A pyramid of three cylinders (statics problem)

[lloydthebartender]

Homework Statement:

I found the question here (https://isaacphysics.org/questions/three_cylinders?board=c197ce8f-5684-418a-9a4b-de5cb43b3da6).

Two rough cylinders, each of mass m and radius r are placed parallel to each other on a rough surface with their curved sides touching. A third equivalent cylinder is balanced on top of these two, with its axis running in the same direction.

For this setup to be stable, what is the minimum coefficient of friction between the horizontal surface and the cylinders? Give your answer to three significant figures.

Relevant Equations:

$F_r=(\mu)R$

Firstly the normal reaction force of the bottom cylinder is $R$ and because all three cylinders are identical $2R=3mg$, and so the frictional force is going to be $F_r={(3mg)/(2}{\mu})$. I don't use this, though.

If the normal reaction force between the top cylinder and one of the bottoms cylinder is $N$, then by the centres of the three cylinders making an equilateral triangle, we can say $N=(1/2)mg$. This reaction force provides a horizontal and vertical force on the bottom cylinder, where the horizontal force is equal to the frictional force. The vertical force adds to the weight of the bottom cylinder.

Because the angle $N$ makes with the vertical is 60 degrees, the vertical component is $(\sqrt3/4)mg$ and the horizontal is $(1/4)mg$. So finally $R=mg+(\sqrt3/4)mg$ and $F_r=(1/4)mg$, and I equate for $\mu$ to get 0.174, which isn't right.

I can't put my finger on where I went wrong so help is appreciated!

 

[lloydthebartender]

Ok so I've spotted at least one mistake (I think) where I calculated the normal reaction force between the top cylinder and the bottom one, $N$ should be half of that value because $mg$ of the top cylinder is split between the two bottom ones. Still don't get the right answer though.

 

[haruspex]

If the normal reaction force between the top cylinder and one of the bottoms cylinder is N, then by the centres of the three cylinders making an equilateral triangle, we can say N=(1/2)mg.

You are overlooking that the frictional force there also has a vertical component.

N should be half of that value

I get that $N=mg\frac{\sqrt 3}4$.
To get that, I started by considering the balance of forces on a lower cylinder.

 

[Lnewqban]

Sorry, I can't see the linked original problem.
Does the coefficient of friction among cylinders have the same value as the coefficint between cylinders and horizontal surface?

 

[ehild]

Because the angle $N$ makes with the vertical is 60 degrees,

Are you sure?

 

[haruspex]

Are you sure?

Just a typo, I assumed, since the subsequent calculations treat it as 60:degrees to the horizontal.

 

[lloydthebartender]

Sorry, I can't see the linked original problem.
Does the coefficient of friction among cylinders have the same value as the coefficint between cylinders and horizontal surface?

The question doesn't specify this but I think so.

Are you sure?

Just a typo, I assumed, since the subsequent calculations treat it as 60:degrees to the horizontal.

It was a typo, my bad. It should say 30 degrees.

You are overlooking that the frictional force there also has a vertical component.

I get that $N=mg\frac{\sqrt 3}4$.
To get that, I started by considering the balance of forces on a lower cylinder.

This is the frictional force between the top and bottom cylinder? I did it again and got $N=mg\frac{\sqrt 3}4$

 

[haruspex]

The question doesn't specify this but I think so.



It was a typo, my bad. It should say 30 degrees.


This is the frictional force between the top and bottom cylinder? I did it again and got $N=mg\frac{\sqrt 3}4$

So do you get the right answer now? If not, please post the rest of your working.

 

[lloydthebartender]

So do you get the right answer now? If not, please post the rest of your working.

$N$ becomes $(3/8)mg$ vertically and $(\sqrt3/8)mg$ horizontally. If the frictional force ($\mu N$) between the upper and bottom sphere points outwards at a 30-degree angle to the equilateral connecting the centres of the spheres, then the vertical component is $(\sqrt3/8)mg$ and $(3/8)mg$, and then equating the sum of horizontal components to the sum of vertical components multiplied by $\mu$ I get a quadratic $\sqrt3 \mu +\mu - \sqrt3/8 = 0$ and if I discard the negative answer I still don't get the right number (should be 0.893)

 

[haruspex]

what is the minimum coefficient of friction between the horizontal surface and the cylinders?
:

Firstly the normal reaction force of the bottom cylinder is $R$ and because all three cylinders are identical $2R=3mg$, and so the frictional force is going to be $F_r={(3mg)/(2}{\mu})$. I don't use this, though.

The question could be interpreted as saying not to worry about the friction coefficient between the cylinders, just assume there is no slipping there. Find the minimum coefficient of friction between each lower cylinder and the horizontal.

 

[lloydthebartender]

The question could be interpreted as saying not to worry about the friction coefficient between the cylinders, just assume there is no slipping there. Find the minimum coefficient of friction between each lower cylinder and the horizontal.

The minimum frictional force between the surface and the bottom cylinder must account for only the horizontal component of the normal force due to the top cylinder so its $F_r= \mu R$ and so $(\sqrt3/8)mg=\mu (1+3/8)mg$ and $\mu=0.157$?

 

[haruspex]

The minimum frictional force between the surface and the bottom cylinder must account for only the horizontal component of the normal force due to the top cylinder so its $F_r= \mu R$ and so $(\sqrt3/8)mg=\mu (1+3/8)mg$ and $\mu=0.157$?

No, you are forgetting the horizontal component of the frictional force between the cylinders.
I get 0.0893...

 

[lloydthebartender]

No, you are forgetting the horizontal component of the frictional force between the cylinders.
I get 0.0893...

The horizontal component of friction between the top and bottom cylinder is $(3/8) mg$ to the left because it makes a 30-degree angle to the side of the equilateral, right?

 

[haruspex]

The horizontal component of friction between the top and bottom cylinder is $(3/8) mg$ to the left because it makes a 30-degree angle to the side of the equilateral, right?

To get that, aren't you using the earlier assumption that the contact between cylinders is about to slip?
I recommend you to start again, just working with the normal and frictional forces, not involving the coefficient.
An interesting place to start is the relationship between the two frictional forces acting on a lower cylinder.

 

[ehild]

I get that $N=mg\frac{\sqrt 3}4$.

Using both force-balance and torque-balance equations, I got $\mu=0.0893$, and N=mg/2 .

 

[haruspex]

Using both force-balance and torque-balance equations, I got $\mu=0.0893$, and N=mg/2 .

Hmm... so do I now. Can't find my original scribblings. Thanks.