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Obviously I want to know how far down my lure is. Thanks for the help

- Thread starter LTLhawk
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Obviously I want to know how far down my lure is. Thanks for the help

Homework Helper

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[tex] depth = L \times \frac{weight \ difference}{tension} [/tex]

Where L is the total length of the line, tension is the tension in the line, and weight difference is the weight of the lure minus the weight of an amount water that has the same volume as the lure.

But I think an easier way would be to measure the angle of the line going down into the water.

When the angle is defined such that it is zero when it lies on the water surface, then:

[tex] L\sin(\theta) = depth \ of \ lure [/tex]

Science Advisor

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[tex] depth = L \times \frac{weight \ difference}{tension} [/tex]

Where L is the total length of the line, tension is the tension in the line, and weight difference is the weight of the lure minus the weight of an amount water that has the same volume as the lure.

But I think an easier way would be to measure the angle of the line going down into the water.

When the angle is defined such that it is zero when it lies on the water surface, then:

[tex] L\sin(\theta) = depth \ of \ lure [/tex]

No .. I don't think that is correct. The buoyant force can probably be neglected in this example ... the upward force on the lure comes from the component of the line tension normal to the water surface. That must balance the gravitational forces, while the parallel component of the tension would balance the viscous drag (friction) force on the lure. So, it seems like there must be some additional information about the shape of the lure and it's cross-sectional area, or some other way of calculating the friction forces on the lure.

Anyway, the depth of the lure will be given by the distance behind the boat time the tangent of the angle between the line and the water's surface. That tangent will be equal to the ratio of the gravitational force to the drag force.

Homework Helper

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Whether the buoyant force can be neglected or not depends on the density of the lure. If its density is much greater than the density of water, then the buoyant force can be neglected.

The weight and drag of the line also contribute to this problem, although I haven't accounted for them in my equation.

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This is good stuff... thanks guys.

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