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#55
Sep2907, 11:56 AM

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#56
Sep2907, 11:57 AM

P: 36




#57
Sep2907, 11:59 AM

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#58
Sep2907, 12:00 PM

P: 36




#60
Sep2907, 12:21 PM

P: 36



#61
Sep2907, 12:48 PM

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300*L=L/2*T*Sin50 becomes 300=1/2*T*Sin50 if you cancel the L on both sides. (Which means dividing both sides by L.) 


#62
Sep2907, 01:34 PM

P: 36



#63
Sep2907, 01:44 PM

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P: 41,579

If you pick the left side of the bridge as your pivot, you will have the three axial loads producing the clockwise torques and the force F2 producing the counterclockwise torques. Find those torques just like in all the previous problems. 


#64
Sep2907, 01:58 PM

P: 36

Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?



#65
Sep3007, 08:13 AM

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I encourage you to study the examples discussed on the webpage I linked in post #37. 


#66
Sep3007, 09:07 AM

P: 36

Meaning that ths distance from the pivot is 5.25M+5.25M? 


#67
Sep3007, 09:16 AM

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No, the distance of F2 from the left pivot point is: 7m + 3.5m + 5.25m + 5.25m = 21m. (You didn't even have to do that addition: the problem states that the bridge supports are 21m apart.)



#68
Sep3007, 09:20 AM

P: 36




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