1. **State the problem:** We have a 260-lb force applied at point A on a rolled-steel section. We want to replace this force with an equivalent force-couple system at point C. 2. **Given data:** - Force magnitude $P = 260$ lb - Point A is located 4 inches above and 2.5 inches to the right of point C. - Point B is 2 inches below point C. - The force $P$ acts along the line through points A and B. 3. **Find:** - The equivalent force $F$ at point C (magnitude and direction angle $\theta$) - The couple moment at point C 4. **Step 1: Find the vector from C to A and C to B** $$\vec{r}_{CA} = 2.5\hat{i} + 4\hat{j}$$ $$\vec{r}_{CB} = 0\hat{i} - 2\hat{j}$$ 5. **Step 2: Find the direction vector of the force $P$ along line AB** $$\vec{r}_{AB} = \vec{r}_{CB} - \vec{r}_{CA} = (0 - 2.5)\hat{i} + (-2 - 4)\hat{j} = -2.5\hat{i} - 6\hat{j}$$ 6. **Step 3: Find the unit vector along $P$** $$|\vec{r}_{AB}| = \sqrt{(-2.5)^2 + (-6)^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5$$ $$\hat{u}_P = \frac{\vec{r}_{AB}}{|\vec{r}_{AB}|} = \frac{-2.5}{6.5}\hat{i} + \frac{-6}{6.5}\hat{j} = -0.3846\hat{i} - 0.9231\hat{j}$$ 7. **Step 4: Express force $P$ as a vector** $$\vec{P} = P \hat{u}_P = 260(-0.3846\hat{i} - 0.9231\hat{j}) = -100\hat{i} - 240\hat{j}$$ 8. **Step 5: The equivalent force at C is the same vector $\vec{F} = \vec{P}$** 9. **Step 6: Calculate the moment (couple) at C due to force at A** $$\vec{M}_C = \vec{r}_{CA} \times \vec{P}$$ Using 2D cross product (scalar moment about z-axis): $$M_C = x_A F_y - y_A F_x = 2.5(-240) - 4(-100) = -600 + 400 = -200 \text{ lb-in}$$ 10. **Step 7: Find magnitude and angle of force $F$ at C** Magnitude: $$|\vec{F}| = \sqrt{(-100)^2 + (-240)^2} = \sqrt{10000 + 57600} = \sqrt{67600} = 260 \text{ lb}$$ Angle $\theta$ from positive x-axis (to force vector): $$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{-240}{-100}\right) = \tan^{-1}(2.4) = 67.38^\circ$$ Since both components are negative, force is in third quadrant, so add 180°: $$\theta = 67.38^\circ + 180^\circ = 247.38^\circ$$ **Final answer:** - Equivalent force at C: $F = 260$ lb - Angle from positive x-axis: $\theta = 247.38^\circ$ - Couple moment at C: $M_C = -200$ lb-in (clockwise moment)