# The cable at the end of the crane boom 2

The cable at the end of the crane boom exerts a force of 250 lb on the boom as shown. Express F as a Cartesian vector. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

To do this question, you must remember the following equation:

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

(Where $\alpha,\beta,\gamma$ are the angles between the positive x,y, and z axes respectively.)

From the diagram, we know the values of $\alpha$ and $\beta$.

$\alpha = 30^0$

$\beta = 70^0$

Subsituting the values we know into our equation gives us:

$\cos^2(30^0) + \cos^2(70^0) + \cos^2\gamma = 1$
$\cos^2\gamma = 0.133$

(take the square root of both sides)

$\cos\gamma = \pm 0.365$

$\gamma = \cos^{-1}(\pm 0.365)$

$\gamma = 68.6^0$ or $111^0$

From the diagram, we can see that the angle between the z-axis and the force must be larger than $90^0$ because it lies in the negative quadrant. Therefore:

$\gamma = 111^0$

We can now express the force in Cartesian vector form:

$F=250\left\{\cos30^0i+\cos70^0j+\cos111^0k\right\}$

$F\,=\,\left\{217i+85j-90k\right\}$ lb

## 2 thoughts on “The cable at the end of the crane boom”

• Ameed Al Gharib

why is -90k

it should be +90k

• questionsolutions Post author

Cos(111°) = -0.358. Also, looking at the diagram, we can see that the force is below the z-axis, so even without looking at numbers, we can see it’ll be negative.