Absolutely! Let's follow the exact same steps as your solution, but change \( P = 3200\, \text{lb} \) to \( P = 4200\, \text{lb} \). --- ### **Given:** - \( P = 4200\, \text{lb} \) - \( d = 0.75\, \text{in} \) - \( \theta = \tan^{-1}\left(\frac{3}{4}\right) = 36.87^\circ \) --- ## **Step 1: Find force in member AE** From equilibrium in \( y \)-direction: \[ \sum F_y = 0 \implies F_{AE} \sin \theta - P = 0 \] \[ F_{AE} = \frac{P}{\sin \theta} \] \[ F_{AE} = \frac{4200}{\sin 36.87^\circ} \] \[ F_{AE} = \frac{4200}{0.6} \] \[ F_{AE} = 7000\, \text{lb} \] --- ## **Step 2: Force in member CG** Taking moments about point \( F \): \[ \sum M_F = 0: \] \[ F_{AE} \sin \theta \times EF - F_{CG} \sin \theta \times FG = 0 \] Assuming \( EF = FG \) and both are 4 units (as in your example): \[ 7000 \cdot \sin 36.87^\circ \cdot 4 - F_{CG} \cdot \sin 36.87^\circ \cdot 4 = 0 \] \[ F_{CG} = \frac{7000 \cdot \sin 36.87^\circ \cdot 4}{\sin 36.87^\circ \cdot 4} \] \[ F_{CG} = 7000\, \text{lb} \] --- ## **Step 3: Normal Stress in CG** \[ \sigma_{CG} = \frac{F_{CG}}{\frac{\pi}{4} d^2} \] \[ = \frac{7000}{\frac{\pi}{4} (0.75)^2} \] Calculate denominator: \[ \frac{\pi}{4} (0.75)^2 = \frac{\pi}{4} \times 0.5625 = 0.4418 \] \[ \sigma_{CG} = \frac{7000}{0.4418} \] \[ = 15849.3\, \text{psi} \] \[ = 15.85\, \text{ksi} \] --- ## **Final Answer** **The normal stress in CG is \( \boxed{15.85\, \text{ksi}} \).** --- **Summary Table:** | Step | Formula | Value | Result | |--------------|--------------------------------------------|-------------------------|---------------| | \( F_{AE} \) | \( \frac{P}{\sin \theta} \) | \( \frac{4200}{0.6} \) | 7000 lb | | \( F_{CG} \) | \( F_{AE} \) | 7000 lb | 7000 lb | | \( \sigma \) | \( \frac{F_{CG}}{\frac{\pi}{4} d^2} \) | \( \frac{7000}{0.4418} \) | 15.85 ksi | --- **Boxed Final Statement:** \[ \boxed{\text{The normal stress in CG is } 15.85\, \text{ksi}.} \]