# The beam is subjected to the two forces shown 2

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.

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

#### Solution:

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.

$F_{1y}=630\left(\dfrac{7}{25}\right)\,=\,176.4$ lb

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

$F_1\,=\,\left\{0i+176.4j-604.8k\right\}$ lb

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.

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

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

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

Show me the diagram

Let us now express force $F_2$ in Cartesian form.

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

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

$F_R\,=\,F_1 + F_2$

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

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

The magnitude of the resultant force is:

magnitude of $F_R\,=\,\sqrt{(125)^2+(-0.6)^2+(-479.8)^2}\,=\,496$ lb

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

$\alpha=\cos^{-1}\left(\dfrac{125}{496}\right)\,=\,75.4^0$

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

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

$F_1\,=\,\left\{0i+176.4j-604.8k\right\}$ lb

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

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

magnitude of $F_R\,=\,496$ lb

$\alpha=75.4^0$

$\beta=90^0$

$\gamma=165^0$