# Pressure, Buoyant Force problem

• ChunkymonkeyI
In summary: I think you can finish this sentence)In summary, the problem involves a cube of metal with a density of 6.00 kg/dm^3 and a cavity inside it. When completely submerged in water, the cube weighs 2.40 times as much as it does in air. Using the equations for density and weight, we can set up equations to solve for the volume of the cavity. By equating the actual weight of the object to the sum of the constituent masses (metal and cavity), and using the given information about the apparent weight in air and water, we can solve for the fraction of the cube's volume that is taken up by the cavity.

## Homework Statement

A cube of metal(density=6.00 kg/dm^3) has a cavity inside it. It weighs 2.40 times as much in air as it does when completely submerged in water. What fraction of the cube's volume is the cavity?

## Homework Equations

Density=m/v
Fb=Density of fluid times volume of fluid times g
F=mg

## The Attempt at a Solution

Density=m/v (for water) Density=m/v
6.00 kg/dm^3=m/1000 kg 6.00 kg/dm^3/2.40m/V

Idk what 2 do from there because I think I set the equations up wrong could someone please explain 2 me what I should do because what's really bugging me is the amount of information that is given to me?

The information is enough to work out the answer (as long as you also know the density of water). And you could either assume the density of air is approximately zero or you could use the actual value, to get a more accurate answer.

To start this question, you should use what they give you. They are saying that the apparent weight of the cube is 2.4 times as much in air than in water. And you know the equation for the apparent weight. So you can use this to find the actual weight of the cube in terms of the volume of the cube. And then you can use this along with the equation for the actual weight of the cube due to the masses contained, to find the fraction of the cube that is hollow.

I still kinda don't get what ur saying can u show me ur work because I dervived Fw=Density of object times g times V and I made this equation equal to Fa plus 9800 times volume of the fluid and I really need help please

So you wrote:
$$Weight_{object} = F_{apparent} + 9800 V$$
Is this the apparent weight when its underwater? what does the 9800 mean?

Maybe I should go through it step-by-step, since there are a few steps which could get confusing when written all in one paragraph.

So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal.

BruceW said:
So you wrote:
$$Weight_{object} = F_{apparent} + 9800 V$$
Is this the apparent weight when its underwater? what does the 9800 mean?

Maybe I should go through it step-by-step, since there are a few steps which could get confusing when written all in one paragraph.

So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal.

Could u show each step if u don't mind because I'm getting the wrong answer for 3 times now and I think it would be helpful 2 c each steps

BruceW said:
So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal.

This is the first step. Write out this equation. It is essentially using the principle that total mass is sum of the constituent masses. In this case, the constituent masses are that of the cavity and the metal bit.
And just use symbols for now. (I find it is easier to use symbols in calculations until you get to the end of the problem). So you could use something like $V_m$ for the volume taken up by the metal (for example).

BruceW said:
This is the first step. Write out this equation. It is essentially using the principle that total mass is sum of the constituent masses. In this case, the constituent masses are that of the cavity and the metal bit.
And just use symbols for now. (I find it is easier to use symbols in calculations until you get to the end of the problem). So you could use something like $V_m$ for the volume taken up by the metal (for example).

This is what I did:
Fa=Fw-Fb
Fa=m(object)g-(density of fluid)(volume of fluid)(g)
Fa=(Density of the object)(Volume of the object)(g)-(Density of fluid)(volume of the fluid)(g)
Fa=g((density of object)(volume of the object)-(density of fluid)(volume of fluid))
m(a)=(density of object)(volume of object) minus (density of fluid)(volume of fluid)
m(a)=(6000 kg/m^3)(V of object) minus (1000 kg/m^3)(Volume of fluid)
Idk what 2 do from there but the only other thing Ik is that since its submerged the volume of the fluid is equal to the volume of the object and the apparent mass=m/2.40 but idk what 2 do from there please show me the steps and the math out :)

ChunkymonkeyI said:
This is what I did:
Fa=Fw-Fb
Fa=m(object)g-(density of fluid)(volume of fluid)(g)
Fa=(Density of the object)(Volume of the object)(g)-(Density of fluid)(volume of the fluid)(g)
Fa=g((density of object)(volume of the object)-(density of fluid)(volume of fluid))
m(a)=(density of object)(volume of object) minus (density of fluid)(volume of fluid)
m(a)=(6000 kg/m^3)(V of object) minus (1000 kg/m^3)(Volume of fluid)

Almost right. The object has an air cavity inside it, so if we assume the air density is negligible, the mass of the object equals the density of metal times the volume which the metal actually takes up. In other words, the volume of the cavity isn't contributing to the mass of the object. If you change this, you will then have the apparent weight in water. To get the apparent weight in air, its just the same equation, but the fluid is now air.

## 1. What is pressure?

Pressure is defined as the force per unit area. In other words, it is the amount of force applied over a specific area. The unit of pressure is typically measured in Pascals (Pa) or Newtons per square meter (N/m^2).

## 2. How does pressure affect buoyancy?

Pressure plays a crucial role in determining the buoyant force acting on an object. The greater the pressure exerted by a fluid on an object, the greater the buoyant force will be. This is because the pressure of a fluid increases with depth and therefore, the deeper an object is submerged, the higher the pressure and the greater the buoyant force.

## 3. What is the formula for calculating buoyant force?

The formula for calculating buoyant force is F(b) = ρVg, where ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity. This formula is known as Archimedes' principle and it states that the buoyant force acting on an object is equal to the weight of the fluid it displaces.

## 4. How do you solve a buoyant force problem?

To solve a buoyant force problem, you must first identify the known and unknown variables, such as the density of the fluid, volume of the displaced fluid, and the weight or mass of the object. Then, use the formula F(b) = ρVg to calculate the buoyant force. Finally, compare the buoyant force to the weight of the object to determine if it will float or sink.

## 5. What factors affect the pressure and buoyant force of an object?

The pressure and buoyant force of an object are affected by several factors, including the density and volume of the fluid, the depth at which the object is submerged, and the shape and size of the object. Additionally, the acceleration due to gravity and the atmospheric pressure can also have an impact on the pressure and buoyant force of an object.