Hydrostatic pressure of liquid in a cart on an inclined ramp

[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.

 

[Lnewqban]

Welcome!

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

This link may help you:

 

[A13235378]

Welcome!

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

This link may help you:

Hello!

I used non-inertial frame:

I can't found
the denominator value

what am i missing

 

[TSny]

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?

 

[A13235378]

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

 

[TSny]

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 $\vec g$.

 

[etotheipi]

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 $\vec{g} - \vec{a}$?

 

[kuruman]

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

 

[TSny]

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.

 

[etotheipi]

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]

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.

 

[TSny]

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.

 

[TSny]

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

 

[kuruman]

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

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 $\mu=\tan\theta$.