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July 31, 2015

Homework Help: Physics

Posted by Jennifer on Tuesday, December 7, 2010 at 11:12pm.

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)-10.5-200=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?

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