Problem

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

Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4–13, and (b) summing the moment of each force about point O. Take F = {25k} N.

Step-by-Step Solution

Solution 1

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

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

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