# find the tension in each cable...

• Aug 24th 2010, 06:49 PM
slapmaxwell1
find the tension in each cable...
ok so the problem wants me to find the tension in each cable, supporting the given load. not too bad of a problem...i have this triangle shaped object with the vertices at A B and C...to get the respected vectors, i call the side CA, U and the side marked CB, V. so there CA = Ucos 130 + Usin 130 and then CB = Vcos 30 + Vsin30, once i break these vectors into horizontal and vertical components i dont know how to solve for U and V...any help would be appreciated. thanks in advance.Attachment 18711
• Aug 24th 2010, 07:43 PM
Quote:

Originally Posted by slapmaxwell1
ok so the problem wants me to find the tension in each cable, supporting the given load. not too bad of a problem...i have this triangle shaped object with the vertices at A B and C...to get the respected vectors, i call the side CA, U and the side marked CB, V. so there CA = Ucos 130 + Usin 130 and then CB = Vcos 30 + Vsin30, once i break these vectors into horizontal and vertical components i dont know how to solve for U and V...any help would be appreciated. thanks in advance.Attachment 18711

If the system is in equilibrium, the nett vertical and horizontal forces are 0. Call the tensions in AC=p and CB=q

Vertically,

we have p sin 50 + q sin 30 = 3000 (9.81) ---1

Horizontally, we have

p cos 50 = q sin 30 -----2

Solve the simultaneous equations.
• Aug 25th 2010, 03:15 AM
slapmaxwell1
thanks for your reply, but i still do not know how to solve the equations. i apologize for my messy handwriting, i think this is why we have different numbers you can see the original problem at calcchat.com (please select free access) select 9th edition calculus 11.1 problem number 89....i have the same setup as the book, i just dont see how they solved the equations..ill try to rework it and then come back. thanks again.
• Aug 26th 2010, 05:22 AM
slapmaxwell1
i figured it out...once i got it i saw how easy it was..sorry for the wasted post...lol