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Lab Manual: Atwood’s Machine, Lab Reports of Physics

The University of Tulsa (TU)Physics

Students will able to learn about Newton 2nd law and also will be able to do the procedure of this Atwood’s Machine lab experiment.

Typology: Lab Reports

2020/2021

Uploaded on 05/11/2021

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bg1
1
Atwood’sMachine
Equipment
Qty Equipment PartNumber
1 MassandHangerSet ME‐8979
1 PhotogatewithPully ME‐6838A
1 UniversalTableClamp ME‐9376B
1 LargeRod ME‐8736
1 SmallRod ME‐8977
1 DoublerodClamp ME‐9873
1 String
Background
Newton’s2ndLaw(NSL)statesthattheaccelerationamassexperiencesisproportionaltothenetforce
appliedtoit,andinverselyproportionaltoitsinertialmass(𝑎

).AnAtwood’sMachineisasimple
deviceconsistingofapulley,withtwomassesconnectedbyastringthatrunsoverthepulley.Foran
‘idealAtwood’sMachine’weassumethepulleyismassless,andfrictionless,thatthestringis
unstretchable,thereforeaconstantlength,andalsomassless.
ConsiderthefollowingdiagramofanidealAtwood’smachine.Oneof
thestandardwaystoapplyNSListodrawFreeBodyDiagramsforthe
massesinthesystem,thenwriteForceSummationEquationsfor
eachFreeBodyDiagram.Wewillusethestandardpracticeoflabeling
massesfromsmallesttolargest,thereforem2>m1.ForanAtwood’s
Machinethereareonlyforcesactingonthemassesinthevertical
directionsowewillonlyneedtowriteForceSummationEquations
forthey‐direction.WeobtainthefollowingFreeBodyDiagramsfor
thetwomasses.Eachofthemasseshavetwoforcesactingonit.Each
hasitsownweight(m1g,orm2g)pointingdownwards,andeachhasthe
tension(T)inthestringpointingupwards.Bytheassumptionofanideal
stringthetensionisthesamethroughoutthestring.Usingthestandard
conventionthatupwardsisthepositivedirection,anddownwardsisthe
negativedirection,wecannowwritetheForceSummationEquationfor
eachmass.
𝑇𝑚
𝑔𝑚
𝑎
𝑇𝑚
𝑔𝑚
𝑎
IntheForceSummationEquations,astheyarewrittenhere,theletters𝑇, 𝑔,and 𝑎onlyrepresentthe
magnitudesoftheforcesactingonthemasses,ortheaccelerationsofthemasses.Thedirectionsof
thesevectorsareindicatedbythe+/‐signsinfrontofeachterm.Intheseequationsthe+signsarenot
actuallywrittenout,buttheyshouldbeunderstoodtobethere.Understandingthiswecanseethatm1
isbeingacceleratedupwardsattheexactsamemagnitudethatm2isbeingaccelerateddownwards.The
rev 09/2019
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Atwood’s Machine

Equipment

Qty Equipment Part Number 1 Mass and Hanger Set ME‐ 1 Photogate with Pully ME‐6838A 1 Universal Table Clamp ME‐9376B 1 Large Rod ME‐ 1 Small Rod ME‐ 1 Double rod Clamp ME‐ 1 String

Background

Newton’s 2 nd^ Law (NSL) states that the acceleration a mass experiences is proportional to the net force

applied to it, and inversely proportional to its inertial mass (𝑎 ൌ ி೙೐೟ ௠ ). An Atwood’s Machine is a simple device consisting of a pulley, with two masses connected by a string that runs over the pulley. For an ‘ideal Atwood’s Machine’ we assume the pulley is massless, and frictionless, that the string is unstretchable, therefore a constant length, and also massless.

Consider the following diagram of an ideal Atwood’s machine. One of the standard ways to apply NSL is to draw Free Body Diagrams for the masses in the system, then write Force Summation Equations for each Free Body Diagram. We will use the standard practice of labeling masses from smallest to largest, therefore m 2 > m 1. For an Atwood’s Machine there are only forces acting on the masses in the vertical direction so we will only need to write Force Summation Equations for the y‐direction. We obtain the following Free Body Diagrams for the two masses. Each of the masses have two forces acting on it. Each has its own weight ( m 1 g, or m 2 g ) pointing downwards, and each has the tension ( T ) in the string pointing upwards. By the assumption of an ideal string the tension is the same throughout the string. Using the standard convention that upwards is the positive direction, and downwards is the negative direction, we can now write the Force Summation Equation for each mass.

In the Force Summation Equations, as they are written here, the letters 𝑇, 𝑔, and 𝑎 only represent the magnitudes of the forces acting on the masses, or the accelerations of the masses. The directions of these vectors are indicated by the +/‐ signs in front of each term. In these equations the + signs are not actually written out, but they should be understood to be there. Understanding this we can see that m (^1) is being accelerated upwards at the exact same magnitude that m 2 is being accelerated downwards. The

rev 09 /201 9

reason m 2 is being accelerated downwards is due to m 2 having a larger weight than m 1 , and therefore there is a greater downwards acting force on m 2 than m 1. To solve for the magnitude of the acceleration that both masses will experience, we can simply use the substitution method by solving one equation for the tension T, then substituting that into the other equation. Let’s use the question for mass 1 to solve for the tension, then insert that into the equation for mass 2, then solve for the magnitude of the acceleration.

Here we see that the magnitude of the acceleration the two masses experience is given by the ratio of the difference of the two masses and the sum of the two masses all times gravitational acceleration. Since that ratio will always be less than 1, the acceleration will always be less than gravitational acceleration. As the ratio gets closer to 1, then the value of the acceleration of the masses approaches the value of gravitational acceleration. However, as the value of this ratio gets closer to zero, then the value of the acceleration approaches zero as well.

Also, comparing this equation to NSL we get.

Here we see that the net force acting on each mass is equal to gravitational acceleration times the difference of the two masses. From the above algebra we can clearly see that 𝐹௡௘௧ ൌ 𝑎ሺ𝑚ଵ ൅ 𝑚ଶ ሻ as well.

Procedure: Constant Total Mass

  1. To one hook add 80.0 g, and to the other add 10.0 g. Each hook itself has a mass of 5.0 g so that means we now have 85.0 g on one end of the string and 15.0 g on the other end. The larger mass with be called m 2 , and the smaller mass will be called m 1. (The masses of the hooks themselves MUST be taken into consideration.)
  2. Slowly pull down on m 1 so that m 2 is about near the pulley, and m 1 is near the table top.
  3. At the bottom left of the screen click on the big red circle to start recording data.  Wait about 1 second and then let go of m 1.  After m 2 stops falling downwards click on the big red square at the bottom left of the screen to stop recording data.
  4. At the top left of Page #1 click on the Highlight Data icon. The Highlight box should appear on the graph.  Move and scale the highlight box such that only the data points that represent the moments when the masses were falling are highlighted. (You might have to scale the graph as well.)  Click on the down arrow next to the applied selected curve fit icon to open the list of best fit curves.  Select the linear fit.  The magnitude of the slope of the line is the magnitude of the acceleration the masses experienced. (The slope will be positive or negative depending on the direction of rotation of the pulley.)  Record, in the Constant Total Mass Table, the experimental acceleration (a (^) ex ) for Run 1.
  5. Repeat sets 1 through 4 for the rest of the mass combinations listed in the Constant Total Mass Table.  Keep in mind that each of the hooks have a mass of 5.0 g, and it must be taken into consideration.

Procedure: Constant Net Force

  1. To one hook add 60.0 g, and to the other add 30.0 g. Each hook itself has a mass of 5.0 g so that means we now have 65.0 g on one end of the string and 35.0 g on the other end. The larger mass with be called m 2 , and the smaller mass will be called m 1. (The masses of the hooks themselves MUST be taken into consideration.)
  2. Slowly pull down on m 1 so that m 2 is about near the pulley, and m 1 is near the table top.
  3. At the bottom left of the screen click on the big red circle to start recording data.  Wait about 1 second and then let go of m 1.  After m 2 stops falling downwards click on the big red square at the bottom left of the screen to stop recording data.
  4. At the top left of Page #1 click on the Highlight Data icon. The Highlight box should appear on the graph.

 Move and scale the highlight box such that only the data points that represent the moments when the masses were falling are highlighted. (You might have to scale the graph as well.)  Click on the down arrow next to the applied selected curve fit icon to open the list of best fit curves.  Select the linear fit.  The magnitude of the slope of the line is the magnitude of the acceleration the masses experienced. (The slope will be positive or negative depending on the direction of rotation of the pulley.)  Record, in the Constant Net Force Table, the experimental acceleration (a (^) ex ) for Run 6.

  1. Repeat sets 1 through 4 for the rest of the mass combinations listed in the Constant Net Force Table.  Keep in mind that each of the hooks have a mass of 5.0 g, and this mass must be included in the values of m 1 and m 2.

Constant Net Force Table (10 points)

Run m 1 (kg) m 2 (kg) m 1 +m 2 aex (m/s^2 ) Fnet (N) ath (m/s^2 ) % Error 1 0.0350 0. 2 0.0400 0. 3 0.0450 0. 4 0.0500 0. 5 0.0550 0.

Complete the above chart. Use the experimental acceleration to calculate the net force, and find the % error between the theoretical acceleration and the experimental acceleration for each run.

1. What is a real world application of an Atwood's Machine? (5 points)

2. What are some reasons that would account for the percent error calculated above?

(5 points)

3. For the Constant Total Mass data (Table 1), plot a graph of F net vs. a exp , and find the slope

of the best fit line for the data. (12 points)

(a) What are the units of the slope? (3 points)

(b) What physical quantity does the slope of the best-fit line represent? (5 points)

(c) Calculate the % Error between this quantity and your slope. (3 points)

5. For the Constant Net Force data (Table 2), plot a graph of aexp vs 1/M tot, and find the

slope of the best fit line for the data. (12 points)

(a) What are the units of the slope? (3 points)

(b) What physical quantity does the slope of the best-fit line represent? (5 points)

(c) Calculate the % Error between this quantity and your slope. (3 points)