Calculate the resulatant downward force of a submarine on the seafloor

[Richie Smash]

Homework Statement

Hi I have uploaded a picture of a pressure question.
It says that there is a submarine 325 m deep, with the sea water of density 1150Kg m^-3 and acceleration of gravity 10m/s and the volume of the submarine is 0.2m³ with mass 480Kg
They first ask for the upthrust.
Then they say the normal force of the sea floor against the submarine is 1800N, what is the resultant downward force?
Then they ask what mass of water must be emptied from the ballasts for the submarine to float?

Homework Equations

Pressure =pgh
Upthrust = m(fluid disaplaced) *g

The Attempt at a Solution

I know that Upthrust is equal to the density of the fluid times gravity times the volume of the object.
Thus I have worked out the upthrust on the submarine to be 2300N.

Now they ask for the resultant force.
Normally I would think it would be 1800 as well, but I realized there are more force acting downwards, the first being the weight of the submarine itself, which is 4800N, now i must subtract the upthrust from this to find the remaining downard force I believe, so I would get 2500N is what is remaining acting in the downwards direction.

So I have 1800 going upwards from the seafloor, and 2500 going downwards from the submarine, would the resultant force be 700N?

But then I have to remember the pressure of the sea acting on the submarine itself which is 375 m* g*p(of the water)

I feel I'm somewhere alng the correct lines.

 

[Alloymouse]

Homework Statement

Hi I have uploaded a picture of a pressure question.
It says that there is a submarine 325 m deep, with the sea water of density 1150Kg m^-3 and acceleration of gravity 10m/s and the volume of the submarine is 0.2m³ with mass 480Kg
They first ask for the upthrust.
Then they say the normal force of the sea floor against the submarine is 1800N, what is the resultant downward force?
Then they ask what mass of water must be emptied from the ballasts for the submarine to float?

Homework Equations

Pressure =pgh
Upthrust = m(fluid disaplaced) *g

The Attempt at a Solution

I know that Upthrust is equal to the density of the fluid times gravity times the volume of the object.
Thus I have worked out the upthrust on the submarine to be 2300N.

Now they ask for the resultant force.
Normally I would think it would be 1800 as well, but I realized there are more force acting downwards, the first being the weight of the submarine itself, which is 4800N, now i must subtract the upthrust from this to find the remaining downard force I believe, so I would get 2500N is what is remaining acting in the downwards direction.

So I have 1800 going upwards from the seafloor, and 2500 going downwards from the submarine, would the resultant force be 700N?

But then I have to remember the pressure of the sea acting on the submarine itself which is 375 m* g*p(of the water)

I feel I'm somewhere alng the correct lines.

Regarding "pressure of the sea acting on the submarine itself", you don't have to consider that as it is already considered in the upthrust itself: it's the pressure difference of the water at different heights.

Generally it looks sound. Although I find it kind of weird that there's still a net 700N that's pushing onto the sea floor as though the sea floor can be pushed down.

 

[Richie Smash]

So you're saying what I've speculated seems correct?

 

[Alloymouse]

So you're saying what I've speculated seems correct?

Yes. I just woke up so I'll take a closer look later when I'm more sane lol. Or wait for someone else to come along to confirm / reject

 

[Richie Smash]

so would that be 250 KG of water required for the submarine to float yes?

 

[Alloymouse]

so would that be 250 KG of water required for the submarine to float yes?

Yep, that is right :)

 

[haruspex]

weird that there's still a net 700N that's pushing onto the sea floor as though the sea floor can be pushed down.

It's worse than that. It implies the sub is accelerating downward at 700/480=1.46m/s².
The net force is surely 0.

The question as posed makes no sense. You can either ignore the 1800N information and arrive at 250kg, or ignore the .2m³ and get 1800N/g = 180kg.
I note that if 250kg of ballast can be expelled then it occupies a volume greater than the sub itself, so I suggest the 0.2m³ is suspect. Perhaps one needs to pay attention to the number of significant figures, i.e. it is anything < .25m³.

I did wonder if the mass given for the sub excludes the contents of its ballast tanks, but that makes matters worse.

 

[Alloymouse]

It's worse than that. It implies the sub is accelerating downward at 700/480=1.46m/s².
The net force is surely 0.

The question as posed makes no sense. You can either ignore the 1800N information and arrive at 250kg, or ignore the .2m³ and get 1800N/g = 180kg.
I note that if 250kg of ballast can be expelled then it occupies a volume greater than the sub itself, so I suggest the 0.2m³ is suspect. Perhaps one needs to pay attention to the number of significant figures, i.e. it is anything < .25m³.

I did wonder if the mass given for the sub excludes the contents of its ballast tanks, but that makes matters worse.

Okay so I wasn't mad to think the 700N was off phew

@Richie Smash Where did you get this question? Do you have a source?

 

[Richie Smash]

Yes its from CXC examinations past papers January 2007 paper 3, but I doubt you'll find that online, I got it emailed from a teacher.. so My asnwer is wrong then?

 

[Alloymouse]

Yes its from CXC examinations past papers January 2007 paper 3, but I doubt you'll find that online, I got it emailed from a teacher.. so My asnwer is wrong then?

Considering stuff is kinda weird here... I suggest you either look at the answer key if you have one to find out what they're trying to say, or follow @haruspex advice. Alternatively, you might want to do other similar questions instead.

 

[Richie Smash]

Umm, there's no answer key, I'll have class ina few hours where the answer will be revealed, so I will let you guys know tonight

 

[Richie Smash]

Ok Actually, He didn't review this paper in class yesterday, He's actually doing it next thursday, so I'll find out and see