Determine the magnitude of the moment of force F about segment OA of the pipe assembly in Fig. 4–24a.

Question

Determine the magnitude of the moment of force [latex]\mathbf{F}[/latex] about segment OA of the pipe assembly in Fig. 4–24a.

Accepted Answer

The moment of [latex]\mathbf{F}[/latex] about the OA axis is determined from [latex]M_{OA} = \mathbf{u}_{OA}· (\mathbf{r} × \mathbf{F})[/latex], where [latex]\mathbf{r}[/latex] is a position vector extending from any point on the OA axis to any point on the line of action of [latex]\mathbf{F}[/latex] . As indicated in Fig. 4–24 b , either [latex]\mathbf{r}_{OD}, \mathbf{r}_{OC}, \mathbf{r}_{AD}[/latex], or [latex]\mathbf{r}_{AC}[/latex] can be used; however, [latex]\mathbf{r}_{OD}[/latex] will be considered since it will simplify the calculation.

The unit vector [latex]\mathbf{u}_{OA}[/latex] , which specifies the direction of the OA axis, is

[latex]\mathbf{u}_{OA} = \frac{\mathbf{r}_{OA}}{r_{OA}} = \frac{\left\{0.3\mathbf{i} + 0.4\mathbf{j} ight\} m}{\sqrt{(0.3 m)^{2} + (0.4)^{2}} } = 0.6\mathbf{i} + 0.8\mathbf{j}[/latex]

and the position vector [latex]\mathbf{r}_{OD}[/latex] is

[latex]\mathbf{r}_{OD} = \left\{0.5\mathbf{i} + 0.5\mathbf{k} ight\} m[/latex]

The force [latex]\mathbf{F}[/latex] expressed as a Cartesian vector is

[latex]\mathbf{F} = F\left(\frac{\mathbf{r}_{CD}}{r_{CD}} ight) [/latex]

[latex]= (300 N) \left[\frac{\left\{0.4\mathbf{i} - 0.4\mathbf{j} + 0.2\mathbf{k} ight\} m}{\sqrt{(0.4 m)^{2} + (- 0.4 m)^{2} + (0.2 m)^{2} }} ight][/latex]

[latex]= \left\{200\mathbf{i} - 200\mathbf{j} + 100\mathbf{k} ight\} N [/latex]

Therefore,

[latex]M_{OA} = \mathbf{u}_{OA}·(\mathbf{r}_{OD} imes \mathbf{F}) [/latex]

[latex]= \left|\begin{matrix} 0.6 & 0.8 & 0 \ 0.5 & 0 & 0.5 \200 & - 200 & 100 \end{matrix} ight|[/latex]

= 0.6 [0(100) - (0.5)(-200)] - 0.8[0.5( 100) - (0.5)(200)] + 0

= 100 N · m

Question 4.9: Determine the magnitude of the moment of force F about segme...

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