A non-uniform is 3.8-m long and has a weight of 560.-n.

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The problem involves a 3.8-meter non-uniform bar weighing 560 N, balanced at its center with a 340 N weight suspended 0.70 m from one end. To find the center of gravity, the sum of torques must be calculated, equating the torques on either side of the pivot point. An attempt was made to solve the equation by multiplying the bar's length and weight, but the method used was incorrect. The unknown distance to the center of gravity needs to be determined for a proper solution. Understanding the principles of torque and balance is essential for solving this problem effectively.
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Homework Statement



A non-uniform is 3.8-m long and has a weight of 560.-n. The bar is balanced in a horizontal position when it is supported at its geometric center and a 340.-n is suspended 0.70-m from the bar's light end. Sketch this system and find the bar's center of gravity.

Homework Equations



Sum of torques.

Autosum of Tc=Tcc

The Attempt at a Solution



I have no clue. We've done examples like this; however, not like this one. :/
 
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I did try, however, multiplying (3.8-m) (560.-n)= (340.-n)(0.70mx) That answer was totally wrong.

x- unknown
 

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