# Static Equilibrium and tension of a cord

A non-uniform bar of weight W = 12.5 N is suspended at rest in a horizontal position by two massless cords. The angle φ between cord 2 and the vertical is equal to 27.5°. The bar has a length L = 6.0 m and the distance of its center of gravity from the left-hand end of the bar is 4.4 m. What is the tension in cord 2?
img:http://i242.photobucket.com/albums/ff106/jtdla/prob01a-2.gif [Broken]

My lab partner and I set up this equation:
Fx: T2*cos (27.5)-T1*cos(theta)=0
Fy: T2*sin(27.5)+T1*sin(theta)-mg=0

Torque: -mgL-T2*cos(27.5)=0

I'm not sure if that L is 4.4 or 6, nor do I know if I am starting this properly.

Last edited by a moderator:

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
Eqn for Fx and Fy are correct, but torque is incorrect; firstly, dimensionally its wrong and secondly, which is cord 1? left side one or right side one?

Cord 1 is on the left

k. then L should be 4.4

so is the torque equation correct if i use 4.4

dimensionality???

I'm afraid I don't understand

Torque: -mgL-T2*cos(27.5)=0
T2*cos(..) has dimensions of force and mgL has dimensions of work.

so would torque be the length of the bar*-mg

torque is : -mgL + T2cos(27.5)*6 (6 is the perpendicular distance of force vector from the point at which you are taking torque)

Last edited: