The bolt is subjected to the force F

The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and α=60° and γ=45°, determine the magnitudes of its components.

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

Solution:

To solve this question, we will have to use the following equation.

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

We will now isolate for $\text{cos}\beta$ .

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

As stated in the question, we will now substitute the values given to us.

$\text{cos}\beta =\sqrt{1-\text{cos}^{2}60^0-\text{cos}^{2}45^0}$

Solve for $\beta$ .

$\beta=120^0$

The next step is to figure out the magnitudes of F’s components knowing that the magnitude of F is 80 N as stated in the question.

$F_x=\mid80\text{cos}60^0\mid= 40N$

$F_y=\mid80\text{cos}120^0\mid= 40N$

$F_z=\mid80\text{cos}45^0\mid= 56.6N$

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-63.

Solution:

To solve this question, we will have to use the following equation.

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

We will now isolate for \text{cos}\beta.

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

As stated in the question, we will now substitute the values given to us.

\text{cos}\beta =\sqrt{1-\text{cos}^{2}60^0-\text{cos}^{2}45^0}

Solve for \beta.

\beta=120^0