The pole is subjected to the force F

The pole is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 3 kN, β = 30°, and γ = 75°, determine the magnitudes of its three components.

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

Solution:

Let us first substitute the values given in the question into our diagram.

To calculate the value of $\alpha$ we have to remember the following formula.

$\text{cos}^2\alpha +\text{cos}^2\beta +\text{cos}^2\gamma =1$

Let’s substitute the values we know. Thus, we have:

$\text{cos}^2\alpha +\text{cos}^{2}(30^0)+\text{cos}^{2}(75^0) =1$

Now we will solve for $\alpha$ by isolating for $\alpha$ :

$\text{cos}^2\alpha=1-\text{cos}^{2}(30^0)-\text{cos}^{2}(75^0)$

$\text{cos}^2\alpha=0.183$

(Take the square root of both sides)

$\text{cos}\,\alpha=0.4278$

$\alpha=\text{cos}^{-1}(0.4278)$

$\alpha=64.67^0$

To find the magnitude of the components, we simply substitute the angle values like so:

$F_x=3\text{cos}\,64.67^0\,=\,1.28$ kN

$F_y=3\text{cos}\,30^0\,=\,2.60$ kN

$F_z=3\text{cos}\,75^0\,=\,0.766$ kN

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-84.

Solution:

Let us first substitute the values given in the question into our diagram.

To calculate the value of \alpha we have to remember the following formula.

\text{cos}^2\alpha +\text{cos}^2\beta +\text{cos}^2\gamma =1

Let’s substitute the values we know. Thus, we have:

\text{cos}^2\alpha +\text{cos}^{2}(30^0)+\text{cos}^{2}(75^0) =1

Now we will solve for \alpha by isolating for \alpha:

\text{cos}^2\alpha=1-\text{cos}^{2}(30^0)-\text{cos}^{2}(75^0)

\text{cos}^2\alpha=0.183

(Take the square root of both sides)

\text{cos}\,\alpha=0.4278

\alpha=\text{cos}^{-1}(0.4278)

\alpha=64.67^0