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

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SUMMARY

The discussion centers on calculating the center of gravity for a non-uniform bar that is 3.8 meters long and weighs 560 Newtons. The bar is balanced at its geometric center with an additional weight of 340 Newtons suspended 0.70 meters from the light end. The correct approach involves applying the principle of torque, specifically the equation for the sum of torques (ΣTc = Tcc). The user initially attempted a calculation that was incorrect, indicating a misunderstanding of torque application in this context.

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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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