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.

#### Solution:

Show me the final answer↓

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

(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.

Subsituting the values we know into our equation gives us:

\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:

We can now express the force in Cartesian vector form:

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

why is -90k

it should be +90k

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.