# Hydrostatic pressure of liquid in a cart on an inclined ramp

• A13235378
In summary: It would also be fun to rig up a magnetic braking system of variable ##\mu## and have the accelerometer roll down at constant velocity when ##\mu=\tan\theta##.
A13235378
Homework Statement
A tank car partially filled with a homogeneous liquid travels at a constant acceleration along a straight slope of the road, inclined from a angle θ to the horizon and which has a kinetic friction coefficient equal to μ. Determine the alpha angle that the free surface of the liquid, supposed to be in equilibrium, forms with the horizon.
Relevant Equations
P= mg
F= μN

Sorry for my English, I'm Spanish. I am not able to make the force diagram for the liquid. I am still in high school and started now little hydrostatic. If someone can give me a light, I would be very .grateful.

Last edited by a moderator:
Welcome!

The acceleration of the car going downhill will be reduced some from gravity-action-only due to friction.

A13235378
Lnewqban said:
Welcome!

The acceleration of the car going downhill will be reduced some from gravity-action-only due to friction.

Hello!

I used non-inertial frame:
I can't found
the denominator value

what am i missing

A13235378 said:

You drew a green vector that is perpendicular to the vector ##\large \vec a##. You then indicate in the drawing that the angle between the green vector and the vertical is equal to ##\large \alpha##. But this is not correct. Did you mean to draw the green vector perpendicular to the surface of the fluid instead of perpendicular to ##\large \vec a##?

The principle that you can use is that the components of ##\large \vec a## and ##\large \vec g## that are parallel to the surface of the fluid must add to zero. It looks like maybe this is what you were trying to express. What is the angle that ##\large \vec a## makes to the surface of the fluid?

Yes! The green vector is perpendicular to the surface of the fluid. Its correct, right? So i just changed the balance equation?

A13235378 said:
Yes! The green vector is perpendicular to the surface of the fluid. Its correct, right? So i just changed the balance equation?
Yes, if the green vector is meant to denote the vector sum of ##\vec a## and ##\vec g##.

TSny said:
Yes, if the green vector is meant to denote the vector sum of ##\vec a## and ##\vec g##.

I'm sure we all have the same idea, but shouldn't it be ##\vec{g} - \vec{a}##?

Watch what happens in this video that shows the cart's speed increase and then decrease, both at constant acceleration.

TSny
etotheipi said:
I'm sure we all have the same idea, but shouldn't it be ##\vec{g} - \vec{a}##?
The OP is working in the frame accelerating with the cart. His vector ##\vec a## is in the direction of the fictitious force, which is opposite to the direction that the cart is accelerating in the inertial frame.

TSny said:
The OP is working in the frame accelerating with the cart. His vector ##\vec a## is in the direction of the fictitious force rather than the direction that the cart is accelerating in the inertial frame.

Thanks for clarifying, usually I thought we would write the fictitious force ##-m\vec{a}## if ##\vec{a}## is the acceleration of this frame w.r.t. some inertial frame, which would give an effective gravitational acceleration of ##\vec{g} - \vec{a}##. But I suppose OP has defined ##\vec{a} \rightarrow -\vec{a}## as the negative of this, which I guess is fine (but maybe a little less natural?)

TSny
etotheipi said:
Thanks for clarifying, usually I thought we would write the fictitious force ##-m\vec{a}## if ##\vec{a}## is the acceleration of this frame w.r.t. some inertial frame, which would give an effective gravitational acceleration of ##\vec{g} - \vec{a}##. But I suppose OP has defined ##\vec{a} \rightarrow -\vec{a}## as the negative of this, which I guess is fine (but maybe a little less natural?)
Yes, I think we're all together.

etotheipi
kuruman said:
Watch what happens in this video that shows the cart's speed increase and then decrease, both at constant acceleration.
It's fun to note the orientation of the fluid surface while the cart is rolling freely down the slope.

A13235378 said:
This should read ##\sin \alpha = \large \frac{\sin \theta - \mu \cos \theta}{\sqrt{1+\mu^2}}##.

TSny said:
This should read ##\sin \alpha = \large \frac{\sin \theta - \mu \cos \theta}{\sqrt{1+\mu^2}}##.
TSny said:
It's fun to note the orientation of the fluid surface while the cart is rolling freely down the slope.
It would also be fun to rig up a magnetic braking system of variable ##\mu## and have the accelerometer roll down at constant velocity when ##\mu=\tan\theta##.

TSny

## 1. What is hydrostatic pressure?

Hydrostatic pressure is the force exerted by a liquid at rest due to the weight of the liquid above it.

## 2. How does the hydrostatic pressure of a liquid change on an inclined ramp?

The hydrostatic pressure of a liquid on an inclined ramp increases as the height of the liquid column increases, and decreases as the angle of the ramp increases.

## 3. What factors affect the hydrostatic pressure of a liquid on an inclined ramp?

The factors that affect the hydrostatic pressure of a liquid on an inclined ramp are the height of the liquid column, the density of the liquid, and the angle of the ramp.

## 4. How can the hydrostatic pressure of a liquid on an inclined ramp be calculated?

The hydrostatic pressure of a liquid on an inclined ramp can be calculated using the formula P = ρghsinθ, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid column, and θ is the angle of the ramp.

## 5. What is the practical application of understanding the hydrostatic pressure of a liquid on an inclined ramp?

Understanding the hydrostatic pressure of a liquid on an inclined ramp is important in various fields such as engineering, physics, and fluid mechanics. It can help in designing and constructing structures that can withstand the pressure of liquids, as well as in predicting the behavior of liquids in different scenarios.

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