“Why don’t you fix your little problem and light this candle.”
–America’s first astronaut, Alan Shepard, on the launch pad after some delays.
Rest position of a rod in a bowl in the absence of friction
Andrew Mosedale
Background sufficient to solve this problem is provided by reading R.C. Hibbeler’s Engineering Mechanics: Statics through Chapter 5. Also, you’ll need to use your math
skills from previous courses. As statics problems go, this one is pretty hard.
Suppose a uniform rod of length L = 3R and weight W is placed in a bowl of radius
R. Further suppose that friction is negligible. What will be the equilibrium position of
the rod?
First thoughts.
Sensible notation and careful figure-drawing are essential to problem-solving.
1. How can we specify the position of the rod? Specifically, is there a single variable
sufficient to fix the position?
2. What is the relevant geometry of this problem? How can we amend the figure to
improve our understanding?
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An Amended Figure.
The angle θ is sufficient to specify the position of the rod. Do you see from the geometry
that all angles marked θ in the figure are equal? This is, perhaps, the crucial step in
solving this problem.
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The Free-body diagram.
We would now like to draw a free-body diagram of the rod, which we presume is in
equilibrium. We set the rod free, throwing away everything that is not the rod. We very
carefully imagine the forces on the rod exerted by everything we threw away. There
are three forces:
1. The force that the bowl exerts on the rod at A. Because there is no friction, this
force must be directed along the mutual normal of the contacting surfaces. In this
case, the direction is normal to the surface of the bowl (and thus along the radius
drawn in black in the previous figure and reproduced here–see how important
good figures are?)
2. The force that the bowl exerts on the rod at B. Again, because there is no friction,
this force must be directed along the mutual normal. In this case, the direction
is normal to the rod.
3. The gravitational force on the rod at C (its midpoint). This resultant will be
directed downward. Note that the distance between A and C is L2 .
Also, notice that we need not submit to the tyranny of the page. We do not have to
choose the x-direction to the right and the y-direction upward. We choose to orient
them parallel and perpendicular to the rod. For convenience.
Here is the free-body diagram:
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Equations of Equilibrium.
Our problem represents the equilibrium of a rigid body in a plane. Thus, there are up
to three independent equations of equilibrium.
In our case, there are exactly three. They are:
X
Fx = 0 = F cos(θ) − W sin(θ)
(1a)
X
Fy = 0 = F sin(θ) + N − W cos(θ)
(1b)
X
MA = 0 = −W cos(θ)
L
+ N 2R cos(θ)
2
(1c)
Note carefully how these equations are derived from the free-body diagram. Without
the free-body diagram, as is so often the case, we are lost. Also, note that in developing
these equations, we resolve F and W into components aligned with our conveniently
defined coordinate axes.
Let us take stock. What are the unknowns? F , N and θ. Why is not W an unknown?
Answer: it is a given–see the first sentence of the problem statement. Thus, we have
three unknowns. We also have three independent equations.
We should feel now the courage to go on.
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Algebra and Trigonometry.
This problem rewards those who have diligently studied algebra and trigonometry.
Solve equation (1a) for F :
F = W tan(θ)
Solve equation (1b) for N :
N = W cos(θ) − F sin(θ)
Solve equation (1c) for W :
W =
4R
N
L
Now back-substitute and simplify to find:
L
= cos(θ) − sin(θ) tan(θ)
4R
(2)
Equation (2) is nice. It is one equation in one unknown, θ, and this is what we have
been looking for. Sadly, in its current form it appears a bit difficult to solve because
we cannot isolate θ.
Let us see if we cannot humble this equation with a little trigonometry. Our goal is
express the equation in terms of a single trigonometric function instead of the current
three trigonometric functions (sine, cosine and tangent).
Try multiplying the left and right sides of equation (2) by cos(θ). The result is:
L
cos(θ) = cos2 (θ) − sin2 (θ)
4R
Now we are down to only two trigonometric functions. We can eliminate sin2 (θ), by
using the well-known trigonometric identity:
sin2 (θ) = 1 − cos2 (θ)
The result is:
L
cos(θ) = 2 cos2 (θ) − 1
4R
We are almost home and dry. We now have just a single trigonometric function in our
equation. Let us rearrange:
2 cos2 (θ) −
L
cos(θ) − 1 = 0
4R
This is a quadratic equation. To make this clearer, let us define x = cos(θ). The
equation is then:
L
2x2 −
x−1=0
(3)
4R
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This is a general result. To specialize for our problem, we know L = 3R, so equation
(3) becomes
3
(4)
2x2 − x − 1 = 0
4
Solving the quadratic equation, we find that x1 = 0.9190 and x2 = −0.5440. We discard
the second root because it would give a negative angle, which is preposterous. Recalling
the definition of x, we say
θ = arccos(x1 ) = 23.21◦
The rod will slant at 23.21◦ to the horizontal.
So we are done, right?
No. Given all the work we have invested to this point, we deserve to explore the general
behavior of our solution.
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Further Exploration.
Some questions that may occur to you at this point include the following:
1. What would happen if the rod were longer or shorter? We’ve solved for only a
single value, L = 3R. However, we had the foresight to find equation (3), which
is valid for general values of rod length, L.
2. Are there limits? In other words, can the rod be too short or too long for our
solution? If there are limits, what happens at (or beyond) these limits?
3. Can we program the solution on a computer to explore these questions? Can we
use the computer to illustrate our insights? I hope you will get in the habit of
using computers and of programming. Computer programming is an essential tool
for the modern engineer in any discipline. Embrace programming enthusiastically,
not reluctantly.
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Results of Further Exploration.
1. If the rod were longer than the given length (3R), it would lie flatter (θ would be
smaller). An interesting figure would be to plot θ versus L/R. This plot would
show the trend I have described.
It is left as a programming exercise to the reader to prepare such a plot.
2. There are limits. The easier limit to visualize is the maximum possible length
of the rod, Lmax = 4R. A rod near this length would lie almost horizontal. A
rod any longer would tip out of the bowl. The harder limit to appreciate is the
minimum possible length of the rod, Lmin = 2 (2/3)1/2 R ≈ 1.633R. A rod near
this length would be fairly steep. If the rod were any shorter, it would slip into
the bowl and wind up lying horizontal within the bowl.
It is left as an exercise to the reader to prove these limits.
3. It is certainly possible to program these results–I did. I used MATLAB. As with
any programming environment, there is a learning curve. Consequently, there is
some frustration. You can easily imagine doing various tasks, but you cannot
quite get them done. Patience and persistence will eventually reward you with
many new capabilities.
The reader is encouraged to explore this problem via programming.
The figure above shows ten rods of varying length between about 1.63R and 4R in their
equilibrium positions in a bowl. The rods are color-coded by length as indicated by the
colorscale on the right. This figure was produced with MATLAB.
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