Jan 20, 2008
Problem 5/73
I don't know if I am just making a mistake somewhere, but I have worked through this problem a few times and I keep getting to a result where Va = Wab*20"*sin(60) - Wbc*30"*sin(theta2) and this is not giving me the right answer. I am using Wab as theta dot, which I think is correct..? Anyone know what I might be doing wrong?
Subscribe to:
Post Comments (Atom)
3 comments:
I figured out what I was doing wrong.. Cosine does not equal opposite over hypotenuse. I should have stayed awake during geometry class..
how did you find an equation for r(b/c), or is there another way to approach this problem than the example on pg. 40?
My suggestion is to define an angle for link BC measured CW from line AC (call this angle "beta"). You can then write r_B/C = -30*cos(beta)*i + 30*sin(beta)*j. Carry through all of the calculations similar to that of the example on page 40. At the end, find the numerical value for the angle of beta using the law of sines: sin(theta)/30 = sin(beta)/20.
[For those of you in Division 2, we are referring above to an example on page 40 of the lecture notes for Divisions 1 and 3. That example is similar to Problem 5/73 if you were to make the length of BC equal to the length of AB.]
Post a Comment