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1. Oct 29, 2016

### inund8

1. The problem statement, all variables and given/known data

2. Relevant equations

ΣFy = 0
ΣFx = 0
TanΘ = sinΘ/cosΘ
3. The attempt at a solution

See above, I get Θ=78.23° and weight is 28.96lb There's a bit of work not shown here b/c it was super messy, but basically I took the eqns that equalled sinΘ and cosΘ and divided them so that TanΘ = sinΘ/cosΘ. I did check my math, but somehow my checks tend to often be wrong. Thanks all!

2. Oct 29, 2016

### Staff: Mentor

Starting from your equations for $\sin \theta$ and $\cos \theta$, I get $\tan \theta = 4.8$, which gives an approximate value for $\theta$ that agrees with what you show (78.23°).
(Later edit: fixed my typo of 73.23°)

I didn't verify the work before your sine and cosine equations, so it's possible you have an error in one or both of them. I didn't verify your value for WA, either.

Last edited: Oct 30, 2016
3. Oct 29, 2016

### inund8

Are there any problems with my assumptions? Mostly TBCy = WA, and the cosθ and sinθ eqns.

4. Oct 30, 2016

### David Lewis

θ - 90o = 0.5 * arctan 5/12.
Tension in BD = 2 * weight * cos (0.5 * arctan 5/12) = 100 lbs.

Last edited: Oct 30, 2016
5. Oct 30, 2016

### inund8

Mark44 - My value was Θ=78.23°. Just a typo right?

So your eqns evaluate to 51.07lb and 78.69°. While I think that .5° discrepancy could be chalked up to rounding, I feel like 2.5+ lb difference cannot be. I guess I made a mistake assigning WA = TBC⋅12/13. WA should be TBDsinΘ = WA(25/13)?

6. Oct 30, 2016

### David Lewis

During the intermediate steps in your calculations, keep more significant figures than are justified by the precision of the given data. Then round off the final answer to the appropriate number of significant figures as the last step.

7. Oct 30, 2016

### inund8

I've been carrying 1 extra sf, generally. Should I be reworking my answer?

8. Oct 30, 2016

### Staff: Mentor

Yes. I read one thing and typed another. 78.23° is what I got (rounded to 2 dec. places). I have edited my earlier post.

9. Oct 30, 2016

### inund8

That happens! I'm redoing it, since I'm pretty sure my sin θ eqn is off

10. Oct 30, 2016

### inund8

Alright I got it! Thanks all! Once I changed the sin eqn to be sinΘ = WA(25/13)TBD it all worked fine. The 1/13 lb difference won't show didn't show in the angle, but it compounded and showed in the final answer!