# Cable Tension

• Jun 5th 2011, 05:20 PM
NeoSonata
Cable Tension
Cable tension question

Triangle with angle:

A is 50 degree
B is 30 degree

2000lb attached to it.

My setup:

T1= cos50
T2=cos30

T1sin50 + t2 sin30 = 2000

solving for T1

T2= T1 cos50/cos30

T1 ( sin50 + (cos50/cos30)sin30) = 2000

Is my set up correct?
• Jun 6th 2011, 02:22 PM
topsquark
Quote:

Originally Posted by NeoSonata
Cable tension question

Triangle with angle:

A is 50 degree
B is 30 degree

2000lb attached to it.

What is the orientation of the triangle? Where is the 2000 lb attached to it?

Quote:

T1= cos50
T2=cos30
These cannot possibly be correct. Did you mean something along the lines of T1 = 2000*cos(50)? etc.

-Dan
• Jun 7th 2011, 12:40 AM
NeoSonata
Triangle is facing down w/ 2000lb attached to it.

A 50 degree __ B 30 degee
\/
2000lb

|t1|cos50 = |t2|cos30

Equation is
|t1|sin50 + |t2|sin30 = 2000

|t2| = |t1| cos50/cos30

plug this in to the quation above

|t1| ( sin50 + (cos50/cos30)sin30 ) = 2000

|t1| = 1758.77

Is t1 correct?
• Jun 7th 2011, 12:09 PM
topsquark
Quote:

Originally Posted by NeoSonata
Triangle is facing down w/ 2000lb attached to it.

A 50 degree __ B 30 degee
\/
2000lb

|t1|cos50 = |t2|cos30

Equation is
|t1|sin50 + |t2|sin30 = 2000

|t2| = |t1| cos50/cos30

plug this in to the quation above

|t1| ( sin50 + (cos50/cos30)sin30 ) = 2000

|t1| = 1758.77

Is t1 correct?

Looks good to me. :)

-Dan
• Jun 7th 2011, 12:18 PM
e^(i*pi)
Are you using lb as a unit of mass or force?

It will need to be the latter to work properly

(I come from Europe so I'm used to kg :p)