Resolution for forces

Question

I'm stuck trying to find the resultant of the force system. Here it is said to find the angle θ the resultant manes with the horizontal. But θ here is narked relative to F₂. How do I find the resultant and θ

Answer 1

In order for the Resultant force to be vertical you only need the sum of the horizontal components of $F_1$ and $F_2$ to be zero

$$F_{1x} - F_{2x} =0 $$ $$F_{1}\cos(70) - F_{2}\cos\theta =0 $$

$$\cos\theta=\frac{F_{1}}{F_{2}}\cos(70) - F_{2} $$

$$\theta=\arccos\left(\frac{F_{1}}{F_{2}}\cos(70) - F_{2}\right) $$

if you substitute you should get $$\theta=50[deg]$$

And the resultant R should be :

$$F_{1y} + F_{2y} =R $$ $$F_{1}\sin(70)+ F_{2}\sin(50) =R $$ $$R \approx 1077 [lb]$$

Answer 2

I don't have time to type this up nicely - thought it would be quicker to write out by hand but in hindsight my handwriting is terrible. Let me know if you can't read anything!

Answer 3

A problem with meaningful solution must have a key with enough conditions/constraints, that make it soluble. As a problem solver, we need to find/identify the key and the associated conditions/constraints. Sometimes work a problem in the backward manner can be helpful. The steps are as follows:

1. Expand the unknown (R = ?) to a more detailed expression. That is, in an orthogonal grid system,

R = (Fx^2 + Fy^2)^1/2; in which, Fx = F1cos70 + F2cos(\theta), and Fy = F1sin70 + F2sin(\theta)

2. Identify/find the key. In this problem, the angle (\theta) is the key, and the statement that says "Determine the angle (\theta) which makes the resultant (R) of the two forces vertical, is the furnished condition/restraint. Laterally the statement tells us to find the angle (\theta) that will result in: sum Fx (horizontal resultant force) = 0, so the relationship can be expressed as,

Fx = F1cos70 + F2cos(\theta) = 0

Now you have the key to solve the problem. Hope this helps.