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.

The cable at the end of the crane boom

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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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

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-65.

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