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**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

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

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,657

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?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

Of course. I'm just happy that you give it a try.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,657

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)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

OK! THIS IS ALL SOLVED NOW! No more problems!

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