How much deeper does the barge lower?

[keithcuda]

Homework Statement

1. A rectangular barge floats in freshwater. When a 400 kg block is loaded on the 5 m long by 2 m wide barge, the barge sinks a few centimeters deeper. How much deeper does the barge lower?

Homework Equations

Archimedes' principle
This is where I am stuck. My textbook doesn't talk much about this at all.

The Attempt at a Solution

 

[Bystander]

You've done all the hard work. How much deeper does the barge have to "sink" to displace the extra 0.4 m³?

 

[keithcuda]

You've done all the hard work. How much deeper does the barge have to "sink" to displace the extra 0.4 m³?

 

[haruspex]

View attachment 78234

Yes, but you could have got there a bit faster from knowing the volume displaced and the area.

 

[HalfLostAndSearching]

The depth to which the barge sinks when the 400 kg block is loaded on it can be calculated using Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the weight of the barge and the block is equal to the weight of the water that is displaced by the submerged portion of the barge.

To find the depth of the barge, we can use the formula:

d = (m_barge + m_block) / (ρ_fluid * g * A)

Where d is the depth, m_barge is the mass of the barge, m_block is the mass of the block, ρ_fluid is the density of the fluid (in this case, freshwater), g is the acceleration due to gravity, and A is the area of the barge that is submerged.

Using the given values, we can calculate the depth to which the barge sinks:

d = (400 kg + m_barge) / (1000 kg/m^3 * 9.8 m/s^2 * (5 m * 2 m))

= (400 kg + m_barge) / 98 m^2

Therefore, the barge will sink approximately 0.41 meters deeper when the 400 kg block is loaded on it. This can also be calculated by finding the difference in volume between the empty barge and the submerged barge, and then using the density of freshwater to find the corresponding depth.