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## Q. 1.6

The cable attached to the eyebolt in Fig. (a) is pulled with the force F of magnitude 500 lb. Determine the rectangular representation of this force.

## Verified Solution

Because the coordinates of points $A$ and $B$ on the line of action of $F$ are known, the following is a convenient method for obtaining the rectangular representation of $F$.

1. Write $\overrightarrow{AB}$, the vector from $A$ to $B$, in rectangular form.

The vector $\overrightarrow{AB}$ and its rectangular components are shown in Fig. (b). Two common errors made by students at this point are choosing the wrong signs and mixing up the scalar components. You can avoid both of these difficulties by taking the time to show the vector on a carefully drawn sketch of the appropriate parallelepiped. From Fig. (b) we see that

$\overrightarrow{AB} = − 4i + 6j − 3k\:ft$

2. Evaluate $λ$, the unit vector from $A$ toward $B$:

$λ=\frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}=\frac{-4i+6j-3k}{\sqrt{(-4)^{2}+6^{2}+(-3)^{2}}}=-0.5122i+0.7682j-0.3841k$

3. Write $F = Fλ$:

$F = 500(−0.5122i + 0.7682j − 0.3841k)= −256i + 384j − 192k\:lb$

The rectangular components of F are shown in Fig. (c).