What is the Force Exerted by a Falling Rod on a Hinged Support?

[member 428835]

Homework Statement

Consider a rod of length $L$ that is attached to a hinge making an angle $\theta$ with the horizontal. There is a force $\vec{F}$ acting vertically on the rod that creates constant angular velocity $-\omega$. If there is fluid under the rod with constant density, what force is the rod exerting on the hinge?

Homework Equations

Continuity. Navier-Stokes?

The Attempt at a Solution

So I'm not really sure where to start. I let the rod have a distance coming out of the page, call this $b$. Then total volume of fluid in the triangular region under the rod and above the horizontal is $V(t) = b L^2 \cos \theta \sin \theta/2 = b L^2 \sin (2 \theta) /4 \implies V'(t) = -b L^2 \omega\cos (2 \theta) /2$, which is at least negative, implying fluid is leaving the region (good).

Now I know torque $\vec{\tau} = \vec{r} \times \vec{F}$, so if we use polar coordinates we have $\vec{\tau} = L \hat{r} \times F\hat{y}$. Now we know the position vector in polar coordinates is $\vec{R} = r\cos\theta \hat{x}+r\sin\theta\hat{y} \implies \hat{r} = \cos\theta\hat{x}+\sin\theta\hat{y}$. Then $\vec{\tau} = L \hat{r} \times F\hat{y} = L(\cos\theta\hat{x}+\sin\theta\hat{y}) \times -F\hat{y} = F\hat{y} \times L(\cos\theta\hat{x}+\sin\theta\hat{y}) = F\hat{y} \times L\cos\theta\hat{x} +F\hat{y} \times \sin\theta\hat{y} = -F L \cos\theta \hat{z}$ where $z$ is out of the page. I'm not really sure how to consolidate this info. I'm thinking that the fluid would exert a sort of pressure force normal to the rod and in the positive $\theta$ direction. Then if we summed forces on the rod we would have the downward force $-F \hat{y}$, the pressure force $F_p \hat{\theta}$ and the reacting force from the hinge to the rod $F_h \hat{x}$. Since the hinge is not accelerating, these forces should balance, right?

 

[haruspex]

the fluid would exert a sort of pressure force normal to the rod and in the positive θθ\theta direction. Then if we summed forces on the rod we would have the downward force $-F \hat{y}$, the pressure force $F_p \hat{\theta}$ and the reacting force from the hinge to the rod $F_h \hat{x}$. Since the hinge is not accelerating, these forces should balance, right?

Yes, that all seems reasonable.
I tried to solve it by assuming a drag force f(x)=Axⁿ, where x is the distance from the hinge (which I take to be at the top). I got an answer, but I had to assume where the force F acts on the rod. I took that to be at the bottom.
By the way, you titled it "rod falling", but I assume gravity is to be ignored.

 

[member 428835]

Yes, that all seems reasonable.
I tried to solve it by assuming a drag force f(x)=Axⁿ, where x is the distance from the hinge (which I take to be at the top). I got an answer, but I had to assume where the force F acts on the rod. I took that to be at the bottom.
By the way, you titled it "rod falling", but I assume gravity is to be ignored.

Yes, I have ignored gravity, though I suppose we wouldn't necessarily have to. Would you mind posting your work, or giving me some insight into how you incorporated drag?

Additionally, I assume to find the pressure force the fluid exerts on the rod I would have to use navier-stokes. Do you agree? I'm thinking of using Navier-Stokes in polar coordinates, although if we assume $\theta$ is small, then perhaps we could assume the horizontal motion is greater than vertical, and hence use Cartesian Navier-Stokes? What are your thoughts?

Thanks for the reply!

 

[haruspex]

to find the pressure force the fluid exerts on the rod I would have to use navier-stokes. Do you agree?

I have no idea... I have never studied anything that advanced in fluid dynamics.
We are not given much information - not the density, not the viscosity, not the width of the rod... - so I assumed a fairly simple treatment is expected.
If we take the drag to be normal to the rod and proportional to some power, n, of the linear speed, we can get expressions for the net force from the drag and its net torque about the hinge. That allows us to write the usual force balance equations, except that we need to make an assumption regarding the point of application of F.
On that basis, I got an answer in terms of F, θ and n.

 

[member 428835]

Perhaps we could see what @Chestermiller thinks? This is for a high-level fluids class (well, it's from a set of notes). This makes me think we'll need the pressure exerted by the fluid, but again, I could be wrong.

Regarding density, we can assume the fluid has constant density $\rho$, and viscous/surface tension is negligible. No viscosity implies zero drag I think (though perhaps the fluid drag is not the drag you're referring to?). I suppose we can assume the rod has no mass (though it would be cool, once solving this, to see how the mass would impact the solution, along with say, friction at the hinge.)

Either way, let's assume I am able to find the pressure force of the fluid. Then would we simply have $F_h = F_p \sin \theta$ and then $F_p \cos \theta = F$?

 

[haruspex]

No viscosity implies zero drag

I meant it in the general sense of whatever the fluid does to impede the rod.

 

[member 428835]

Ohhhhh, so you're saying whether it's pressure or something else like drag? Also, I edited post #5. I have a final question there I was hoping you could address?

 

[haruspex]

assume I am able to find the pressure force of the fluid. Then would we simply have $F_h = F_p \sin \theta$ and then $F_p \cos \theta = F$?

No. You do not know the direction of the hinge reaction. You will also need to consider torque.

 

[member 428835]

Ok, that's actually something I was wondering. So would we apply conservation of angular momentum, which for this case, is sum of torque's is zero since the rod is moving at constant angular velocity?

 

[haruspex]

Ok, that's actually something I was wondering. So would we apply conservation of angular momentum, which for this case, is sum of torque's is zero since the rod is moving at constant angular velocity?

Yes.

 

[member 428835]

Ok cool. So we would have the torque I explained in my initial post, which is the torque resulting from the downward force. Then we would also have the pressure torque. Now if I can come up with an expression for pressure as a function of position, how would we go about this torque since it's acting on all of the rod? Would it by an integral of some sorts, say $\iint_S p \hat{n}\, dS$ where $\hat{n}$ is unit-normal to the plane (so $\hat{\theta}$)?

 

[haruspex]

Ok cool. So we would have the torque I explained in my initial post, which is the torque resulting from the downward force. Then we would also have the pressure torque. Now if I can come up with an expression for pressure as a function of position, how would we go about this torque since it's acting on all of the rod? Would it by an integral of some sorts, say $\iint_S p \hat{n}\, dS$ where $\hat{n}$ is unit-normal to the plane (so $\hat{\theta}$)?

Yes, but I took it as a single integral over the length of the rod. Since the force is normal to the rod, this is quite easy.

 

[member 428835]

Yes, but I took it as a single integral over the length of the rod. Since the force is normal to the rod, this is quite easy.

Right, if we had pressure $p$ then $$\vec{F_p} = \int_0^b \int_0^L p \hat{\theta} \,dl \, dw = b \int_0^L p \hat{\theta} \, dl $$ right? This gives us a force, but torque is $\vec{r} \times \vec{F}$. What would the $\vec{r}$ be; not $L$ right, since $\vec{F_p}$ is distributed over all of $L$, right?

 

[haruspex]

Right, if we had pressure $p$ then $$\vec{F_p} = \int_0^b \int_0^L p \hat{\theta} \,dl \, dw = b \int_0^L p \hat{\theta} \, dl $$ right? This gives us a force, but torque is $\vec{r} \times \vec{F}$. What would the $\vec{r}$ be; not $L$ right, since $\vec{F_p}$ is distributed over all of $L$, right?

Just put a factor l in the integrand.

 

[Chestermiller]

I really don't understand the geometry. Can you please draw a diagram.

 

[member 428835]

Just put a factor l in the integrand.

Oooooh ok, so you're saying sum all the torques from pressure, like $$\tau_p = b\int_0^L l \hat{r} \times p \hat{\theta} \, dl = b\int_0^L l p \hat{z} \, dl$$ where $\tau_p$ is the torque due to the pressure.

And Chet, I've attached the picture for your consideration.

 

[haruspex]

Oooooh ok, so you're saying sum all the torques from pressure, like $$\tau_p = b\int_0^L l \hat{r} \times p \hat{\theta} \, dl = b\int_0^L l p \hat{z} \, dl$$ where $\tau_p$ is the torque due to the pressure.
.

yes. And I assumed p=Alⁿ, for some A, n.

 

[member 428835]

yes. And I assumed p=Alⁿ, for some A, n.

Gotcha, this makes a lot of sense! So if gravity were to be present would we do a similar analysis as we did on pressure (only $\hat{r}$ would be crossed with $-\hat{y}$ (I am calling $y$ the vertical direction so $z$ is out of the page)? At any rate, if we know the pressure then the torque balance would give us $\tau_p + \tau_f+\tau_r = 0$ where we have the pressure torque, the force torque, and the reaction torque respectively. Then we could solve for the reaction torque $\tau_r$, which would be in the $z$ direction. But how would we get a force from this?

Hey Chet, any idea how to compute the pressure?

 

[haruspex]

Then we could solve for the reaction torque

It's a hinge.

 

[member 428835]

It's a hinge.

Sorry, hinge torque then? But do you agree with the gravity assessment?

 

[haruspex]

Sorry, hinge torque then

You are missing my point. Hinges do not exert torques, other than perhaps frictional.

do you agree with the gravity assessment?

I would prefer to ignore gravity, for now at least. E.g., for all we know, the rod and fluid have the same density.

 

[member 428835]

You are missing my point. Hinges do not exert torques, other than perhaps frictional.

I thought there was something more than just wording. This makes sense since there is not rotating arm. So we just have a balance of those two torques?

 

[haruspex]

I thought there was something more than just wording. This makes sense since there is not rotating arm. So we just have a balance of those two torques?

Yes. But I should have written, hinges do not exert torques about themselves. If you take moments about some other axis then the reaction force of the hinge may have a torque about that.

 

[member 428835]

I'm kind of confused then how we would compute the force the hinge exerts on the ground, assuming we have the torque balance.

 

[haruspex]

I'm kind of confused then how we would compute the force the hinge exerts on the ground, assuming we have the torque balance.

On the ground? Do you mean, on the rod?

 

[member 428835]

On the ground? Do you mean, on the rod?

No, on the ground. When there is resistance to the rod, it seems that is would want to be displaced, kinda of like the falling ladder calculus problem only if something were there holding the bottom leg in place. Wouldn't the hinge be applying a force on the ground to avoid horizontal displacement?

 

[haruspex]

No, on the ground. When there is resistance to the rod, it seems that is would want to be displaced, kinda of like the falling ladder calculus problem only if something were there holding the bottom leg in place. Wouldn't the hinge be applying a force on the ground to avoid horizontal displacement?

Yes, but that is of no interest, is it? You can treat the hinge as a fixed point, by whatever means. We only care about forces on the rod.

 

[member 428835]

The initial question posed was asking what force the hinge exerts on the ground.

 

[haruspex]

The initial question posed was asking what force the hinge exerts on the ground.

Fair enough... that was so long ago.
Anyway, that is just equal and opposite to the force it exerts on the rod, so we can concentrate on finding that.

 

[member 428835]

Fair enough... that was so long ago.
Anyway, that is just equal and opposite to the force it exerts on the rod, so we can concentrate on finding that.

hahaha yea I know. And I agree. So how would we proceed here? We balance these two torques but how will this help us resolve that force?

 

[Chestermiller]

For drag flow past a circular cylinder, the drag coefficient is nearly constant on the order of about 1.0 for Reynolds numbers on the order of about 1000 to 100,000. So, the drag force per unit length on the rod is $$f=\frac{1}{2}\rho v^2 D$$where D is the diameter and v is the velocity across the rod (the component of velocity normal to the rod). So $$f=k(\omega r)^2=k'r^2$$where r is the distance from the hinge. So, to a good approximation, Haruspex's exponent is 2. The moment of the drag force is $$M=FL\cos{\theta}=\int_0^L{k'r^3dr}=k'\frac{L^4}{4}$$So, $$k'=\frac{4F}{L^3}\cos{\theta}$$This means that the local normal force per unit length exerted by the fluid on the rod is:
$$f=\frac{4Fr^2}{L^3}\cos{\theta}$$So, what is the normal force that the fluid exerts on the rod?

 

[haruspex]

hahaha yea I know. And I agree. So how would we proceed here? We balance these two torques but how will this help us resolve that force?

When you have found the total normal force on the rod from the fluid (as per Chet's post) you can similarly find the total torque from the fluid.
This gives you effectively three balance equations, two being the two dimensional balance of linear forces, and one from the balance of torques. That means you can find both components of the hinge reaction even without knowing the A factor in the force from the fluid.
As Chet says, n=2 is reasonable.

 

[Chestermiller]

When you have found the total normal force on the rod from the fluid (as per Chet's post) you can similarly find the total torque from the fluid.
This gives you effectively three balance equations, two being the two dimensional balance of linear forces, and one from the balance of torques. That means you can find both components of the honge reaction even without knowing the A factor in the force from the fluid.
As Chet says, n=2 is reasonable.

I used the total torque to get the value of the constant k'. All he needs to do now is determine the total normal component of the fluid force.

 

[haruspex]

I used the total torque to get the value of the constant k'. All he needs to do now is determine the total normal component of the fluid force.

Ah yes, I didn't read your post in enough detail. I think it's actually a bit simpler my way, to write out all the equations assuming the drag is at distance r from the hinge is Ar². The A's cancel without ever having to determine it; in particular, the dependency on the angle.

 

[member 428835]

This means that the local normal force per unit length exerted by the fluid on the rod is:
$$f=\frac{4Fr^2}{L^3}\cos{\theta}$$So, what is the normal force that the fluid exerts on the rod?

$$F_n=\frac{4Fr^3}{3L^3}\cos{\theta}$$ Thank you both for the input!

 

[Chestermiller]

$$F_n=\frac{4Fr^3}{3L^3}\cos{\theta}$$ Thank you both for the input!

Good, except lose the r^3 and the L^3

 

[kuruman]

The moment of the drag force is $M=FL\cos{\theta}=\int_0^L{k'r^3dr}=k'\frac{L^4}{4}$

Question: Why is the upper limit of integration L? The drawing attached to post #16 shows the fluid extending only partially along the full length of the rod. Or is it just a drawing?

 

[Chestermiller]

Question: Why is the upper limit of integration L? The drawing attached to post #16 shows the fluid extending only partially along the full length of the rod. Or is it just a drawing?

That's not how I interpreted the diagram. I don't know what the correct interpretation is. Josh?

 

[member 428835]

Yea, my drawing is imperfect. Fluid is coming out of the entire triangular region.

 

[TSny]

The initial question posed was asking what force the hinge exerts on the ground.

There is no mention of the ground in the OP. Have you stated the problem word-for-word as given?

Not that I think I can contribute anything here, but I just wanted to understand the statement of the problem. I can't make any sense out of it whatsoever.