Let's solve the problem step by step. --- ## **Part (a): Moment of the Force about the \(a-a\) Axis** ### **Step 1: Write the Force Vector \( \vec{F} \)** Given: - \( F = 30 \) N - Direction cosines: \( \alpha = 60^\circ \), \( \beta = 60^\circ \), \( \gamma = 45^\circ \) Direction cosines mean: \[ \hat{F} = \cos\alpha \, \hat{i} + \cos\beta \, \hat{j} + \cos\gamma \, \hat{k} \] \[ \hat{F} = \cos 60^\circ \, \hat{i} + \cos 60^\circ \, \hat{j} + \cos 45^\circ \, \hat{k} \] \[ \hat{F} = (0.5) \hat{i} + (0.5) \hat{j} + (0.7071) \hat{k} \] So, \[ \vec{F} = 30 \left[ 0.5 \, \hat{i} + 0.5 \, \hat{j} + 0.7071 \, \hat{k} \right] \] \[ \vec{F} = (15) \hat{i} + (15) \hat{j} + (21.213) \hat{k} \ \text{N} \] --- ### **Step 2: Position Vector from \(a\) to the Point of Application** From diagram: - From \(a\), go 100 mm along \(x\), 100 mm along \(y\), and 50 mm along \(z\). - So, \[ \vec{r} = (0.1) \hat{i} + (0.1) \hat{j} + (0.05) \hat{k} \ \text{m} \] --- ### **Step 3: Calculate Moment about Point \(a\)** \[ \vec{M}_a = \vec{r} \times \vec{F} \] Write the determinant: \[ \vec{M}_a = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.1 & 0.1 & 0.05 \\ 15 & 15 & 21.213 \end{vmatrix} \] Expanding: \[ \vec{M}_a = \hat{i} \begin{vmatrix} 0.1 & 0.05 \\ 15 & 21.213 \end{vmatrix} - \hat{j} \begin{vmatrix} 0.1 & 0.05 \\ 15 & 21.213 \end{vmatrix} + \hat{k} \begin{vmatrix} 0.1 & 0.1 \\ 15 & 15 \end{vmatrix} \] Calculate each term: #### **\(\hat{i}\) component:** \[ \hat{i}: (0.1)(21.213) - (0.05)(15) = 2.1213 - 0.75 = 1.3713 \] #### **\(\hat{j}\) component:** \[ \hat{j}: (0.1)(21.213) - (0.05)(15) = 2.1213 - 0.75 = 1.3713 \] #### **\(\hat{k}\) component:** \[ \hat{k}: (0.1)(15) - (0.1)(15) = 1.5 - 1.5 = 0 \] But actually, for \(\hat{j}\), the sign is negative: \[ - \hat{j} \left[ (0.1)(21.213) - (0.05)(15) \right] = -1.3713 \hat{j} \] So, \[ \vec{M}_a = (0.1 \times 21.213 - 0.05 \times 15) \hat{i} - (0.1 \times 21.213 - 0.05 \times 15) \hat{j} + (0.1 \times 15 - 0.1 \times 15) \hat{k} \] \[ = 1.3713 \hat{i} - 1.3713 \hat{j} + 0 \hat{k} \] \[ \vec{M}_a = 1.3713 (\hat{i} - \hat{j}) \ \text{N}\cdot\text{m} \] --- ### **Step 4: Moment about the \(a-a\) Axis** The axis \(a-a\) is along the pipe, which, from the diagram, appears to be along the \(x\)-axis. Take the component of \(\vec{M}_a\) along the \(x\)-axis: \[ M_{a-a} = \vec{M}_a \cdot \hat{e}_a \] where \(\hat{e}_a = \hat{i}\). \[ M_{a-a} = 1.3713 \] **Final Answer (a):** \[ \boxed{M_{a-a} = 1.37 \ \text{N}\cdot\text{m}} \] --- ## **Part (b): Maximum Moment about \(a-a\) Axis and Direction Angles of \(\vec{F}\)** The maximum moment is produced when \(\vec{F}\) is perpendicular to the position vector \(\vec{r}\) and lies in the plane perpendicular to the \(a-a\) axis. - The maximum moment is: \[ M_{max} = F \cdot d \] where \(d\) is the perpendicular distance from the axis to the line of action of the force. From the diagram, the perpendicular distance from the \(a-a\) axis to the force application point is: \[ d = \sqrt{(0.1)^2 + (0.05)^2} = \sqrt{0.01 + 0.0025} = \sqrt{0.0125} = 0.1118 \text{ m} \] \[ M_{max} = 30 \times 0.1118 = 3.35 \ \text{N}\cdot\text{m} \] **Direction Angles:** - The force should be perpendicular to the \(a-a\) axis (\(\hat{i}\)), so pick direction cosines that satisfy \(l = 0\), \(m^2 + n^2 = 1\). - For example: \(m = \cos 45^\circ = 0.707\), \(n = \cos 45^\circ = 0.707\), \(l = 0\). --- **Final Answers:** ### **a. Moment about the \(a-a\) axis:** \[ \boxed{1.37 \ \text{N}\cdot\text{m}} \] ### **b. Maximum moment and direction angles:** \[ \boxed{3.35 \ \text{N}\cdot\text{m}} \] **Direction angles:** \( \alpha = 90^\circ, \beta = 45^\circ, \gamma = 45^\circ \) (or any combination with \(l = 0\)).