Physics 133 - Summer 2002
Experiment #2: Newton's Second Law
In this experiment we will verify the validity of Newton's Second Law
by studying the motion of the system depicted in Fig. 1. By measuring
the kinematical value of the acceleration and comparing it to the value
predicted by the Second Law we will obtain another value for the
acceleration of gravity, g.
Dynamics is the science that studies the causes of motion;
Newton's Second Law is its fundamental principle. Newton's Law is a
cause-effect relationship: any dynamical system subjected to external
forces will obey
On the left-hand side we have the cause, the total or
net force acting on the system; on the right we have the effect:
the system accelerates in the direction of the net force, and
the amount of acceleration is directly proportional to the amount of
force; the constant of proportionality defines the system's mass.
Figure 1. Experimental setup
If you did the homework, you should already be acquainted with the
system whose motion we will study. We have two masses tied together
with a long piece of tape, which we can take to be massless and rigid.
One of the masses rests on a frictionless horizontal surface; the
ribbon is draped over a frictionless ``pulley'' in such a way that the
other mass hangs freely.
Newton's Second Law predicts that the system will move with uniform
acceleration, the value of which we can find using the following argument:
On one hand, the only unbalanced external force that acts on-and
thus drives-the system is the weight mg of the hanging mass; on
the other hand, the inertial mass of the entire system is m+M, and
so the acceleration must be
Both masses will move with the same acceleration because the tape
doesn't stretch. We wish to compare the acceleration thus predicted
to one that we can measure directly. One of the equations that
describe the motion of a body with uniform acceleration tells us
its speed after it has covered a distance x:
a = ||
If the object is initially at rest, the square of the speed is then
simply v2=2ax, whence a=v2/2x. If we compare this last
expression to Eq. (2), we find the equality
which tells us that if we plot v2/2x vs. m/(m+M) we will obtain
a straight line with slope g and no intercept.
A linear air track is a hollow beam pierced by many small holes that
emit gentle streams of air. The glider of mass M rests on the beam
and, when the air is turned on, floats on a thin cushion of
air and moves with negligible friction. Affixed to the top of the
glider is a black card of length L. On the same end of the track as
the ``pulley'' is a light source shining on a photogate, shown in
Fig. 1 by a circle.
An electronic clock has been set so that it will start whenever the
light beam is interrupted by the passing black card and will stop when
the beam is restored. The time of interruption, t, is displayed in
milliseconds on an LED device. (Don't forget to reset the timer
before each data run.) During this time the glider advances a
distance equal to L, the length of the card. The quantity L/t is
a good approximation to v, the glider's velocity at the point where the
light beam is interrupted by the very center of the car; at this
same point, the acceleration is approximately given by v2/2x.
Start by finding the mass of the glider using a scale. Write
down the value (and don't forget the units ):
Measure L, the length of the card, using a ruler.
Move the glider on the track towards the photogate until
you have it right at the point where the timer is about to start
running. Record the position of the glider (use the pin on the
glider's side as your pointer). Refer to Fig. 1: you have found
Keep moving the glider towards the pulley until the timer
is about to stop running. Record the position; now you have
Subtract b from a. The number you obtain, denoted
in Fig. 1 by c, should be equal to L as found in Step 2 (why?).
Does it? If the values are not identical, which one do you trust
the most? Why? Would you rather use an average of the two? Why?
By averaging b and a you can find the position of
point d. We want x = 50 cm throughout the experiment; assuming
that the photogate won't move, we can release the glider from
the same initial position every time. What is this position?
Now you are ready to take data. Hang a 5-gram mass from
the tape. Make sure there are no obstacles on its way. Slide the
glider to x0 and hold it still for a few seconds. Let it go.
Record the readout from the timer; don't forget to convert it to
seconds. Vary M and m and fill the data table: the first column
is Ma, the mass added to the glider (zero in this first case); the
third is the hanging mass; the second is M=M0+Ma. The mass ratio
will have the same value regardless of whether you use grams or
kilograms, as long as you are consistent.
Make a graph showing v2/2x on the y-axis and m/(m + M)
on the x-axis. Use all of the points. (Don't forget to label
your axes; whenever something has units, display them. Give your graph a
title. ) Perform a linear fit. Write down the values you obtained
for the slope m and the intercept b.
Repeat Step 1 but just use the points where m is smallest.
Write down the slope m and intercept b that you obtain.
m= b =
Repeat Step 1 using the points where m is largest.
m= b =
Repeat Step 1 using the points where M is smallest.
Repeat Step 1 using the points where M is largest.
m= b =
The hanging mass m has two forces acting on it: its
weight mg and the tension T of the string. Newton's Second Law
then says that the equation of motion for this mass is mg-T=ma. In
the space below, draw the mass and the forces that act on it.
Three forces act on mass M: its weight, Mg, the
tension T of the string, and the normal force N exerted
by the air track that keeps the mass from falling through.
We now have two equations of motion: Mg=N and T=Ma.
In the space below, draw the mass and the forces acting on it.
Why have we taken the tension to be the same in both
cases? Why have we assumed that the acceleration is the same
for both masses?
Write down the value of g that you found in the
preceding section. Express it in m/s2.
Look at your graphs from the preceding section and
select the one whose slope is lowest. Which one is it?
What is the slope? Call this slope (s1) and write it down.
Find g-s1, using g from Step 4.
s1= g-s1 =
Now select the graph whose slope is highest.
Which one is it? What is the slope? Call this slope (s2) and write
it down. Find s2-g, using g from Step 4.
s2= s2-g =
The values calculated in the last two steps of the
preceding section give us an estimate of the uncertainty in our value
of g. We can say that g=[`g]±Dg, where
[`g] is the value that includes all points; as for the
uncertainty, given g-s1 and s2-g we can take Dg to be
the larger of the two. Write down yet again the value you obtained
for g, but now add the uncertainty (which has units).
The uncertainty in the measurement can also be
expressed as a fraction of the final value: g=[`g]±p%,
where p=Dg /[`g]×100. The uncertainty
is now dimensionless.
Compare your final g to the accepted value,
g » 9.8 m/s2. Is your measured value too low or too high?
What would be the effect of friction and air drag? Could these
explain why your result was too low or too high?
Which of the plots from the section on Data
Analysis gives the value of g that is most ``wrong''? Which
one gives the ``best'' value? Could this information help us
know how to get a more accurate value for g?
We have not discussed the intercept of any of
the plots. We know that they should be very close to
zero. Is this the case? What does this intercept mean?
Data Sheet for Experiment #2: Newton's Second Law
File translated from
On 21 Jun 2001, 19:12.