The members of a truss are pin connected at joint O. Determine the magnitudes of F_1 and F_2 for equilibrium. Set ϴ = 60°.

#### Solution:

### Let us first draw a free body diagram and label our forces like so:

### As stated in the question, we set ϴ = 60° in our diagram.

### We will also pick which sides are positive. In this case, we will assume \rightarrow^+ is positive and \uparrow+ is positive.

### Now, we will write each force in unit vector form as follows:

### F_1=(F_1\text{cos}60^0)i\,+\,(-F_1\text{sin}60^0)j

### F_2=(F_2\text{sin}60^0)i\,+\,(F_2\text{cos}60^0)j

### F_3=(-5\text{cos}30^0)i\,+\,(-5\text{sin}30^0)j

### F_3=(-7\frac{4}{5})i\,+\,(-7\frac{3}{5})j

##### (Note the negative signs. When the component of a force is towards the negative quadrant, we set it as negative, since we assumed that \rightarrow^+ is positive and \uparrow+ is positive)

### For the forces to be in equilibrium, the x and y (i and j in unit vector form) forces must add up to zero. Mathematically, we can write this as:

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

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

### Let us now only look at the x forces (unit vector i):

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

### 0\,=\,0.5F_1\,+\,0.866F_2\,-\,4.33\,-\,5.6

##### (Here, we simplified all the trigonometric values) Simplify further:

### 0\,=\,0.5F_1\,+\,0.866F_2\,-\,9.93 ——- Eq.(1)

### Now, we will look at the y forces (unit vector j):

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

### 0\,=\,-0.866F_1\,+\,0.5F_2\,+\,2.5\,-\,4.2

##### (Again, we simplified all the trigonometric values) Simplify further:

### 0\,=\,-0.866F_1\,+\,0.5F_2\,-\,1.7 ——- Eq.(2)