Experiment 8. Acceleration of a constant mass vs. net applied force gives a linear graph which can be extrapolated to obtain the acceleration of gravity.

See Introductory Video Experiments video clip #15, CD EPS01.  Be sure you look under the heading Introductory Video Experiments.

As shown in the video clip, we will relate the net force accelerating a constant mass to the acceleration of the mass.

• We will do this for a system consisting of a low-friction dynamics cart on a uniform incline set to allow the cart to travel down the incline at a constant velocity, and a number of paper clips.   Different numbers of clips will be suspended over a pulley so that the gravitational force on the suspended mass might provides a variety of net accelerating forces on the system.

We first obtain the data necessary to determine the acceleration of the system vs. the number of suspended clips.

• As instructed on the clip, for each number of suspended clips we time the cart as it accelerates from rest thru the given 69-cm distance.

We determine the acceleration of the system for each number of suspended clips.

• For each number of suspended clips, from the time to accelerate from rest and the distance, determine the average velocity, final velocity and acceleration of the system.

Next, plot the acceleration of the system vs. the number of suspended clips and interpret the results.

• Plot an accurate graph of acceleration vs. number of suspended clips and determine how well a straight line can be made to fit the data.
• Explain why, if the system is set up as described, the x-intercept of the graph should be very close to the origin.
• The x intercept will not go exactly through the origin, unless you have allowed the existence of the origin to influence the location of your line.  Explain the meaning of the x intercept.

Plot acceleration vs. the net number of suspended clips.  The net number of suspended clips is the number of suspended clips, minus the number required to overcome friction.

• Assuming that gravity exerts the same force on each suspended clip, how well do your results confirm Newton's Second Law, which states that the acceleration of a system of constant mass is directly proportional to the net force on the system?
• Describe another experiment that would be necessary to completely validate Newton's Second Law.

Use your results to predict the acceleration of the entire system when the net force is equal to that of the gravitational force on the mass of the system:

• Each clip has a mass of about 1.2 grams.  The cart has a mass of about 300 grams.  There were a total of 22 clips on the cart and suspended from the pulley.  How many clips would be required to equal the mass of the entire system?
• According to your graph of acceleration vs. net number of accelerating clips, what acceleration would be expected if the system was subjected to the force of gravity on this number of clips?
• If the system is simply dropped and allowed to fall freely to the floor, it will accelerate in response to the force exerted by gravity on its mass. What will this acceleration be?

See Introductory Video Experiments video clip #12, CD 0.  Be sure you look under the heading Introductory Video Experiments.

Note that on the video clip the equilibrium position is not visible.  This was an accident that occurred as a result of some automatic cropping of the video image by the video capture program; we attempted to get the best possible closeup of the setup and simply got too close for the capture program.   However we have decided to leave the clips as they are and ask the user to extrapolate the equilibrium position of the pendulum.  The user should obtain data only for 5, 10, 15, 20, 25 and 30 suspended masses, inasmuch as this is possible, but should also observe whether the addition of a single weight gives a clear difference in position.

• Create a table for position measurement vs. suspended mass, and add a column for the displacement from equilibrium.  Use common sense to determine what the equilibrium position must have been, based on the positions for 5, 10, 15, 20, 25 and 30 suspended weights.
• Construct an accurate graph of y = pendulum displacement from equilibrium vs. x = suspended mass. 

• Your graph should, within experimental uncertainties, give you a straight line through the origin. Use the slope of this graph to estimate the displacement from equilibrium corresponding to a suspended mass of 50 grams, and to a suspended mass of 100 grams.

Now check out the proportionalities involving forces and distances.

• The force exerted on the pendulum bob by the tension in the thread is in fact somewhat less than the force exerted on the suspended mass by gravity. This is due to the friction associated with the pulley. This friction reduces the force by approximately .5%, so that the actual force exerted on the pendulum will be approximately .995 of the force exerted by gravity on the suspended weight. Thus the effective suspended masses can be regarded as .995 of the actual suspended masses.
• Make a table of pendulum displacement from equilibrium vs. effective suspended masses.
• Based on this table, create another table of [ pendulum displacement / length of pendulum ] vs. [ effective suspended mass / mass of pendululm bob ].

• If we let x = pendulum displacement at measurement position, L = length of pendulum string to measurement position, m = effective suspended mass and M = mass of pendulum bob, then this last table depicts (x / L) vs. (m / M).
• Does your table, within experimental uncertainty, confirm the hypothesis that pendulum displacement as a proportion of pendulum string length (i.e., x/L) is identical to the ratio of masses (m / M)?

• Construct a graph of effective suspended mass m vs. pendulum displacement x. Fit a straight line to your graph and determine its slope `dm / `dx.
• If it is in fact true that x / L = m / M, then x / m = L / M so that the slope `dx / `dm must be equal to the unchanging ratio L / M.

• How well do your results confirm the hypothesis that x / L = m / M?

Use the proportionality x / L = m / M to predict various displacement and suspended masses.

• Use the proportionality to predict the displacement x corresponding to a suspended mass of 70 grams. Recall that m stands for the effective suspended mass, which for this pulley with its .5% coefficient of friction is .995 of the actual suspended mass.
• Use the proportionality to predict the effective mass m that will lead to a displacement of 2.3 cm from the equilibrium position.

Justify the following statements: 

• Since the weight of an object is equal to the force exerted on it by gravity, which is proportional to mass, then if F is the force of gravity on the effective mass and W the weight of the pendulum bob, m / M = F / W and therefore x / L = F / W.  That is, the displacement is in the same proportion to the length of the pendulum as the applied force to the weight of the pendulum.
• If a force F is applied to the pendulum in the horizontal direction, then the displacement of the pendulum from equilibrium will be F / W times the length of the pendulum.
• The length and displacements of the pendulum as measured in this experiment were not the actual length and actual displacements of the pendulum.  The actual length should have been measured to the center of the sphere, and the displacements should have been measured at this position.  However, the ratios x / L obtained in the experiment are identical to the x / L ratios that would have been obtained had these quentities been measured at the center of the sphere.

Analysis of Errors

• Estimate the errors in your effective masses and in your measurements of pendulum position, and represent your errors on the appropriate graph.
• Discuss the implications of your errors.

Note that the results obtained here are only valid as long as the displacement of the pendulum from equilibrium is small compared to its length.

• If displacement is less that .1 of the length, then predictions of forces or displacements are accurate to within +-.5%, or +-.005 of the actual quantities.

See Introductory Video Experiments video clips #13, CD EPS01.  Be sure you look under the heading Introductory Video Experiments.

By pulling back a pendulum of known length and mass using two attached strings with a rubber band between them, we see that the additional force required to pull the pendulum back further and further results in a greater and greater stretch of the rubber band. By measuring the length of the rubber band vs. the displacement of the pendulum we infer and graph force vs. stretch for the rubber band. Repeating the experiment with two identical rubber bands in series, then again with the two rubber bands in parallel, we compare the forces exerted by these combinations with the force of a single rubber band. We store our calibration of the rubber bands in a convenient calibration program. We repeat this experiment for two different types of rubber bands.

We can measure forces using a rubber band and a scale by which to measure the stretch of the rubber band. The force vs. stretch curve is not linear for a rubber band. For a spring which is light compared to its strength, a force vs. stretch curve would be very nearly but not precisely linear; however we choose here to use the rubber band in part because it is readily available and cheap, and in part for the very reason that it is nonlinear to a degree that requires us to acknowledge it.

We begin by using the rubber band to displace a simple pendulum of known mass and length through a variety of displacements.

Begin, as instructed in the video clip, by observing rubber band length vs. pendulum position.

• Construct a table of rubber band length vs. pendulum position. Add a column for rubber band stretch and one for force.
• In the column for rubber band stretch, fill in the magnitude of the stretch of the rubber band in excess of its length when the slack has just been removed.
• In the column for force, determine the magnitude of the force corresponding to the appropriate pendulum displacement. (The weight of the pendulum is equal to the force exerted on it by gravity, which is W = m g; this weight will be in Newtons when m is in kg and g in m/s^2.  The pendulum has a mass of 1180 grams or 1.18 Kg).

Sketch a graph of force vs. rubber band stretch.

• Construct an accurate graph of the force exerted by the rubber band, in Newtons, vs. rubber band stretch, in meters.

• Note that in Video Experiment 9 you determined that the suspended mass m necessary to displace the pendulum is in the same proportion to the mass of the pendulum as the displacement to the length of the pendulum.

  It follows that the force F necessary to displace the pendulum is in the same proportion to the weight of the pendulum as the displacement from equilibrium to the length of the pendulum.

  This force is provided by the rubber band. So F / W = x / L and F = (x/L) * W.

  The weight W is the product of the mass and the acceleration of gravity, as mentioned above.

• Submit a description of this graph, and speculate on the reasons for the was its shape deviates from linearity.
• Keep your graph for future reference.

The graph you have obtained is a calibration curve for the rubber band.  Any rubber band can be calibrated in a similar manner.

Your graph might look something like that shown below:

We calibrate another rubber band by hanging weights.

See Introductory Video Experiments video clip #14, CD EPS01. 

Observe the lengths of the suspended rubber band vs. the total mass suspended from the rubber band (each mass except the first, which is 10 grams, is 50 grams)

Make a table of force in Newtons vs. stretch in meters.

• Construct a graph of force vs. stretch.

Answer the following questions:

• According to your curve, what force would be associated with each of the following stretches:  .3 cm, 1.4 cm, 2.7 cm?
• According to your curve and pendulum proportionalities, what stretch would be associated with each of the following pendulum displacements:  5 cm, 15 cm, 25 cm and 35 cm?

Analyze the errors in this experiment.

See Introductory Video Experiments video clips #19, 20, CD EPS01.

The 'effective' slope of a ramp is the slope in excess of the (small) slope necessary to overcome friction -- the slope at which the friction cart moves down the ramp at a constant velocity.  

By observing the slopes at which various numbers of suspended washers exert a force equal to the component of the gravitational force in the direction down the incline, we test whether for small slopes the net force exerted on a cart coasting down a ramp is proportional to the 'effective' slope of the ramp.

We begin by determining the slope at which the cart moves down the ramp at constant velocity.

• This was not shown on the video clip, but the for the cart used the constant-velocity slope is approximately .01.

We next determine the forces required to prevent the cart from accelerating down the ramp for various slopes in excess of the constant-velocity slope.

• Observe the slopes associated with the different numbers of paper clips.
• Organize your data for number of paper clips vs. slope into a table.

We convert the data to force vs. slope information.

• From your data table, obtain the force in Newtons and the incline slope corresponding to each trial.  Each paper clip has a mass of approximately 1.2 grams.
• For each slope, subtract the constant-velocity slope to obtain the effective slope.
• Present your force vs. effective slope data in a table.

We now plot force vs. effective slope and interpret the results.

• Construct an accurate graph of force in Newtons vs. effective slope.
• Sketch the straight line that best fits the data.
• Determine whether this straight line indicates a force vs. effective slope proportionality.

• Recall that a proportionality is indicated when, within the limits of experimental error, the straight line that fits the data passes through the origin and the data points.

Compare the slope of the graph with the weight of the cart, and draw conclusions about accelerating forces and accelerations.

• From the mass of the cart (approximately 550 grams) determine its weight. Is the weight, within experimental error, equal to the slope of the graph?
• Can we as a result conclude that, for small slopes at least, the force required to restrain the cart from accelerating down the incline is equal to the weight of the cart multiplied by the effective slope of the incline?
• Can we conclude that for small slopes, the net force accelerating a cart coasting down that slope is equal to the weight of the cart multiplied by the effective slope of the incline?
• How can we therefore conclude that the acceleration of a low-friction cart down a small slope is equal to the acceleration of gravity multiplied by the effective slope of the incline?