The members of a truss are pin connected

The members of a truss are pin connected at joint O. Determine the magnitude of $F_1$ and its angle ϴ for equilibrium. Set $F_2$ = 6 kN.

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

Solution:

In the previous solution, we solved the question using unit vectors. In this solution, we will avoid using them.

(unit vectors are not necessary to solve these questions, however, they help a lot when dealing with questions involving the z-axis).

Let us first draw the free body diagram with the values given to us in the question like so:

We will also assume $\rightarrow^+$ is positive and $\uparrow+$ is positive.

Now, let us write the forces in the x-axis domain as follows:

$\sum \text{F}_\text{x}=0$

$6\text{sin}\,60^0\,+\,F_1\text{cos}\,\theta \,-\,5\text{cos}\,30^0\,-\,7(\frac{4}{5})\,=\,0$

(isolate for $F_1$ )

$F_1=\frac{4.734}{\text{cos}\theta}$

Now, we can write the forces in the y-axis domain as follows:

$\sum \text{F}_\text{y}=0$

$6\text{cos}\,60^0\,-\,F_1\text{sin}\,\theta \,+\,5\text{sin}\,30^0\,-\,7(\frac{3}{5})\,=\,0$

(simplify by solving the trigonometric values)

$F_1\text{sin}\,\theta \,=\,1.3$

We can now simultaneously solve both equations by substituting the value of $F_1$ from the first equation into the second equation.

$\frac{4.734}{(\text{cos}\theta)}\text{sin}\theta=1.3$

(Remember $\frac{\text{sin}\theta}{\text{cos}\theta}=\text{tan}\theta$ )

$\text{tan}\theta=\frac{1.3}{4.734}$

$\theta=\text{tan}^{-1}(\frac{1.3}{4.734})$

$\theta=15.36^0$

Now that we have the value of $\theta$ we can substitute this back into our equation, $F_1=\frac{4.734}{\text{cos}\theta}$

Doing so gives us:

$F_1=\frac{4.734}{\text{cos}(15.36)^0}$

$F_1=4.91$ kN

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-2

Solution:

In the previous solution, we solved the question using unit vectors. In this solution, we will avoid using them.

(unit vectors are not necessary to solve these questions, however, they help a lot when dealing with questions involving the z-axis).

Let us first draw the free body diagram with the values given to us in the question like so:

We will also assume \rightarrow^+ is positive and \uparrow+ is positive.

Now, let us write the forces in the x-axis domain as follows:

\sum \text{F}_\text{x}=0

6\text{sin}\,60^0\,+\,F_1\text{cos}\,\theta \,-\,5\text{cos}\,30^0\,-\,7(\frac{4}{5})\,=\,0

(isolate for F_1)

F_1=\frac{4.734}{\text{cos}\theta}

Now, we can write the forces in the y-axis domain as follows:

\sum \text{F}_\text{y}=0

6\text{cos}\,60^0\,-\,F_1\text{sin}\,\theta \,+\,5\text{sin}\,30^0\,-\,7(\frac{3}{5})\,=\,0

(simplify by solving the trigonometric values)

F_1\text{sin}\,\theta \,=\,1.3

We can now simultaneously solve both equations by substituting the value of F_1 from the first equation into the second equation.

\frac{4.734}{(\text{cos}\theta)}\text{sin}\theta=1.3

(Remember \frac{\text{sin}\theta}{\text{cos}\theta}=\text{tan}\theta)

\text{tan}\theta=\frac{1.3}{4.734}

\theta=\text{tan}^{-1}(\frac{1.3}{4.734})

\theta=15.36^0

Now that we have the value of \theta we can substitute this back into our equation, F_1=\frac{4.734}{\text{cos}\theta}