Recent content by Brendan Webb

MODERATOR'S NOTE: NO TEMPLATE BECAUSE THREAD WAS SUBMITTED TO ONE OF THE REGULAR FORUMS Summary: Calculate the air change rate required to keep a certain temperature In question d above, you were asked to calculate how long the air flow in the system would be active in a day. Now factor in...

Thanks!

Homework Statement Consider an equilateral triangle of side 15.6cm. A charge of +2.0μC is placed at one vertex and charges -4μC each are placed at the two. Determine the electric field at the centre of the triangle. Homework Equations E=kQ/r^2 The Attempt at a Solution I am hoping someone...

Thanks for the reply, I believe I solved the problem (or at least I sent in my assignment with my interpreted answer). Next time I post I will include the diagram and make my thoughts extra clear. Cheers

If the unbraced length of a beam is 3 m and the maximum bending moment in this unbraced segment is 312 kN-m, and also the maximum moment in the braced segment of the beam is 350 kN-m, select an economical section just based on the moment requirements. Use Table A.2M included in the course...

Do i have the ΣMb equation right if I wrote it equal to 0? Ok so for the Fy = 0 equation it would be -Ay -P1 - P2 +By = 0 For the Fx = 0 equation it would be -Ax + Bx = 0 (However, is Ax and Bx the total force across the cable or are the broken up into three components?)

I have come to the point in the equation where I need the moment equation to solve for Ay. I am not sure how to get By in term of Ay though. The moment equation I get for Ay is: ∑Mb = 8(-P2) - 20(-P1) - 32(Ay) + 15Ax I know P1 = 270 - P2 However I am not sure if I even have Ay and Ax right...

ahh yes, I was having a hard time understanding. Thanks!

Hey steve, how did you decide to use the 1.6m when dividing by the torque force

Good to know. Thanks and all the best on your next courses!

Oops misread response. Is the test easier in comparison to the assignment questions?

Steve - We must be doing the exact same course through Athabasca university because every problem I have troubles with I come look at here and I often see you have posted in the thread. Although it looks like you might be done by now. Thanks for the help

yes I can, and did. not only was I forgetting to make it 4v^2, I was forgetting to divide the v^2 by 0.5 to give me 2v^2 for a grand total of 6v^2 and -8v giving me the exact same proportions and answer on the quadratic equation shown by v2' (3v2' -4). Wow i wasted 2 hours stuck in a mind trap...

Every time I try to substitute the formulas of v1' = 2 - (2)v2 into 2 = 0.5v1'^2 + v2' I get what is shown on the image which doesn't give the right answer when solved through the quadratic formula, Where am i going wrong?

Thank you both for your time and patience! I finally understand! :) Cheers