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give answer in 2 step with explanation at the end of each step and final answer at the end:PA Incoch cae, determine he reslant moment of he oes acing shot 3. 1 2 0 [oN ’ od A Ey J ax on 5 Pe wy A Ton ® a - HON a — 5 = < < Si fox © Fons

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give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded ImagePA Incoch cae, determine he reslant moment of he oes acing shot 3. 1 2 0 [oN ’ od A Ey J ax on 5 Pe wy A Ton ® a - HON a — 5 = < < Si fox © Fons

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Asked by: Rahul
Created at: 2025-11-19 06:09:01

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Let's solve **Problem P3-3(a)** step by step as requested: --- ## **Step 1: Write Position Vectors and Force Vectors** Choose the origin (O) at the bottom left corner of the plate. **Coordinates of force application points:** - **200 N (upwards, at A):** (, , ) - **50 N (upwards, at B):** (, 3, ) - **100 N (upwards, at C):** (2, 3, ) - **300 N (upwards, at D):** (2, , ) **Force vectors:** All forces act in the **positive z-direction**: - \(\vec{F}_1 = 200\hat{k}\) at (,,) - \(\vec{F}_2 = 50\hat{k}\) at (,3,) - \(\vec{F}_3 = 100\hat{k}\) at (2,3,) - \(\vec{F}_4 = 300\hat{k}\) at (2,,) --- **Explanation:** We identified the coordinates where each force is applied and wrote their vectors. --- ## **Step 2: Calculate the Moment of Each Force About the Origin** The moment of a force about the origin is: \[ \vec{M} = \vec{r} \times \vec{F} \] Calculate for each force: ### **Force at A (origin):** \[ \vec{r}_1 = (, , ), \ \vec{F}_1 = (, , 200) \] \[ \vec{M}_1 = \vec{r}_1 \times \vec{F}_1 = (,,) \] ### **Force at B:** \[ \vec{r}_2 = (, 3, ), \ \vec{F}_2 = (, , 50) \] \[ \vec{M}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ & 3 & \\ & & 50 \end{vmatrix} = (3 \cdot 50)\hat{i} - ( \cdot 50)\hat{j} + ()\hat{k} = 150\hat{i} \] ### **Force at C:** \[ \vec{r}_3 = (2, 3, ), \ \vec{F}_3 = (, , 100) \] \[ \vec{M}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \\ & & 100 \end{vmatrix} = (3 \cdot 100)\hat{i} - (2 \cdot 100)\hat{j} = 300\hat{i} - 200\hat{j} \] ### **Force at D:** \[ \vec{r}_4 = (2, , ), \ \vec{F}_4 = (, , 300) \] \[ \vec{M}_4 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & & \\ & & 300 \end{vmatrix} = ( - )\hat{i} - (2 \cdot 300)\hat{j} + ()\hat{k} = -600\hat{j} \] --- **Explanation:** We calculated the moment for each force using the cross product formula. --- ## **Step 3: Sum the Moments to Get the Resultant** Sum the \( \hat{i} \) and \( \hat{j} \) components: \[ \vec{M}_{\text{res}} = ( + 150 + 300 + )\hat{i} + ( + - 200 - 600)\hat{j} + \hat{k} \] \[ \vec{M}_{\text{res}} = 450\hat{i} - 800\hat{j} \ \text{N}\cdot\text{m} \] --- ### **Final Answer:** \[ \boxed{ \vec{M}_{\text{res}} = 450\hat{i} - 800\hat{j} \ \text{N}\cdot\text{m} } \] - Moment about x-axis (\(M_x\)): \(450\) N·m - Moment about y-axis (\(M_y\)): \(-800\) N·m - Moment about z-axis (\(M_z\)): \(\) N·m --- If you want solutions for (b) or (c), let me know!

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