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)_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.
(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.
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:
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.