Question: ENGR& 214 Ch4_6: Moment of Couples 1. Given: Two couples act on the beam with the geometry shown. Find: The magnitude of $F$ so that the resultant couple moment is $1.5$ kN·m clockwise. Graph/top-right: a horizontal beam fixed at the left, with an upward force $F$ at point $A$ and a downward force $F$ at point $B$, and a $2$ kN leftward force near the top right balanced by a $2$ kN rightward force near the lower right; the horizontal separation is $0.9$ m and the vertical separation is $0.3$ m. 2. Given: A $450$ N force couple acting on the pipe assembly. Find: The couple moment in Cartesian vector notation. Graph/bottom-right: a $3D$ pipe assembly with point $O$ at the base, vertical pipe to point $A$, and slanted pipe from $A$ to $B$; two $450$ N forces act along opposite directions at $A$ and $B$, with direction triangles marked $3$-$4$-$5$, and dimensions $0.3$ m vertically and $0.4$ m along the pipe.solve step by step

1. **Problem 1: Find the magnitude of $F$ for the resultant couple moment of $1.5$ kN·m clockwise.** 2. The beam has two couples: one from forces $F$ at points $A$ and $B$, and another from $2$ kN forces separated by $0.3$ m vertically. 3. The moment of a couple is $M = F \times d$, where $d$ is the perpendicular distance between the forces. 4. For the $2$ kN couple, moment is $M_1 = 2 \times 0.3 = 0.6$ kN·m (counterclockwise). 5. For the $F$ couple, horizontal separation is $0.9$ m, so moment is $M_2 = F \times 0.9$ (direction depends on force orientation). 6. The resultant moment is $M_R = M_2 - M_1 = 1.5$ kN·m clockwise, so $M_2 = 1.5 + 0.6 = 2.1$ kN·m. 7. Solve for $F$: $$F = \frac{M_2}{0.9} = \frac{2.1}{0.9} = 2.333... \approx 2.33 \text{ kN}.$$ --- 8. **Problem 2: Find the couple moment in Cartesian vector notation for the $450$ N force couple on the pipe assembly.** 9. Forces are equal and opposite $450$ N at points $A$ and $B$. 10. Use the position vectors: $\vec{r}_A = 0.3 \hat{j}$ m (vertical), $\vec{r}_B = 0.4 \hat{i} + 0.3 \hat{j}$ m (along pipe). 11. Force directions from $3$-$4$-$5$ triangle: unit vector along pipe is $\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$. 12. Force at $A$: $\vec{F}_A = 450 \hat{j}$ N upward. 13. Force at $B$: $\vec{F}_B = -450 \left( \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j} \right) = -270 \hat{i} - 360 \hat{j}$ N. 14. Couple moment $\vec{M} = \vec{r}_A \times \vec{F}_A + \vec{r}_B \times \vec{F}_B$. 15. Calculate cross products: $$\vec{r}_A \times \vec{F}_A = (0.3 \hat{j}) \times (450 \hat{j}) = 0$$ (parallel vectors). $$\vec{r}_B \times \vec{F}_B = (0.4 \hat{i} + 0.3 \hat{j}) \times (-270 \hat{i} - 360 \hat{j})$$ 16. Expand: $$= 0.4 \hat{i} \times (-270 \hat{i}) + 0.4 \hat{i} \times (-360 \hat{j}) + 0.3 \hat{j} \times (-270 \hat{i}) + 0.3 \hat{j} \times (-360 \hat{j})$$ 17. Use cross product rules: $\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{j} \times \hat{j} = 0$. 18. So: $$= 0 + 0.4 \times (-360) \hat{k} + 0.3 \times (-270)(-\hat{k}) + 0$$ $$= -144 \hat{k} + 81 \hat{k} = (-144 + 81) \hat{k} = -63 \hat{k} \text{ N·m}.$$ 19. The total couple moment is: $$\vec{M} = 0 + (-63 \hat{k}) = -63 \hat{k} \text{ N·m}.$$ 20. The negative $\hat{k}$ direction indicates clockwise moment about the $z$-axis. **Final answers:** - $F \approx 2.33$ kN for problem 1. - Couple moment vector $\vec{M} = -63 \hat{k}$ N·m for problem 2.