Equivalent System Point A 6B2Bff

1. **State the problem:** We need to find the equivalent system at point $A$ for the pipe system, which means finding the resultant force and the resultant moment at point $A$ due to the forces $F_1$ and $F_2$. 2. **Given data:** - Force $F_1 = -20\hat{i} - 10\hat{j} + 25\hat{k}$ lb applied at some point $P_1$. - Force $F_2 = -10\hat{i} + 25\hat{j} + 20\hat{k}$ lb applied at point $P_2$ (top of vertical segment). - Point $A$ is the reference point for the equivalent system. 3. **Assumptions and positions:** We need the position vectors of the points where $F_1$ and $F_2$ are applied relative to point $A$. Let: - $\vec{r}_1 = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ be the position vector from $A$ to the point of application of $F_1$. - $\vec{r}_2 = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$ be the position vector from $A$ to the point of application of $F_2$. Since $F_2$ is applied at the top of the vertical segment at $A$, $\vec{r}_2 = \vec{0}$. 4. **Resultant force at $A$:** $$\vec{R} = F_1 + F_2 = (-20\hat{i} - 10\hat{j} + 25\hat{k}) + (-10\hat{i} + 25\hat{j} + 20\hat{k}) = (-30\hat{i} + 15\hat{j} + 45\hat{k})$$ 5. **Resultant moment at $A$:** The moment at $A$ due to each force is given by: $$\vec{M}_A = \vec{r}_1 \times F_1 + \vec{r}_2 \times F_2$$ Since $\vec{r}_2 = 0$, $$\vec{M}_A = \vec{r}_1 \times F_1$$ 6. **Finding $\vec{r}_1$:** From the problem, the horizontal segment extends from $O$ to $A$, and $F_1$ is applied near the middle-left segment. Assume the position of $F_1$ relative to $A$ is: $$\vec{r}_1 = -dx\hat{i} - dy\hat{j} + 0\hat{k}$$ where $dx$ and $dy$ are distances from $A$ to the point of application of $F_1$ along $x$ and $y$ axes respectively. 7. **Calculate the moment:** $$\vec{M}_A = \vec{r}_1 \times F_1 = (-dx\hat{i} - dy\hat{j}) \times (-20\hat{i} - 10\hat{j} + 25\hat{k})$$ Calculate the cross product: - $\hat{i} \times \hat{i} = 0$ - $\hat{i} \times \hat{j} = \hat{k}$ - $\hat{i} \times \hat{k} = -\hat{j}$ - $\hat{j} \times \hat{i} = -\hat{k}$ - $\hat{j} \times \hat{j} = 0$ - $\hat{j} \times \hat{k} = \hat{i}$ So, $$\vec{M}_A = -dx\hat{i} \times (-20\hat{i} - 10\hat{j} + 25\hat{k}) - dy\hat{j} \times (-20\hat{i} - 10\hat{j} + 25\hat{k})$$ Calculate each term: - $-dx\hat{i} \times (-20\hat{i}) = -dx \times -20 (\hat{i} \times \hat{i}) = 0$ - $-dx\hat{i} \times (-10\hat{j}) = -dx \times -10 (\hat{i} \times \hat{j}) = 10dx \hat{k}$ - $-dx\hat{i} \times (25\hat{k}) = -dx \times 25 (\hat{i} \times \hat{k}) = -25dx (-\hat{j}) = 25dx \hat{j}$ - $-dy\hat{j} \times (-20\hat{i}) = -dy \times -20 (\hat{j} \times \hat{i}) = 20dy (-\hat{k}) = -20dy \hat{k}$ - $-dy\hat{j} \times (-10\hat{j}) = -dy \times -10 (\hat{j} \times \hat{j}) = 0$ - $-dy\hat{j} \times (25\hat{k}) = -dy \times 25 (\hat{j} \times \hat{k}) = -25dy \hat{i}$ Sum all components: $$\vec{M}_A = (0 + 0 - 25dy)\hat{i} + (25dx + 0 + 0)\hat{j} + (10dx - 20dy + 0)\hat{k} = (-25dy)\hat{i} + (25dx)\hat{j} + (10dx - 20dy)\hat{k}$$ 8. **Final equivalent system at $A$:** - Resultant force: $$\boxed{\vec{R} = -30\hat{i} + 15\hat{j} + 45\hat{k}}$$ - Resultant moment: $$\boxed{\vec{M}_A = -25dy\hat{i} + 25dx\hat{j} + (10dx - 20dy)\hat{k}}$$ 9. **Summary:** The equivalent system at point $A$ consists of the resultant force $\vec{R}$ and the resultant moment $\vec{M}_A$ as above, expressed in Cartesian vector form. Note: To find numerical values for the moment, the distances $dx$ and $dy$ must be known or given.