Question with Archimedes' principles

[john8765]

A cylindrical fishing float is 15cm long, with an average cross section of 3cm^2. It is made of polystyrene and has negligible mass. A lead weight of 30g is attached to the bottom of the float using a thin nylon monofilament line.

Question: Calculate how far the float will sink into the water.

Treat g as equal to 10N kg-1



I know Archimedes said upthrust acting on a body is equal to weight of fluid displaced but can't see how to apply that in this case. Thanks for any help.

 

[DukeLuke]

Can you find an expression for the volume of the water displaced based on the shape of the float? In this case the density of water can be used to find the mass of the water displaced.

 

[john8765]

Can you find an expression for the volume of the water displaced based on the shape of the float? In this case the density of water can be used to find the mass of the water displaced.

Thanks for your help. I've come up with an answer of 0.1m?

 

[DukeLuke]

You're welcome, I get the same answer.

 

[rwisz]

Based on Archimedes' principle, the buoyant force acting on an object is equal to the weight of the fluid it displaces. In this case, the cylindrical fishing float will displace a volume of water that is equal to its own volume. Since the float is made of polystyrene, which has a density of approximately 1 kg/m^3, its volume can be calculated by multiplying its length (15cm) by its average cross section (3cm^2) which is 0.03m^3.

Therefore, the buoyant force acting on the float is equal to 0.03m^3 x 10N/kg = 0.3N.

Now, taking into account the weight of the lead weight, which is 30g or 0.03kg, the total weight of the float and weight combined is 0.03kg + 0.03kg = 0.06kg.

Using the formula F = ma, where F is the force, m is the mass, and a is the acceleration, we can calculate the acceleration of the float as it sinks into the water.

Since the only force acting on the float is the buoyant force (0.3N) and the weight of the float and weight combined is 0.06kg, we can set up the following equation:

0.3N = 0.06kg x a

Solving for a, we get a = 5m/s^2.

Now, using the equation d = 1/2at^2, where d is the distance, a is the acceleration, and t is the time, we can calculate the distance the float will sink into the water in a given amount of time.

Assuming the float starts from rest, we can set t = 1 second.

Therefore, d = 1/2 x 5m/s^2 x (1s)^2 = 2.5m

This means that in 1 second, the float will sink 2.5m into the water.

In conclusion, the float will sink 2.5m into the water when the lead weight is attached to the bottom using a thin nylon monofilament line. It is important to note that this calculation is based on ideal conditions and may vary in real-life situations due to factors such as water currents and the angle at which the weight is attached