# Resolving Forces

1. Jan 14, 2008

### america8371

1. The problem statement, all variables and given/known data
A. Resolve the 360-lb force into components along the cables AB and AC. Use $$\alpha$$ = 55 and $$\beta$$ = 30.

B. The supporting cables AB and AC are oriented so that the components of the 360-lb force along AB and AC are 185 lb and 200 lb, respectively. Determine the angles $$\alpha$$ and $$\beta$$.

2. Relevant equations
I think I need to use
Parallelogram law: A + B = C

3. The attempt at a solution
I tried to rearrange the forces so that I could use the parallelogram law but i didn't know where to go from there. I seemed pretty simple when I first looked at it but I guess I just don't know where to start.

Last edited: Jan 14, 2008
2. Jan 14, 2008

### rock.freak667

There are two additional forces in the diagram that you need to know; the tension in the cables.Those are the forces you need to resolve

Last edited: Jan 14, 2008
3. Jan 14, 2008

### america8371

Yeah I tried
F$$_{AB}$$ = 360(cos 55) = 206.5
and
F$$_{AC}$$ = 360(cos 30) = 311.77
but I didn't think they were correct.

4. Jan 14, 2008

### belliott4488

Can you draw the vector diagram with all three forces that act at pt. A? Once you do that, then draw the separate horizontal and vertical (x and y) components for each vector. Now, since all three forces must sum to zero (since the point A is not moving), you can invoke the fact that all the horizontal components must add to zero, and all the vertical components must add to zero. Make sense?

5. Jan 14, 2008

### america8371

Yeah it kind of makes sense, but wouldn't i need at least one angle to find the x and y components in the second question?

6. Jan 14, 2008

### belliott4488

Nope - the same relations should still work, just in the other direction. Basically you have two knowns and two unknowns in both cases: two known angles and two unknown force magnitudes in the first question, then two known force magnitudes and two unknown angles in the second question. You can go either direction, so long as you have two independent equations, which is what you should get from balancing the vector components in the x- and y-directions.

7. Jan 14, 2008

### stewartcs

The relevant equation you'll need is Newton's Second Law.

$$F_{net} = ma$$

Caveat: Make sure you use the correct angles when resolving the vectors.

CS

8. Jan 15, 2008

### america8371

Am I taking the right step in getting the x and y components?

F$$_{x}$$ = 200(sin$$\beta$$)
F$$_{y}$$ = 200(cos$$\beta$$)

F$$_{x}$$ = 200(sin$$\alpha$$)
F$$_{y}$$ = 200(cos$$\alpha$$)

9. Jan 16, 2008

### belliott4488

Almost .... you don't want 200 lbs in all cases. That's the force along the AC direction in part B. of the question, right? What about the 185 lbs acting along AB?

Also, I haven't seen where you've answered part A. yet. In that one, you know the vertical force downward at pt. A (360 lbs), so that tells you the sum of the vertical components of the two forces along the lines. What about the horizontal components? What should they sum to?

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