Physics
posted by Jennifer .
A 30 N fishing pole is 2.00 m long and has its center of gravity .350 m from the heavy end. A fisherman holds the end of the pole in his left hand as he lifts a 100 N fish. If his right hand is .800 m from the heavy end, how much force does he have to exert to maintain equilibrium?
I have:
F(_1) + F(_2) = 30N + 100N = 130N
I took a moment at F(1)
Then, the sum of the torques equals zero:
F(_2)(.8)30N(.350m)100N(2m)=0
(.8)F(_2)10.5200=0
(.8)F(_2)=210.5
F(_2)=263.125
If I solve for F(_1) I get a negative torque. I am not sure of this. Could we have a negative torque? Is the questions only asking for the force of the right hand?

Assuming the fishing pole is horizontal...
Yes, the fisherman's left hand is exerting a downward force. You are correct. 
Since the moments were calculated from the end the fisherman is holding in his left hand, that force does not generate any moment.

Thanks Quidditch!

You are very welcome!