# Newton Second Law ```Newton’s Second Law
Laboratory Report 3
Done by:
-Omar Alrawi 201104429
-Husam Al-Jawaheri 201103311
- Mohammed Abunahia 201000021
- Ammar Mansawi 20085761
Date:
19/3/2012
Course name &amp; CRN:
Experimental General Physics for Engineers I, PHYS 192 - 25269
Instructor:
Dr. Hocine Merabet
Abstract:
To measure the acceleration of an object using newton’s second law and compare it with the
theoretical value.
Theory:
Newton’s second law states that the acceleration of an object is proportional to the sum of the
forces applied on that object:
𝑎⃗ =
1
∑ 𝐹⃗
𝑚
If we apply the law to our following system:
We choose downward as positive direction then we get:
1- For 𝑚1 : 𝑇1 − 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚1 𝑎
2- For 𝑚2 : 𝑚2 𝑔 − 𝑇2 = 𝑚2 𝑎
Now by neglecting the mass of the string we get that tension is equal at the ends and by
applying newton’s second law to the string:
𝑇2 − 𝑇1 = 𝑚𝑎 = 0𝑎 = 0
𝑇2 = 𝑇1
Using above equations we get:
𝑚2 𝑔 − 𝑚1 𝑎 − 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚2 𝑎
And therefore:
𝑎=
𝑚2 𝑔 − 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑚1 + 𝑚2
We can see that the acceleration in the last equation is constant therefore the distance can be
calculated using:
1
𝑥 = 𝑥0 + 𝑣0 𝑡 + 2 𝑎𝑡 2
Rewriting the equation:
𝑥 − 𝑥0
1
= 𝑣0 + 𝑎𝑡
𝑡
2
Which is of the form 𝑦 = 𝐼 + 𝑆𝑥 where S is the slope of the line and intercept I. Therefore,
plotting
𝑥−𝑥0
𝑡
against t gives us a straight line which we can calculate its slope and thus the
1
acceleration 𝑆 = 2 𝑎 which we will be using in our calculations.
Equipment needed
Two photo-light sensors connected to a digital timer, pulley, track &amp; cart, a mass.
Experimental procedure:
1.
2.
3.
4.
5.
We made sure that the time measuring device worked properly.
We put a small mass at the end of the string.
We set a distance of 25 cm between the two sensors.
We calculated the time taken for the object to cover the distance between the sensors.
We repeated step 3 and 4 for distances of 30, 35, 40, 45, and 50 3 times for each.
Data:
By following our procedure we obtained the data in the following table:
Distance between the
two sensors ∆𝑋(𝑐𝑚)
25
30
35
40
45
50
𝑡1 (𝑠)
0.876
0.971
1.071
1.157
1.228
1.324
Time 𝑡(𝑠)
𝑡2 (𝑠)
0.885
0.984
1.048
1.140
1.196
1.306
𝑡3 (𝑠)
0.854
0.956
1.089
1.137
1.231
1.311
Average
Time 𝑡𝑚 (𝑠)
0.871
0.970
1.064
1.144
1.218
1.313
∆𝑋(𝑐𝑚)
𝑡𝑚 (𝑠)
28.70
30.92
32.74
34.96
36.94
38.08
Data analysis:
1. We estimated the error for ∆𝑥 which is the error on the ruler to be 1mm.
2. We estimated the error on 𝑡𝑚 using standard deviation method on the third row which seems to
give biggest error.
(1.071 − 1.064)2 + (1.048 − 1.064)2 + (1.089 − 1.064)2
𝜎𝑡 = √
= 0.021 𝑠
3−1
𝜎𝑡𝑚 =
𝜎𝑡
0.021
= 0.012 s
√3
√3
3. Our estimate for the error on 𝑡𝑚 is 0.01 s (we put only 1 s.f. in the error).
4. Then estimate the error on
∆𝑋(𝑐𝑚)
𝑡𝑚 (𝑠)
=
:
We have
∆(∆𝑋) 2
∆𝑡𝑚 2
∆𝑌 = 𝑌√(
) +(
)
∆𝑋
𝑡𝑚
0.01 2
0.012 2
∆𝑌 = 32.74√(
) +(
) = 0.37 𝑐𝑚/𝑠
35
1.064
5. Plot of
∆𝑋
𝑡𝑚
versus 𝑡𝑚
From the liner fit we have:
𝑆𝑙𝑜𝑝𝑒 = 23.12 ∓ 1.1 𝑐𝑚/𝑠^2
𝐼𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 8.4 ∓1.2 𝑐𝑚/𝑠
The error on the slope using excel linest function:
S
dS
R^2
23.11735
1.104083
0.990958
438.4024
67.79195
8.402753153 I
1.220312613 dI
0.393235404 dY
4
0.618536333
1
𝑎,
2
6. From the theory we have 𝑆 =
therefore we can calculate the acceleration using 𝑎 = 2𝑆
∆𝑎 = 2∆𝑆
So we get:
𝑎 = 2 𝑥 23.1
𝑐𝑚
𝑐𝑚
𝑚
= 46.2 2 = 0.462 2
2
𝑠
𝑠
𝑠
∆𝑎 = 2 x 1.1
Therefore, ∆𝑎 = 0.02
𝑚
𝑠2
𝑐𝑚
𝑐𝑚
𝑚
= 2.2 2 = 0.022 2
2
𝑠
𝑠
𝑠
(1 s.f. for the error)
And our result for 𝑎 is:
𝑎 = (0.462 ∓ 0.02)
𝑚
𝑠2
To compare with the theoretical acceleration we calculate it using (neglecting friction):
𝑎=
𝑚2 𝑔
𝑚1 + 𝑚2
We measured 𝑚1 = 211 ∓ 0.1 𝑔 and 𝑚2 = 21 ∓ 0.02 𝑔.
𝑎=
0.0211𝑥9.81
𝑚
= 0.89 2
0.21 + 0.0211
𝑠
Calculating error on the acceleration:
2
2
0.1x0.02
0.1 + 0.02
𝑚
√
∆𝑎 = 0.89 (
) +(
) = 0.46 2
(0.21 + 0.0211)
0.0211 ∗ 9.81
𝑠
Therefore, using 𝑥 2 method to compare the 2 measurements:
(0.462 − 0.89)2
𝜒 =
= 0.86
(0.02)2 + (0.46)2
2
The value of 𝜒 2 is less than 3, so we conclude that our measurement is compatible with the
predicted value. We can also see our error which is 0.02
𝑚
𝑠2
represents 4.32% of the measured value,
0.462 𝑚/𝑠 2 , which is quite precise measurement (within 5% error bound).
Conclusion:
We have measured the acceleration on an object caused by force applied on it using newton’s second
law, and obtained the value 𝑎 = (0.462 ∓ 0.02)𝑚/𝑠 2 which is compatible with the predicted value 𝑎 =
(0.89 ∓ 0.5)𝑚/𝑠 2 .
Further questions:
1. What are the effects of friction and air resistance, i.e. how do they affect your value
of the measured acceleration?
The effect of air resistance is the force of air exerted on the object when it's moving, and
the effect of friction is the force that generate when an object moves on rough surfaces.
However, these forces affect the calculation because when the object move there will be a
force in the other direction which is the friction force and air resistance will slow the speed
of the object.
2. You did estimate the upper limit of the static friction; by how much you expect it to
change your value of the measured acceleration? (The kinematic friction is always less
that the static one).
If we work around the formula of the acceleration we get that:
𝑚2 𝑔 − 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑎=
𝑚1 + 𝑚2
That means the more the friction increases the less acceleration there will be and it is expected
to have a big influence on the final result of the measured acceleration.
3. How does the acceleration obtained in this experiment compare with 𝒈? Does is make
sense to you?
The acceleration that we obtained in this experiment is lower than the gravity which
perfectly makes sense because whenever the angle increases from 0 to 90 the acceleration
will increase until it reaches maximum value at 90 which is 9.8 which is the gravity.
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