Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Sol...

Given \(\mathbf{F}=\{25 \mathbf{k}\} \mathrm{N}\)

Coordinates of \(A\) are \((0.65 \mathrm{~m}, 0.4 \mathrm{~m}, 0 \mathrm{~m})\)

Coordinates of \(B\) are \((0.3 \mathrm{~m}, 0.2 \mathrm{~m}, 0 \mathrm{~m})\)

Coordinates of \(O\) are \((0 \mathrm{~m}, 0 \mathrm{~m}, 0 \mathrm{~m})\)

The position vector directed from the point \(A\) to \(B\) is expressed as

$$ \begin{aligned} &\mathbf{r}_{A B}=\mathbf{i}(0.3-0.65) \mathrm{m}+\mathrm{j}(0.2-0.4) \mathrm{m}+\mathrm{k}(0-0) \mathrm{m} \\ &\mathbf{r}_{A B}=-(0.35 \mathrm{~m}) \mathrm{i}-(0.2 \mathrm{~m}) \mathrm{j} \end{aligned} $$

(a)

From the Cartesian vector formulation, we have the moment of the force \(\mathbf{F}\) about point \(O\) as

$$ \mathbf{M}=\mathbf{r}_{A B} \times \mathbf{F} $$

Accordingly, we have

$$ \begin{aligned} &\mathbf{M}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -0.35 & -0.2 & 0 \\ 0 & 0 & 25 \end{array}\right| \mathrm{N} \cdot \mathrm{m} \\ &\mathbf{M}=\mathbf{i}[(-0.2 \times 25)-(0 \times 0)] \mathrm{N} \cdot \mathrm{m}-\mathrm{j}[(-0.35 \times 25)-(0 \times 0)] \mathrm{N} \cdot \mathrm{m} \\ &+\mathbf{k}[(-0.35 \times 0)-(-0.2 \times 0)] \mathrm{N} \cdot \mathrm{m} \\ &\mathbf{M}=(-5 \mathrm{~N} \cdot \mathrm{m}) \mathbf{i}+(8.75 \mathrm{~N} \cdot \mathrm{m}) \mathrm{j} \end{aligned} $$

Moment of the force about point \(O\) is

$$ \mathbf{M}=(-5 N \cdot m) i+(8.75 N \cdot m) j $$

(b)

The position vector directed from the point \(O\) to \(A\) is expressed as

$$ \begin{aligned} &\mathbf{r}_{o s}=\mathbf{i}(0.65-0) \mathrm{m}+\mathrm{j}(0.4-0) \mathrm{m}+\mathrm{k}(0-0) \mathrm{m} \\ &\mathbf{r}_{o s}=(0.65 \mathrm{~m}) \mathbf{i}+(0.4 \mathrm{~m}) \mathrm{j} \end{aligned} $$

The position vector directed from the point \(O\) to \(B\) is expressed as

$$ \begin{aligned} &\mathbf{r}_{O N}=\mathbf{i}(0.3-0) \mathrm{m}+\mathrm{j}(0.2-0) \mathrm{m}+\mathrm{k}(0-0) \mathrm{m} \\ &\mathbf{r}_{O s}=(0.3 \mathrm{~m}) \mathbf{i}+(0.2 \mathrm{~m}) \mathbf{j} \end{aligned} $$

The summation of the moment of each force about point \(O\) is expressed as

$$ M=\left(\mathbf{r}_{o d} \times-\mathbf{F}\right)+\left(\mathbf{r}_{O B} \times \mathbf{F}\right) $$

Accordingly, we have

$$ \begin{aligned} &\mathbf{M}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0.65 & 0.4 & 0 \\ 0 & 0 & -25 \end{array}\right| \mathbf{N} \cdot \mathbf{m}+\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0.3 & 0.2 & 0 \\ 0 & 0 & 25 \end{array}\right| \mathbf{N} \cdot \mathbf{m} \\ &\mathbf{M}=\mathbf{i}[(0.4 \times-25)-(0 \times 0)] \mathrm{N} \cdot \mathrm{m}-\mathbf{j}[(0.65 \times-25)-(0 \times 0)] \mathrm{N} \cdot \mathrm{m} \\ &+\mathbf{k}[(0.65 \times 0)-(0.4 \times 0)] \mathrm{N} \cdot \mathrm{m}+\mathrm{i}[(0.2 \times 25)-(0 \times 0)] \mathrm{N} \cdot \mathrm{m} \\ &-\mathbf{j}[(0.3 \times 25)-(0 \times 0)] \mathrm{N} \cdot \mathrm{m}+\mathbf{k}[(0.3 \times 0)-(0.2 \times 0)] \mathrm{N} \cdot \mathrm{m} \end{aligned} $$

$$ \begin{aligned} &\mathbf{M}=(-10 \mathrm{~N} \cdot \mathrm{m}) \mathbf{i}+(16.25 \mathrm{~N} \cdot \mathrm{m}) \mathrm{j}+(5 \mathrm{~N} \cdot \mathrm{m}) \mathrm{i}-(7.5 \mathrm{~N} \cdot \mathrm{m}) \mathrm{j} \\ &\mathbf{M}=(-5 \mathrm{~N} \cdot \mathrm{m}) \mathrm{i}+(8.75 \mathrm{~N} \cdot \mathrm{m}) \mathrm{j} \end{aligned} $$

Moment of the force about point \(O\) is

$$ \mathbf{M}=(-5 \mathrm{~N} \cdot \mathrm{m}) \mathbf{i}+(8.75 \mathrm{~N} \cdot \mathrm{m}) \mathbf{j} $$