Statics Homework: Solving for Weight with ΣFy = 0 and ΣFx = 0 Equations

[inund8]

Homework Statement

Homework Equations

[/B]
ΣFy = 0
ΣFx = 0
TanΘ = sinΘ/cosΘ
https://lh3.googleusercontent.com/QUtPJ9Zyv4SeBRZ3Cizb1QNWCxLzp_dah-xusJupCgOn-nfMzxL95v-ydqO4Hpam77IiXSlyLF6ENSkxmaAuKKq0K0QcZws3YVpSsG-BRamO1Lo-BIacuKQ32MRIKpQ8e6ZP2La0YZz9uFZNgrjMZg6u1IKgEOKv50SFwAAXR8-UIMQc9xSEtZStUKkdgj90J-K_cpDDRywW5XHGSm_UZVfhggDrdS0RQLpw8ui4VrJ3bar6HzALDlW4ILKyWs9mL4hr_u1caTY4K4iKavjayzgn9SBiwsbFFD9hLhjd2e8Ym76elx4P98mal8k6-7fTmjS6HcUOgnS9FnCHJuhqi2JvuB7I1m1udMOGHpfeXypGp6QVb35KyjJYJGp51biD84BcKTWanUMVT3Y5QTRgT6wzmvW7wmwFfEspVS5IYSusMeepZnbM0WiJ3ZX6YK7tWLT_KmbGXuIMXccs0lc-eKHbdH4x8UldG634t3c33Y4Cb5lSATmcgSc-I_R3JKZD3fbrARfKsk5fL2sHQltS4yTf93DIeR_FRbpSRYgM4qw5hG60U-BqCaJZg7Y6_YHBOOAsPMHG54H0LfnfRoNFviY2IEE5SnmMUVWfRYUvIZWaROrm=w1370-h770-no
https://lh3.googleusercontent.com/7tIrg9d4i8YduYP9wkblnarTuCMkB-0wYT5yWAcaoeLV3LNRVlqM9d6ZURzEHIzT9paboOJ99n-ihg3tR8LJxHXFofzYgro957Vu5C9jJCU-Q3eGDpV4_hUP_1m-vZ55i5lsev_mRur89vOm46nkM7g7bX94-oGuj14BpecieJ25gzaGbocpQfKkYp9SyHDhWQrcslUUkf7KQNuXULXOaDABfyEnYljSntEVMyyBjV3RHGeCZ7e_CBYpfbV2aMKYqauLZ-0p_0id8coNO24_mWC51m21I1Nv7t3cPYkMSZ0G4dpcWCL9TtZOLTJ5-xGz9diCiGljF60nkXcodXhKPz4HvhVTTkVfeSIMDuavm4eki8aoBifIWhiE7c4UVpfC0Z4Lw_mwKAccpNZMtXK1Qm2FzSJel_qIUFaZpu35wtQDAcoXAfGM13Hoz4EqFdQbd2gcTiM5fSXOoyw2Q1aNhd4VW6n8R9usVYA9VYhLXkKPETXpa5sQnL_Fb5V2J0GA1nf5zwM_REs4vu6Llq8b4U-IXfIsP_vPzUWQA3eMw-JBTe_y3CyXWPUFniYpdddA-AEyTDy-Zojctqpi2Pe5EYOMxUGs4deRfKrueT8sz71AVQa8=w1370-h770-no
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!

 

[Mark44]

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 W_A, either.

 

[inund8]

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

 

[David Lewis]

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

 

[inund8]

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 (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 W_A, either.

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

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

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 W_A = T_BC⋅12/13. W_A should be T_BDsinΘ = W_A(25/13)?

 

[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.

 

[inund8]

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

 

[Mark44]

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

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.

 

[inund8]

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

 

[inund8]

Alright I got it! Thanks all! Once I changed the sin eqn to be sinΘ = W_A(25/13)TB_D 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!