# The magnitude of the resultant force acting on the bracket

The magnitude of the resultant force acting on the bracket is to be 400 N. Determine the magnitude of F1 if Φ = 30°

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 draw out the vector components as follows:

We drew this image quite large, because it’s really important to see the x and y components of each of the force vectors. We also set Φ to 30° as stated in the question.

Now, we can write the components of each force, $F_1$$F_2$$F_3$, using our diagram.

$(F_1)_x=F_1\text{cos 30}^0=0.8660F_1$

$(F_1)_y=F_1\text{sin30}^0=0.5F_1$

$(F_2)_x=650(\frac{3}{5})=390\,N$

$(F_2)_y=650(\frac{4}{5})=520\,N$

$(F_3)_x=500\text{cos 45}^0=353.55\,N$

$(F_3)_y=500\text{sin 45}^0=353.55\,N$

To find the resultant, we will sum the x and y forces. We simply add the x components together, and we add the y components together, and in this case, we chose up and to the right as the positive side.

$+\rightarrow\sum(F_R)_x=\sum(F_x)$

$(F_R)_x=0.8660F_1-390+353.55$

$(F_R)_x=0.8660F_1-36.45$

$+\uparrow\sum(F_R)_y=\sum(F_y)$

$(F_R)_y=0.5F_1+520-353.55$

$(F_R)_y=0.5F_1+166.45$

The question states that the magnitude of the resultant, $F_R$ is 400 N. Therefore, we can write the following using the Pythagorean theorem.

$F_R=\sqrt{(F_R)_x^2+(F_R)_y^2}$

$400=\sqrt{(0.8660F_1-36.45)^2+(0.5F_1+166.45)^2}$
(simplify)

$F_1^2+103.32F_1-130967.17=0$

Solving for $F_1$ gives us:

$F_1=314\,N$ or $F_1=-417\,N$

Notice that we got a negative value as well. This simply means the direction we chose for $F_1$ is opposite to that of what we chose.