Mech - Girder Section - Picture included

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The discussion revolves around a mechanics problem involving a girder section and the balance of weights on either side. It questions whether the weights are equal or if they should be calculated using dimensions represented by letters. Participants clarify that the weights are not equal and suggest using moments to analyze the situation, even without specific distances. A method is proposed to simplify calculations by assuming equal distances for the triangular sections. The conversation highlights the importance of understanding support structures and encourages using unit assumptions for solving the problem.
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Homework Statement



http://img132.imageshack.us/img132/8958/mechanicsgv3.jpg

The Attempt at a Solution



(i) Is the weight balanced on both sides in a situation like this, in which case it would be:

(6000 +3000) / 2 = 4500N on A and 4500N on G

Or are the weights not equal and infact I am meant to work it out with dimensions in terms of letters, for example A-C?

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(ii) Could anyone tell me where i can learn about these support structures, i brought a mechanics book but it doesn't cover and after looking on the internet i can't find any good sources of information to teach me.
 
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Spadez said:
Or are the weights not equal and infact I am meant to work it out with dimensions in terms of letters, for example A-C?

Hi Spadez! :smile:

No, they're not equal.

Hint: take a moment.
 
Ok, so i take moments but i have to use the distance with respect to the letters, I am not given any distances. Can i still do it like that?
 
hmm … assuming the triangles are all the same, just say "let AC = CE = EG = 1 unit" … the units will cancel in the end :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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