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?
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!
To solve this problem, let's start by breaking it down step by step.
First, let's calculate the torque exerted by the 30 N fishing pole. We need to find the product of the force and the perpendicular distance from the pivot point to the line of action of the force. In this case, the center of gravity of the pole is 0.350 m from the heavy end, so the distance from the pivot point (left hand) to the center of gravity is 0.350 m.
Torque = Force x Distance
Torque1 = 30 N x 0.350 m
Next, let's calculate the torque exerted by the 100 N fish being lifted. The distance from the pivot point (left hand) to the fish is 2.00 m.
Torque2 = 100 N x 2.00 m
Now, let's calculate the total torque exerted on the fishing pole. Since the pole is in equilibrium, the sum of the torques on the pole must be zero.
Total Torque = Torque1 + Torque2 = 0
Substituting the values we calculated earlier:
30 N x 0.350 m + 100 N x 2.00 m = 0
Now, let's determine the force (F2) the fisherman needs to exert with his right hand to maintain equilibrium. The distance from the pivot point (left hand) to his right hand is 0.800 m.
Total Torque = F2 x 0.800 m
Solving for F2:
F2 x 0.800 m = -(30 N x 0.350 m + 100 N x 2.00 m)
Now, let's calculate F2:
F2 = (-(30 N x 0.350 m + 100 N x 2.00 m)) / 0.800 m
Evaluating this expression will give us the force (in Newtons) that the fisherman needs to exert with his right hand to maintain equilibrium.
Note: It's important to pay attention to the signs of the torques. Torques can be positive or negative depending on the direction of rotation they produce. In this case, if the left-hand side of the pole is the pivot point, clockwise torques (negative) and counterclockwise torques (positive) must balance each other to maintain equilibrium.