After the first 6 steps of this problem, I am asked to show the sum of total for
ID: 1860713 • Letter: A
Question
After the first 6 steps of this problem, I am asked to show the sum of total forces F(a)+F(b)+F(c) to equal 0.
The sum of the x-component total force is:
|F(a)|cos40 + |F(b)|cos195 + 0 = 0
The sum of the y-component total force is:
|F(a)|sin40 + |F(b)|sin195 + 60 (kN) = 0
Then right after this last line ^ the next step says "Solving equations (1) and (2)" (which are those previous sum equations), and then give the answers:
The magnitude of force F(a) is |F(a)| = 137 kN
The magnitude of force F(b) is |F(b)| = 109 kN
I'm curious how these two magnitudes were calculated from the sum of forces equations shown above?
Explanation / Answer
Multiplying first eqn by Sin40 and second eqn by cos40 and then subtracting them gives,
Fb*(Cos195*sin40 - sin195*cos40) = -60*Cos40
Solving this, Fb = 109 kN
Putting it in any of the given eqn, we can find Fa.