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spatel600
Apr18-04, 02:09 PM
Please help with this problem:

http://labphysics.uncc.edu/Phys2101/001/Test3A-001-F94.html
#4

Please any help would be greatly appreciated.
AOL SN: spatel600

Please let me know. Thanks. Shaun

Doc Al
Apr18-04, 02:40 PM
This is a straightforward problem in static equilbrium. You have three unknowns, but you can write three equations:
(1) Sum of the vertical components of forces = 0
(2) Sum of the horizontal components of forces = 0
(3) Sum of torques about any axis = 0

Give it a try.

spatel600
Apr18-04, 02:47 PM
I tried that and I got Fx= Tcos() Fy = Fv + Tsin () = mg

Torque = : Lsin (150) T = L/2 (sin 90) mg

I have tried all options and I have been stuck for many hours now. I don't understand. Please help. Thanks

Doc Al
Apr18-04, 02:56 PM
Assuming you are using the correct angles, this looks OK to me. What's the problem?

Start with the torque equation. That will allow you to find T. Then use that to find the rest.

spatel600
Apr18-04, 03:02 PM
The numbers do not seem to be right. Everytime I put the answers in the practice test they are incorrect. Give it a try, maybe the teacher made a mistake.

Doc Al
Apr18-04, 03:04 PM
The numbers do not seem to be right. Everytime I put the answers in the practice test they are incorrect.
Tell me your answers and I'll check them out.

spatel600
Apr18-04, 03:16 PM
My numbers were T= 980 which is 490/sin (150)
Fh = 848.71 which is T cos (30)
Fv = 490 which is mg-Tsin(30)

Doc Al
Apr18-04, 03:26 PM
My numbers were T= 980 which is 490/sin (150)
Fh = 848.71 which is T cos (30)
Fv = 490 which is mg-Tsin(30)
What units are you using for force?

I'll bet I can guess what you're doing wrong: you are calculating mg as 100x9.8 = 980, right?

If so... oops. The weight of the beam is given as 100 lbs; that means mg = 100 lbs.

spatel600
Apr18-04, 03:31 PM
hahah thanks so much. I can't believe i made that stupid mistake. I really appreiciate all you help. Thanks. Shaun