A pyramid of three cylinders (statics problem)

In summary, the normal reaction force between the top and bottom cylinders is half of the weight of the top cylinder and the bottom cylinder has a vertical and horizontal force applied to it. The angle between the normal reaction force and the vertical is 30 degrees. The coefficient of friction between the cylinders and the horizontal surface is assumed to be the same as the coefficient between the cylinders. After considering the balance of forces on a lower cylinder, the minimum coefficient of friction between the horizontal surface and the bottom cylinder is calculated to be 0.157. However, this answer does not take into account the horizontal component of the frictional force between the cylinders.
  • #1
lloydthebartender
50
1
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!
 

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  • #2
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.
 
  • #3
lloydthebartender said:
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.
lloydthebartender said:
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.
 
Last edited:
  • #4
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?
 
  • #5
lloydthebartender said:
Because the angle ##N## makes with the vertical is 60 degrees,
Are you sure?
 
  • #6
ehild said:
Are you sure?
Just a typo, I assumed, since the subsequent calculations treat it as 60:degrees to the horizontal.
 
  • #7
Lnewqban said:
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.

ehild said:
Are you sure?
haruspex said:
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.

haruspex said:
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##
 
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Likes Lnewqban
  • #8
lloydthebartender said:
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.
 
  • #9
haruspex said:
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)
 
  • #10
lloydthebartender said:
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.
 
  • #11
haruspex said:
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##?
 
  • #12
lloydthebartender said:
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...
 
  • #13
haruspex said:
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?
 
  • #14
lloydthebartender said:
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.
 
  • #15
haruspex said:
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 .
 
  • #16
ehild said:
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.
 

Related to A pyramid of three cylinders (statics problem)

1. What is a pyramid of three cylinders in a statics problem?

A pyramid of three cylinders in a statics problem is a geometric structure made up of three cylindrical objects stacked on top of each other in a triangular shape. This type of problem is commonly used in engineering and physics to analyze the forces acting on the cylinders and determine their stability and equilibrium.

2. How do you calculate the weight of a pyramid of three cylinders?

To calculate the weight of a pyramid of three cylinders, you will need to know the mass and dimensions of each cylinder, as well as the gravitational acceleration of the Earth. The total weight can be found by adding the weight of each individual cylinder together using the formula W = mg, where W is the weight, m is the mass, and g is the gravitational acceleration.

3. What forces are acting on a pyramid of three cylinders?

There are several forces acting on a pyramid of three cylinders, including the weight of the cylinders, the normal force from the surface they are resting on, and any external forces that may be applied to the structure. Additionally, there may be frictional forces between the cylinders and the surface, depending on the materials and surface conditions.

4. How do you determine if a pyramid of three cylinders is in equilibrium?

To determine if a pyramid of three cylinders is in equilibrium, you will need to analyze the forces acting on each cylinder and ensure that the net force and net torque are equal to zero. This means that the cylinders are not moving or rotating, and are in a stable position. If the net force or net torque is not equal to zero, the cylinders are not in equilibrium and may be at risk of toppling over.

5. What factors can affect the stability of a pyramid of three cylinders?

The stability of a pyramid of three cylinders can be affected by several factors, including the weight and dimensions of each cylinder, the surface they are resting on, the presence of external forces, and the coefficient of friction between the cylinders and the surface. Additionally, the shape and placement of the cylinders can also impact the stability of the structure.

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