Determine the couple moment acting on the pipe shown in Fig. 4–32a.
Segment AB is directed 30° below the x–y plane.

Question

Accepted Answer

I (VECTOR ANALYSIS)
The moment of the two couple forces can be found about any point . If point O is considered, Fig. 4–32 b , we have

[latex]\mathbf{M} = \mathbf{r}_{A} × (- 25\mathbf{k}) + \mathbf{r}_{B} × (25\mathbf{k})[/latex]
[latex]= (8\mathbf{j}) × (- 25\mathbf{k}) + (6 cos 30°\mathbf{i} + 8\mathbf{j} - 6 sin 30°\mathbf{k}) × (25\mathbf{k})[/latex]
[latex]= - 200\mathbf{i} - 129.9\mathbf{j} + 200\mathbf{i}[/latex]
[latex]= \left\{ - 130\mathbf{j}\right\} lb · in.[/latex]

It is easier to take moments of the couple forces about a point lying on the line of action of one of the forces, e.g., point A , Fig. 4–32 c . In this case the moment of the force at A is zero, so that

[latex]\mathbf{M} = \mathbf{r}_{AB} × (25\mathbf{k})[/latex]
[latex]= (6 cos 30°\mathbf{i} - 6 sin 30°\mathbf{k}) × (25\mathbf{k})[/latex]
[latex]= \left\{ - 130\mathbf{j}\right\} lb · in.[/latex]

II (SCALAR ANALYSIS)
Although this problem is shown in three dimensions, the geometry is simple enough to use the scalar equation M = Fd. The perpendicular distance between the lines of action of the couple forces is d = 6 cos 30° = 5.196 in., Fig. 4–32 d . Hence, taking moments of the forces about either point A or point B yields

M = Fd = 25 lb (5.196 in.) = 129.9 lb · in.

Applying the right-hand rule, [latex]\mathbf{M}[/latex] acts in the [latex]- \mathbf{j}[/latex] direction. Thus,

[latex]\mathbf{M} = \left\{ - 130\mathbf{j}\right\} lb · in.[/latex]

Question 4.12: Determine the couple moment acting on the pipe shown in Fig....

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