Δ
=
L
n
−
L
0
(mm)
Strain
ε)
(
=
L
/
L
Δ
N
/
m
)
(
2
Stress
=
σ)
(
/
A
(
N
/
m
)
F
2
1
9.8 N
13.23 mm
0.11mm
0.0084
3.141
2
19.6 N
13.35 mm
0.23 mm
0.0175
6.28
3
29.4 N
13.46 mm
0.34 mm
0.0259
9.42
4
39.2 N
13.56 mm
0.44 mm
0.0335
12.56
5
49 N
13.67 mm
0.55 mm
0.0419
15.71
6
58.8 N
13.79 mm
0.67 mm
0.0511
18.85
7
68.6 N
13.90 mm
0.78 mm
0.0594
21.98
Table II
Weight (kg)
Applied
Force (N)
(mm)
L
0
L
Δ
=
L
n
−
L
0
(mm)
Strain
ε)
(
=
L
/
L
Δ
N
/
m
)
(
2
Stress
=
σ)
(
/
A
(
N
/
m
)
F
2
7
68.6 N
13.90 mm
0.78 mm
0.0594
21.98
6
58.8 N
13.79 mm
0.67 mm
0.0511
18.85
5
49 N
13.69 mm
0.57 mm
0.0434
15.71
4
39.2 N
13.56 mm
0.44 mm
0.0335
12.56
3
29.4 N
13.46 mm
0.34 mm
0.0259
9.42
2
19.6
13.35 mm
0.23
0.0175
6.28
1
9.8
13.23 mm
0.11
0.0084
3.141
3.2
Calculations:
L
Δ
=
L
n
−
L
0
ε =
L
/
L
Δ
/
A
σ =
F
L
3.23
3.12
Δ
= 1
− 1
.11/13.12
ε = 0
.8/3.12
σ = 9
L
.11
mm
Δ
= 0
.0084
N
/
m
ε = 0
2
.141
N
/
m
σ = 3
2
4.
Analysis and Discussion

By utilizing the equations pertaining to Hooke’s Law, we were able to calculate the
Young Modulus of Elasticity. In order to get the modulus, we needed to find the stress and strain
for each trial on the string. The ratio of stress over strain will give us the modulus, which was the
main objective of the lab.
Discussion and Questions
1.
When the elastic limit is not exceeded, the deformation is recoverable so comparing
both cases of increasing and decreasing the hanging weights, the strain should be
the same for the same stress. That is the Young’s Modulus should be the same for
both cases. Do you observe this?

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