Statics problem

[iwfc87]

Heres a problem I've tried doing for ages but I can't quite get it right

The answer is *apparently* approximately 10.8 degrees.

 

Heres the question and a diagram to help visualise:

 

 

Three smooth cylinders are stacked up in a trough as shown in the diagram below. Each cylinder weighs 400N and has a diameter of 125mm. Calculate the minimum angle theta which will support these cylinders in equilibrium.

[mezarashi]

An interesting problem indeed. It helped me brush up some of my highschool free body analysis skills. There was a little tricky part but otherwise standard procedure. I hope I can explain it well enough to not have to use any diagrams

 

Given the 3 cylinder system, we know that the cylinders are fixed relative to each other, meaning that they will always form an angle of 60 degrees due to the symmetery. This means that the force of cylinder A on B and C is always constant and proportional to the weight of cylinder A. We have:

 

[1. Fa ]

2(Fa)sin60 = Fgrav, where Fgrav = 400 N as given

 

Then Fa = 230.94 N

 

[2. Fn ]

Analyzing any of the two bottom cylinders (either is fine due to symmetry, but I used cylinder C), we can get the equilibrium forces in the x and y directions. Let me know if you have difficulty with drawing out the geometry, then diagrams will be necessary.

 

y: (Fn)cosx = Fgrav + (Fa)sin60, where x denotes theta.

The right side is a constant, and leads to

 

Fn = 600/cosx

 

x: (Fn)sinx = (Fa)cos60 + Fb

Fb as a function of Fn would be:

Fb = (Fn)sinx - (Fa)cos60

 

[3. Fb]

Substituting in equations we get:

Fb = 600sinx/cosx - (Fa)cos60

Fb = 600tanx - (Fa)cos60

 

Three equations but 4 unknowns (Fa, Fb, Fn, x), what do we do? The trick is understanding that at the verge of slipping, or when theta is just perfectly right, then Fb will be zero! You can imagine this if the cylinders are on a horizontal floor. There will be no Fb. If the cylinders are not in equilibrium, then there cannot be an Fb. Cylinder A will undoubtedly push apart cylinders B and C.

 

So with our last condition of Fb = 0, we get:

 

0 = 600tanx - (Fa)cos60, where Fa = 230.94

Solving:

 

x = 10.8 degrees

 

[MetaFrizzics]

By inspection, the answer is <=30o for friction-free smooth cylinders. (original question)

 

An equilateral triangle has 60o corners.

 

(1) If there is no friction, then the cylinders will slip in the manner of slippery soap: Consider arbitrary weights of cylinders first (general case)

The moment the angle between point of contact with the inclined plane and the point of contact with the upper cylinder is >90o in the direction of true horizontal, the lower cylinder(s) pops outward.

 

The 'normal' to a 60o line is...30o. All done.

 

(2) In the case of non-slip friction between the cylinders and surfaces, counter-rotation causes a severe complication in the calculation, but essentially, the answer is 0o ! (The cylinders can rest virtually horizontal!)

 

(3) Furthermore, if the coefficient of friction is unknown but in the range where slippage is possible, you have to solve the problem using Catastrophe Theory, and probabilistic models.

[mezarashi]

(3) Furthermore' date=' if the coefficient of friction is unknown but in the range where slippage is possible, you have to solve the problem using Catastrophe Theory, and probabilistic models.[/quote']

 

Catastrophe Theory... thank god they're considered frictionless

[MetaFrizzics]

(1) We assume first that the top cyclinder can be any arbitrary weight relative to the bottom cylinders:

 

 

The angle of the force requires that the plane be inclined less than 30o regardless of relative weights. Now the (original) question is, when does the downward direct force (potential energy from gravity used) overpower the force required to raise the other two cylinders up the inclined plane the amount needed to get them out of the way? (2x for two upward cylinders)

 

To understand this, and why the angle must be >90o for any motion to occur, it can be modelled by an analogy: imagine the centre cylinder is actually a ball coming down and bouncing off a stationary ball on the horizontal plane:

 

 

(Friction-free case) As long as the ball strikes spot on, the ball resting on the plane does not move at all, regardless of the angle the top ball strikes it at. It is as though the ball struck the plane itself directly.

 

 

What is interesting here is that as the center cylinder moves downward (accelerating) less and less energy is actually required to overcome the other two cylinders: by the time the center cylinder is nearly horizontal with them, very little displacement is needed to allow passage, and hence little potential gravitational energy is transferred to the other cylinders. This is due to the shape of the objects in the plane of interest (the vertical slice shown in diagram). Note that although adjustable, the plane is assumed to be fixed throughout the motion of the cylinders, once the critical angle is established. (original problem).

 

Thus while the center cylinder positively accelerates, the other two cylinders accelerate then decelerate. Due to inertia, they will also travel slightly farther apart than is necessary for the centre cylinder to clear them.

 

What is exciting is that we know that once we have enough energy to begin motion of the centre cylinder, we can be confident that the action will carry through! That is why it should be relatively easy to define the critical angle.

 

In fact, we should be able to calculate the actual acceleration of the central cylinder, the acceleration and range of motion of the outside cylinders, and the force of impact when they roll back together.

[MetaFrizzics]

Interestingly, the problem is only mildly more complicated if only either the surface of the inclined plane is friction-free, or the surfaces between the cylinders are. In that case, friction would cause rotational motion of the outside cylinders. This stores energy in the form of angular momentum due to inertia, and so robs the central cylinder of some potential energy available for lifting the outside cylinders at the start of the problem, while re-injecting that energy later in the process of motion.

 

From this we can deduce that balanced friction reduces the problem to the previous situation (without rotation) only adding an additional load requirement for the mass of the central cylinder.