpha = \arctan(1.9677) \] \[ \alpha \approx 63.5^\circ \] --- ## **Final Answers** ### **Boxed Answers** \[ \boxed{F_3 = 171 \text{ lb}} \] \[ \boxed{\alpha = 63.5^\circ} \] --- ### **Summary of Steps** 1. **Resolve** each force into x and y components. 2. **Set the sum of x and y components to zero** (since resultant is in z-direction). 3. **Solve the system** for \( F_3 \) and \( \alpha \). If you have any follow-up questions, just ask!