Thread: 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?

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?

T1= cos50
T2=cos30

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

-Dan

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?

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

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 )