(a+b)(Fg)-b(Fcosθ) = 0
-->solve for F
Have you done #1 or #3 of the assignment? If you have, please help!
No but I am going crazy trying to do the calculations! Grr...wish I had more time to review before doing this h/w!
Thanks though. If I figure it out i'll post back here, ok
Hey Anna, can you break that down. I'm so confused!
Fg in the eqn she gave is mg...
Did any of you guys find 1 or 3?
So, basically i wrote down the torque equilibrium equation where torque equals 0 and the point of reference being the one shown in the diagram. The distance from the point of reference to where Fg is applied is a PLUS b since a is just the distance from the point where F is applied...NOT from the point of reference. The signs you can find using the right-hand rule..i used positive as counterclockwise..so torque is pointing towards you.
Anna if u are having trouble with number 3, have a look at Ex5.2 it will help.
these are the steps to do number 3. i still have yet to try it though:
first you need all the euations of the forces
so horizontal and vertcal
then you need torque
so once you hv your torque youll hv lots of L
you should factor that out
nd solve for T
then sub T into the horizontal forcees equation
nd solve for Rx
And for Question 2 are we supposed to convert cm to m and when we find b, there's no given units so am i supposed to just multiply it with a and it automatically becomes the same unit as a?
the units are given.. it's cm throughout..you don't have to convert because the units for l get canceled out in the end anyways
i got an answer around 3000 for #3? Anyone else get something similar?
yep, it's supposed to be a huge number for #3
i got around 4000s. Hey for question 2 is it supposed to be a tremendous number as well. I got like 4450
Did anyone figured out #1?
No! I've tried multiple ways to do it.. and yet still not getting the right answer. I think I'm messing up my coordiante system and the angles for the forces..
T(l1)cos30 + Warm(l2)cos30 - WM(l3)cos30
solve for T