The beam is subjected to the two forces shown. Express each force in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

#### Solution:

Show me the final answer↓

Using the diagram, we can express each component corresponding to it’s force.

We will first look at force F_1. Note how it lies on the z-y plane, therefore, it does not have a x-component.

F_{1z}=630\left(\dfrac{24}{25}\right)\,=\,-604.8 lb

(note how our z-component is negative. This is because the force is going downwards. In other words, the z-component lies in the negative z-axis.)

We can now express force F_1 in Cartesian form as follows:

Now, we will look at force F_2. Each component can be found by multiplying the magnitude of the force by the corresponding coordinate direction angles.

F_{2y}\,=\,250\cos135^0\,=\,-177 lb

F_{2z}\,=\,250\cos60^0\,=\,125 lb

Let us now express force F_2 in Cartesian form.

We can find the resultant force by adding the two forces together.

F_R\,=\,\left\{0i+176.4j-604.8k\right\}+\left\{125i-177j+125k\right\}

F_R\,=\,\left\{125i-0.6j-479.8k\right\} lb

The magnitude of the resultant force is:

The coordinate direction angles can be found by taking the cosine inverse of each component of the resultant force divided by it’s magnitude.

\beta=\cos^{-1}\left(\dfrac{-0.6}{496}\right)\,=\,90^0

\gamma=\cos^{-1}\left(\dfrac{-479.8}{496}\right)\,=\,165^0

#### Final Answers:

F_2\,=\,\left\{125i-177j+125k\right\} lb

F_R\,=\,\left\{125i-0.6j-479.8k\right\} lb

magnitude of F_R\,=\,496 lb

\alpha=75.4^0

\beta=90^0

\gamma=165^0

Okay, I see how F1 is on the zy plane and that F2 is on the xyz plane but why is the sum of forces for F2 250 cos(60), 250 cos(135), and 250 cos(60)? Why is sin not involved?

When you are given forces with coordinate direction angles, it’s always cosine only, never sine. Please see the following: https://www.youtube.com/watch?v=igmrd7mYkfs&t=122s