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?

Question

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?

Answer 1

Assuming the fishing pole is horizontal...

Yes, the fisherman's left hand is exerting a downward force. You are correct.

Answer 2

Since the moments were calculated from the end the fisherman is holding in his left hand, that force does not generate any moment.

Answer 3

Thanks Quidditch!

Answer 4

You are very welcome!