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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.

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.

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