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.

The beam is subjected to the two forces shown

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

Solution:

Show me the final answer↓

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

The beam is subjected to the two forces shown

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

 

Final Answers:

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

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-70.

Leave a comment

Your email address will not be published.

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