# 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

$\alpha=75.4^0$

$\beta=90^0$

$\gamma=165^0$

## 2 thoughts on “The beam is subjected to the two forces shown”

• Anthony Mullins

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?