Components Parallel and Normal to a Line Suppose that you pull on the cable OA in Fig. 2.37, exerting a 50-N force F at O. What are the components of F parallel and normal to the cable OB?

Question

Components Parallel and Normal to a Line

Suppose that you pull on the cable OA in Fig. 2.37, exerting a 50-N force F at O. What are the components of F parallel and normal to the cable OB?

Strategy
Resolving F into components parallel and normal to OB (Fig. a), we can determine the components by using Eqs. (2.26) and (2.27).

[latex]U_{p} = (e . U)e .[/latex] (2.26)

[latex]U_{n} = U - U_{p} .[/latex] (2.27)

But to apply them, we must first express F in terms of scalar components and determine the components of a unit vector parallel to OB. We can obtain the components of F by determining the components of the unit vector pointing from O toward A and multiplying them by |F|.

Accepted Answer

The position vectors from O to A and from O to B are (Fig. b)

[latex] r_{OA} = 6i + 6j - 3k (m) .[/latex]

[latex] r_{OB} = 10i - 2j + 3k (m) .[/latex]

Their magnitudes are [latex] \left|r_{OA}\right| = 9 m[/latex] and [latex] \left|r_{OB}\right| = 10.6 m[/latex] . Dividing these vectors by their magnitudes, we obtain unit vectors that point from the origin toward A and B (Fig. c):

[latex] e_{OA} = \frac{r_{OA} }{\left|r_{OA} \right| } = \frac{6i + 6j - 3k}{9} = 0.667i + 0.667j - 0.333k , [/latex]

[latex] e_{OB} = \frac{r_{OB} }{\left|r_{OB} \right| } = \frac{10i - 2j + 3k}{10.6} = 0.941i + 0.188j - 0.282k . [/latex]

The force F in terms of scalar components is

[latex]F = |F|e_{OA} = (50)(0.667i + 0.667j - 0.333k)[/latex]

= 33.3i + 33.3j - 16.7k (N).

Taking the dot product of [latex]e_{OB}[/latex] and F , we obtain

[latex]e_{OB} \cdot F = (0.941)(33.3) + (-0.188) (33.3) + (0.282)(-16.7)[/latex]

= 20.4 N.

The parallel component of F is

[latex]F_{p} = (e_{OB} \cdot F)e_{OB} = (20.4)(0.941i - 0.188j + 0.282k) [/latex]

= 19.2i - 3.8j + 5.8k (N).

and the normal component is

[latex]F_{n} = F - F_{p} = 14.2i + 37.2j - 22.4k (N).[/latex]

Discussion

You can confirm that two vectors are perpendicular by making sure their dot product is zero. In this example,

[latex]F_{p} \cdot F_{n} = (19.2)(14.2) + (-3.8)(37.2) + (5.8)(-22.4) = 0 .[/latex]

Question 2.15: Components Parallel and Normal to a Line Suppose that you pu...

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