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Topics: Newton's laws of motion, Classical mechanics, Percent difference Pages: 5 (1328 words) Published: January 13, 2013
Lab 3: Newton’s Second Law: The Atwood Machine
Introduction:
In the study of physics a lot of the basics were put in place by Isaac Newton. Out of the 3 laws of motion he had declared the second law states that force equals mass times acceleration (F=ma). The Atwood machine is a machine that has a pulley in the air and a string running through the pulley, some kind of mass is suspended by each end of the string. When the suspended masses are unequal, the system will accelerate towards the direction of the larger mass. In this experiment, we used different masses to the velocity of the Atwood system. The data we collect for this experiment are the differences in mass between the two masses, the distance the heavier mass has to fall to hit the ground, and the time it took for the heavier mass to hit the ground. We use this data to determine the acceleration and compare it to the theoretical acceleration calculated according to Newton’s Second Law of Motion.

Example of Atwood Machine

Procedure:
1. Set up the Atwood Machine with the pulley suspended over the ground and a length of string that we determined would be most useful to use to develop accurate results. We then attached weights to each end of the strings. 2. We then measured and recorded the distance that the heavier mass had to fall to hit the ground, which is y on the data table. 3. We then recorded the difference in mass between the two sides of string. 4. We pull the lighter mass to the ground and hold in place, leaving the heavier mass suspended in the air. 5. We then released the lighter mass allowing the heavier mass to fall to the ground. 6. We had used a stopwatch to determine the time it took for the heavier mass to fall to the ground. 7. Record the time in the data table as t. Perform five trials. 8. Repeat steps 3-7 for the three different variations in mass.

Data:
Distance to ground: 1.16m
Set 1:
Mass required to start the system moving: 5.5g
M1: 50g M2: 55.5g
Trial| y (m)| t (s)| ay (m/s2) |
A| 1.16| 2.6| .17|
B| 1.16| 2.8| .15|
C| 1.16| 2.6| .17|
D| 1.16| 2.7| .16|
E| 1.16| 2.6| .17|

Average ay = .164 m/s2

Set 2:
Mass required to start the system moving: 11g
M1: 50g M2: 61g
Trial| y (m)| t (s)| ay (m/s2) |
A| 1.16| 1.76| .37|
B| 1.16| 1.7| .4|
C| 1.16| 1.7| .4|
D| 1.16| 1.85| .34|
E| 1.16| 1.73| .39|

Average ay = .38 m/s2

Set 3:
Mass required to start the system moving: 28.2g
M1: 50g M2: 78.2g
Trial| y (m)| t (s)| ay (m/s2) |
A| 1.16| 1.34| .65|
B| 1.16| 1.4| .59|
C| 1.16| 1.35| .64|
D| 1.16| 1.2| .8|
E| 1.16| 1.37| .62|

Average ay = .66 m/s2
Free Body Diagrams

Solve for Atwood formula
M1: -m1a= m1g- ft M2: m2a= m2g- ft Ft= m1g + m1a
M2a= m2g- (m1g + m1a)
M2a= m2g- m1g - m1a
M2a + m1a = m2g – m1g
A (m2 + m1) = g(m2-m1)
*A= g(m2-m1)/ (m2 + m1)

Theoretical Accelerations:
Set 1:
A= g(m2-m1)/(m2+m1)
A= 10m/s2(55.5g-50g)/(55.5g+50g)= .52m/s2
Set 2:
A= g(m2-m1)/(m2+m1)
A= 10m/s2(61g-50g)/(61g+50g)= .99m/s2
Set 3:
A= g(m2-m1)/(m2+m1)
A= 10m/s2(78.2g-50g)/(78.2g+50g)= 2.2m/s2

Percent Error:
Set 1:
Percent Error= theoretical- measured/ theoretical x 100
.52m/s2- .164m/s2/ .52m/s2 x 100 = 68.46%
Set 2:
Percent Error= theoretical- measured/ theoretical x 100
.99m/s2- .38m/s2/ .99m/s2 x 100 = 61.62%
Set 1:
Percent Error= theoretical- measured/ theoretical x 100
2.2m/s2- .66m/s2/ 2.2m/s2 x 100 = 70%
Conclusion:

Newton’s second law states that force equals mass times acceleration. Because we know that Newton’s second law of motion is correct we can determine a theoretical acceleration for any combination of height and difference in masses using the Atwood machine. We used this theoretical acceleration that we determined by using Newtons...

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