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Topic review (newest first)
- GurraTedden
- 2005-07-24 21:11:51
OK! THIS IS ALL SOLVED NOW! No more problems!
- MathsIsFun
- 2005-07-23 14:09:54
OK, you should be able to solve the bar on its own, then proceed to solve the cylinder later
How about another moment formula?
∑Mb = 1.0(Fa cosθ - Na sinθ) - 0.25 mg cosθ
(better check I have that oriented correctly)
Then you have three equations and three unknowns:
∑Fx = Fa - Nb sinθ + Fb cosθ
∑Fy = Na - mg + Fb sinθ + Nb cosθ
∑Mb = 1.0(Fa cosθ - Na sinθ) - 0.25 mg cosθ
(known: Nb=141.1, mg=24 x 9.81, sinθ=0.6, cosθ=0.8)
- GurraTedden
- 2005-07-23 08:43:56
Of course. I'm just happy that you give it a try.
- MathsIsFun
- 2005-07-23 08:16:34
Hi Gustav,
I have had a brief look at the first problem, and understand it enough to know that I need to sit down quietly with a pad and paper to solve it. Can it wait a day?
- GurraTedden
- 2005-07-23 05:30:59
Uhum, I was just hoping that there's someone knowing something about
mechanics, or just can flick an eye on my calculations for this problem.
MathIsFun and mathysperson I'm sure will recognize one of these two problems.
http://engman.bravehost.com/mechanics/2probs.JPG
On 7.14 you can have a look of how far I've got, but I'm stuck.
http://engman.bravehost.com/mechanics/incomplete_solution714.JPG
7.39 I haven't a clue how to do it. The answer is: angle between 32.8 and 45 degrees. The 45 degrees
seems logic, and I guess that should calculate the intuition way. I'll buy that.
IF this is the wrong forum, if you wanna strictly do math around here, I would really appreciate a similar
forum with mechanics.
/Mr. G