Thursday, February 22, 2024
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The physics of 'sniping' for gold

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There are three forces acting on the wreck. First, the downward force of gravity (FYes) due to contact with the earth. This force depends on both the mass (m) of the object and the gravitational field (G = 9.8 newtons per kilogram on Earth).

Next, we have the buoyancy force (Fb, When an object is immersed in water (or any liquid), there is an upward force from the surrounding water. The magnitude of this force is equal to the weight of the water displaced, such that it is proportional to the volume of the object. Note that both the force of gravity and the force of buoyancy depend on the size of the object.

Finally, we have a drag force (FD) due to the interaction between the moving water and the object. This force depends on both the size of the object and its speed relative to the water. We can use the magnitude of the drag force (in water, not to be confused with air drag) stoke’s lawAccording to the following equation:

Illustration: Rhett Allen

In this expression, R is the radius of the spherical object, μ is the dynamic viscosity, and v is the velocity of the fluid with respect to the object. In water, the kinematic viscosity is approximately 0.89 x 10-3 kilogram per meter per second.

Now we can model the motion of a piece of gold in flowing water versus the motion of a rock. Although there is one small problem. According to Newton’s second law, the net force acting on an object changes the object’s velocity—but as the velocity changes, so does the force.

One way to approach this problem is to break down the motion of each object into smaller time intervals. During each interval, I can assume that the net force is constant (which is approximately true). With a constant force, I can then find the speed and position of the object at the end of the interval. Then I need to repeat the same process for the next interval.