Help Needed: Physics Problem w/ Elevator & Goldfish Tank

[alonzo]

Hi! I have a physics problem that I just don't understand. It is as follows:

Imagine that you are in an elevator with a tank of water and three goldfish. A device in the gravel (of the fish tank) measures gauge pressure at the bottom at the tank.
(a) Suppose it reads 4000 N / m² with the elevator car at rest. Will it increase, decrease, or stay the same when the car accelerates upwards? Explain.
(b) What will it read if the car drops at 9.8 m / s²?

I know about pressure in relation to fluids at rest but I'm not sure what happens when the fluid is accelerated upwards or downwards. I also understand that gauge pressure is absolute pressure minus atmospheric pressure.
Please help - thanx!

 

[Crumbles]

The pressure recorded is due to the water and is given by
P =hdg;
where h is the height of the water, d is the density of the water and g is the acceleration due to gravity.

Now, d, the density =m/v,
where m is the total mass of the water and v is the total volume.

Therefore, P=[(hmg)/v]

Now, you probably realize that the 'mg' in the expression is the force due to gravity on the water.

If you now consider the case where the car is accelerating upwards (say with a force F), then then net force acting on the water will be [mg-F].

So, your new pressure will be P=[(h(mg-F))/v]

Therefore, the pressure registered on the gauge will be less than 4000Pa

Using a similar reasoning you can get to the answer to the second part of your question

 

[Locrian]

I'm getting a different sign than you are Crumbles. If the elevator is accelerating upwards, this should mean the pressure registered on the gauge is greater, not lesser, should it not?

An elevator accelerating upwards should be no different than having a gravity equal to mg_earth + F.

Set me straight if I am incorrect.

 

[ehild]

Hi! I have a physics problem that I just don't understand. It is as follows:

Imagine that you are in an elevator with a tank of water and three goldfish. A device in the gravel (of the fish tank) measures gauge pressure at the bottom at the tank.
(a) Suppose it reads 4000 N / m² with the elevator car at rest. Will it increase, decrease, or stay the same when the car accelerates upwards? Explain.
(b) What will it read if the car drops at 9.8 m / s²?

I know about pressure in relation to fluids at rest but I'm not sure what happens when the fluid is accelerated upwards or downwards. I also understand that gauge pressure is absolute pressure minus atmospheric pressure.
Please help - thanx!

Think. What makes the water in the tank accelerate upward with "a"? A net force, equal to m*a. This force is the resultant of the downward force due to gravity , -m*g, the upward normal force Fn from the bottom of the tank and the downward force due to the atmospheric pressure, -A*Po, where A is the cross-section are of the tank. So m*a = Fn-m*g-A*Po.
The gauge pressure P read by the device at the bottom of the tank is equal to the normal force per unit area minus the atmospheric pressure.
P = Fn/A-Po = m*(a+g)/A.
The pressure increases if the elevator accelerates upward (and so does the tank together with it). It decreases in the opposite case. When the elevator is in the state of free fall, you feel "weightless" and so does the water in the tank: a=-g, P=0. (As there is no air between the water and the bottom of the tank, Fn must balance the outer pressure, and the gauge pressure is zero.)

ehild

 

[Crumbles]

I'm getting a different sign than you are Crumbles. If the elevator is accelerating upwards, this should mean the pressure registered on the gauge is greater, not lesser, should it not?

An elevator accelerating upwards should be no different than having a gravity equal to mg_earth + F.

Set me straight if I am incorrect.

Yes, I apologise, you are actually right, the pressure must be greater when the car is accelerating up. I've just spotted where I went wrong. It was in the sign of the forces:

Your new pressure will be P=[(h(mg-F))/v]

But F is in the opposite direction to mg so it's actually P=[(h(mg-(-F)))/v], which is P=[(h(mg+F))/v] and therefore the pressure will be greater.