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Static Equilibrium and tension of a cord

  • #1
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

  • #2
631
0
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?
 
  • #3
Cord 1 is on the left
 
  • #4
631
0
k. then L should be 4.4
 
  • #5
so is the torque equation correct if i use 4.4
 
  • #6
631
0
dimensionality???
 
  • #7
I'm afraid I don't understand
 
  • #8
631
0
Torque: -mgL-T2*cos(27.5)=0
T2*cos(..) has dimensions of force and mgL has dimensions of work.
 
  • #9
so would torque be the length of the bar*-mg
 
  • #10
631
0
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:

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