NEWTON’S SECOND LAW
Skylur Jameson
25 March 2009
Experiment # 5
Purpose:
The purpose of the experiment is to examine Newton’s Second Law through the
test trials using the Atwood Machine.
Equipment: Rod, clamps, pulley, masses, mass holders, string, meter stick, stopwatch,
masking tape
Discussion:
The lab is aimed at allowing the calculation of Newton’s Second Law using the
Atwood Machine. By setting up two simultaneous equations relating to the
tension in the string and the acceleration of the two masses, we can solve for
acceleration using the formula: ath = (m1–m2 / m1+m2) g. This equation allows us
to predict the theoretical acceleration of the two masses. In order to calculate the
experimental acceleration, aexp, we will need to derive the equation: aexp = 2y / t2.
Procedure:
The experiment will allow us to simulate the motion of an elevator using the
Atwood machine. We will accept that m1 = ME+MH+ML+m and m2 = MC+MH.
Where: ME = mass of the elevator, MH = mass of the holder, MC = mass of the
counter weight, ML = mass of the load, and m = small mass to offset friction in
the pulley. To begin, set up the apparatus so that two elevators of the same weight
can offset each other in equilibrium. Next select three masses so we can measure
ML. The three chosen masses should be 5g, 10g, and 15g. After selecting the
correct masses, add additional weight to the elevators (the addition of ME+MH
should come to equal 150g, 250g, and 350g). Lastly, run three trials at each
weight at each individual mass. The number of total trials should total twentyseven trials. Once all the trials have been concluded, calculate the experimental
acceleration and theoretical acceleration. Once aexp and ath have been found, find
the percentage error for each trial.
Observations: The change of the holder could have affected results. Long meal rod moved
from time to time. The elevators might have not been at equilibrium. The higher
the weight went the slower it fell. The more weight added to the elevator, the
faster it dropped. Elevator swung at times while falling. There were a few miss
trails. Elevator bounced off string a few times. The stopwatch holder could have
been a little slow at starting watch. The apparatus was vibrating at times. Rope
fell off pulley once in a while. Time difference between the stopwatch and
letting go could have affected results.
Conclusion:
In conclusion, the experiment was successful in that it proved Newton’s second
law to be true. We found that the average time for each trial mass did not
exceed the nine second mark. Once we had calculated our time average, we
calculated our experimental and theoretical acceleration. Using all the
information, we calculated the percent error for each trial, and found that the
lowest percent error was 6.6% while the highest percent error was 33.5%. The
only criticism I have for the experiment is that it was quite suseptable to human
error. At times, as is seen in the high percent error, the individual releasing the
elevator started the stopwatch either too early or too late resulting in a high rate
of error. Overall, the experiment was enjoyable and helped our groups
understand of Newton’s second law of motion.
Trial M E + M H (M E=M C) (grams) M L (g)
1
150
5
2
150
10
3
150
15
4
250
5
5
250
10
6
250
15
7
350
5
8
350
10
9
350
15
Y (meters) =
t1 (sec) t2 (sec) t3 (sec)
5.32
5.59
5.31
3.56
3.63
3.53
2.97
2.9
3
7
7.25
7.06
4.62
4.63
4.66
3.69
3.72
3.97
8.19
8.1
8
5.22
5.34
5.4
4.31
4.43
4.5
tavg aex p (m/s 2) ath (m/s 2) % error
5.41
0.126
0.161
28.1
3.57
0.287
0.316
10.1
2.96
0.420
0.467
11.3
7.10
0.073
0.097
33.5
4.64
0.171
0.192
12.7
3.79
0.255
0.286
12.0
8.10
0.056
0.070
24.3
5.32
0.130
0.138
6.6
4.41
0.188
0.206
9.2
1.835
2
Gravity (m/s ) =
9.81
Trial ME + MH (ME=MC ) (g) ML (g) t1 (sec) t2 (sec) t3 (sec)
tavg
aexp (m/s 2 )
ath (m/s 2 )
% error
1
150
5
5.32
5.59
5.31
=AVERAGE(D2,E2,F2)
=(2*$C$12)/(G2^2)
=((C2)/(B2+C2+B2))*$C$13
=ABS((H2-I2)/H2)*100
2
150
10
3.56
3.63
3.53
=AVERAGE(D3,E3,F3)
=(2*$C$12)/(G3^2)
=((B3+C3)-B3)/(B3+C3+B3)*$C$13
=ABS((H3-I3)/H3)*100
3
150
15
2.97
2.9
3
=AVERAGE(D4,E4,F4)
=(2*$C$12)/(G4^2)
=((B4+C4)-B4)/(B4+C4+B4)*$C$13
=ABS((H4-I4)/H4)*100
4
250
5
7
7.25
7.06
=AVERAGE(D5,E5,F5)
=(2*$C$12)/(G5^2)
=((B5+C5)-B5)/(B5+C5+B5)*$C$13
=ABS((H5-I5)/H5)*100
5
250
10
4.62
4.63
4.66
=AVERAGE(D6,E6,F6)
=(2*$C$12)/(G6^2)
=((B6+C6)-B6)/(B6+C6+B6)*$C$13
=ABS((H6-I6)/H6)*100
6
250
15
3.69
3.72
3.97
=AVERAGE(D7,E7,F7)
=(2*$C$12)/(G7^2)
=((B7+C7)-B7)/(B7+C7+B7)*$C$13
=ABS((H7-I7)/H7)*100
7
350
5
8.19
8.1
8
=AVERAGE(D8,E8,F8)
=(2*$C$12)/(G8^2)
=((B8+C8)-B8)/(B8+C8+B8)*$C$13
=ABS((H8-I8)/H8)*100
8
350
10
5.22
5.34
5.4
=AVERAGE(D9,E9,F9)
=(2*$C$12)/(G9^2)
=((B9+C9)-B9)/(B9+C9+B9)*$C$13
=ABS((H9-I9)/H9)*100
9
350
15
4.31
4.43
4.5
=AVERAGE(D10,E10,F10) =(2*$C$12)/(G10^2) =((B10+C10)-B10)/(B10+C10+B10)*$C$13 =ABS((H10-I10)/H10)*100
Y (meters) =
2
Gravity (m/s ) =
1.835
9.81