Theatre Design & Technology - Winter 2008 - 13

FT F Fg F FT Fg F Ffr F F FN Fg F F FT Ffr FN F
Starting from rest, the tension in the cable was slowly increased until the skid started to move. The reading on the load
cell at the moment just prior to movement gave us the static
frictional force (for a more detailed discussion of physics
terms, see the section "The Test Process," below). Once the
skid was moving, we waited for the drive to level off the velocity
so that there was no acceleration. In this case, the reading on
the load cell gave us the kinetic frictional force.
Because the new system incorporated a continuous chain,
some care had to be taken to ensure that we were recording not
only the force it took to get the skid moving, but also any resisting force caused by the chain. Consequently, the stage had two
load cells-one to record the pulling force on the front end of
the skid and one to record the cable tension pulling backward.
The latter can be subtracted from the former to ﬁnd the net pull
force. This will be discussed in detail in the section "The Test Process."
The tension data provided by the load cells was processed
through a signal conditioner (National Instruments SCC SG24
module in a National Instruments SC 2345 chassis) that signiﬁcantly reduces electrical signal noise and then acquired on
a personal computer via a National Instruments PCI 6221 DAQ
card. Once all the tensions were collected, Microsoft Excel was
used to analyze the data and compile ﬁnal results. The macro
creation feature in Excel made this process much quicker.

sured in pounds) and g = 32.2 ft/s 2 is the acceleration of gravity. In this case, it includes the weight of the skid and everything
on the skid.
F N The normal force is the force of the platform pushing
up on the skid. There is no generic formula for the size of the
normal force. It must be determined using Newton's laws of
motion (we'll see how to do this below).
F T, 1 and F T, 2 The tensions in the chain pulling in the
direction of motion (F T, 1) and opposing the motion (F T, 2).
Like normal forces, there is no basic tension formula. In this case,
these forces were measured using load cells during these tests.
F fr, s and F fr, k The static and kinetic frictional forces.
When the skid is stationary, we use the static frictional force;
and when it's moving, we use the kinetic. In equations, the coefﬁcient of friction is represented by the Greek letter µ, pronounced "mew." The magnitude of the static frictional force is
Ffr, s ≤ µs FN (it's as big as it needs to be to prevent motion, up
to a maximum allowed value) and the magnitude of the kinetic
frictional force is Ffr, k = µk FN . For the static case, the easiest
point at which to take data is when it's at a maximum, so for our
tests, Ffr, s = µs FN . In these equations, µs is the coefﬁcient of
static friction and µk is the coefﬁcient of kinetic friction-the
things we're trying to ﬁnd! Solving these equations for µ, we get:
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THE TEST PROCESS
When analyzing a problem using forces, the most direct approach is to use Newton's laws of motion. Most important to
this case is the second law, which states that the sum of forces
(the net force) acting on an object is equal to its mass times
its acceleration (symbolically: F net = ma ). We designed the
experiment so that we would be collecting data only when the acceleration was zero (either before the skid started moving-the
static case, or when the skid was moving at a constant speed-
the kinetic case). In both of these cases, Newton's second law
of motion says that the sum of all the forces acting on our skid,
F net , has to add up to zero, which makes the analysis simpler.
The ﬁrst step is to identify the forces present in the system.
As seen in ﬁgure 5, the test skid feels ﬁve forces:
F g More commonly known as weight, the magnitude
(size) of the force of gravity is given by Fg = mg, where m is
the mass of the object (measured in slugs, if forces are mea→

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Figure 5. Free body diagram showing forces on skid during ﬁnal test.

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THEATRE

DESIGN

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TECHNOLOGY

W I N T E R

2008

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When we add these forces, we cannot just add them as if
they are regular numbers. Forces are vectors-it matters how
big a force is, but it also matters which way it points. Two people applying forces of 100 lbs in the same direction will produce a much different result than the same two people pulling
in opposite directions. The way we deal with this is to break the
vectors into components-how much of the force lies along
the x-axis, and how much lies along the y-axis. When we do
this, we get two equations, Fnet, y = 0 (the sum of the forces in
the vertical direction is zero) and Fnet, x = 0 (the sum of the
forces in the horizontal direction is zero). In general, this can
get complicated, but in this case, where all of the forces are
either horizontal or vertical, it's not that bad.
Vertical Components. From ﬁgure 2 we can see
that F N and F g are the only vertical forces, so when we add
them together (recalling that the acceleration is zero), we get
Fnet, y = FN - Fg = 0.
Note that we have to be careful about direction-the
vertical component of the normal force is positive because it
points upward and the vertical component of gravity is negative
because it points downward. Solving this equation, we ﬁnd that
FN = Fg, or in other words, the magnitude of the normal force
is equal to the total weight of the skid.
Horizontal Components. In the horizontal direction, there are three forces-the two tensions and
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Table of Contents for the Digital Edition of Theatre Design & Technology - Winter 2008
