Solve for the moment of this Torque

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SUMMARY

The discussion centers on calculating the moment of torque in a two-bar system, specifically analyzing the counteracting effects of counterclockwise (CCW) torque produced by 11.27 against the clockwise (CW) torque of 300. Participants highlight the importance of understanding the units of force and dimensions, noting that the effective lever arm for torque T is approximately 0.7 times that of the 300 torque. The conversation emphasizes the necessity of clear problem statements and unit consistency for accurate torque calculations.

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andychung0921
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Homework Statement
I have problem on T, is it T is clockwise so the answer is negative?
Relevant Equations
My answer is: M(A)= (Tv x 25)+(Tn x 15)-(300x40)

0=(T sin(5/13)) +(T cos(12/13))- 10200

T=11.27

Please clarify my answer, thank you
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Welcome to PF.

Can you please provide the complete problem statement? Is that a 2-bar mass of some sort, welded at the join? It looks like the dimensions are in imperial units (feet, inches), what are the force units in? What are the units in the calculations you have shown?
 
Welcome, Andy! :cool:

Why (300x40)?

Intuitively, do you believe that the CCW torque produced by 11.27 can counteract the CW torque produced by 300?

Note that the effective lever of T is about 0.7 the lever of 300.
 
Lnewqban said:
Intuitively, do you believe that 11.27 can counteract 300?
Well, technically, it depends on the units, eh? :wink:
 
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