HOOKE’S LAW
Name: Kabir Singh Mann
Student Id: 30528607
Introduction
Image of Robert Hooke:
Image
source: https://i.ytimg.com/vi/JUx5lpMc-NI/maxresdefault.jpg
Robert
Hooke is an extremely famous English physicist who was born in 1635 and died in
early 1703. He is famously known due to his big contribution to physics by
discovering the law of elasticity or rather more popularly known as Hooke’s Law.
He demonstrated the relationship
between the extension of the spring and the force applied to the spring in his
discovery of the law of elasticity. He stated that the extension of a spring is
directly proportional to the distance needed to compress or extend a spring. In
1678, a solution was found to translate the Latin anagram and when it was
published, the anagram basically meant "as the extensions, so the force."
In other words, it can be said that the extension of spring is directly
proportional to the force applied.
Robert
Hooke is an extremely famous English physicist who was born in 1635 and died in
early 1703. He is famously known due to his big contribution to physics by
discovering the law of elasticity or rather more popularly known as Hooke’s Law.
He demonstrated the relationship
between the extension of the spring and the force applied to the spring in his
discovery of the law of elasticity. He stated that the extension of a spring is
directly proportional to the distance needed to compress or extend a spring. In
1678, a solution was found to translate the Latin anagram and when it was
published, the anagram basically meant "as the extensions, so the force."
In other words, it can be said that the extension of spring is directly
proportional to the force applied.
From this
information, a formula was formed which is F = -kX, where F is the force in the
form of stress or strain acted on the spring, k is the spring constant which
states the stiffness (ability to stretch the spring) of the spring, and X shows
how much is the displacement of the spring whether has it been stretched or compressed
from it’s original position, and the negative (-) value shows that the spring
is being stretched.
Below is a
diagram to further understand the formula, F=kx :
There
is a limit to where we can apply Hooke's law and that is known as the elastic
limit. The elastic deformation is where Hooke's law can be applied as the
spring will return back to original by force(kx) after being expand or
compressed. The plastic deformation is the region after the elastic limit and
this deformation does not obey Hooke's law because after a certain amount of
force is applied, the spring or could not return back to its original shape.
After the plastic deformation region, is the breaking point (ultimate tensile
strength region), where the spring will just break.
Graph of Young's
Modulus which shows each region point of the spring.
From this
information, a formula was formed which is F = -kX, where F is the force in the
form of stress or strain acted on the spring, k is the spring constant which
states the stiffness (ability to stretch the spring) of the spring, and X shows
how much is the displacement of the spring whether has it been stretched or compressed
from it’s original position, and the negative (-) value shows that the spring
is being stretched.
Below is a
diagram to further understand the formula, F=kx :
There
is a limit to where we can apply Hooke's law and that is known as the elastic
limit. The elastic deformation is where Hooke's law can be applied as the
spring will return back to original by force(kx) after being expand or
compressed. The plastic deformation is the region after the elastic limit and
this deformation does not obey Hooke's law because after a certain amount of
force is applied, the spring or could not return back to its original shape.
After the plastic deformation region, is the breaking point (ultimate tensile
strength region), where the spring will just break.
Results & Graph
Table1:
x(N)
y1(mm)
y2(mm)
1.00
3.00
2.2583
2.00
4.50
4.3166
3.00
6.00
6.3749
4.00
7.50
8.4332
5.00
9.00
10.4915
6.00
10.50
12.5498
7.00
13.00
14.6081
8.00
14.00
16.6664
9.00
15.00
18.7247
From the graph, we
can also get the spring constant, k for both the materials y1 and y2 by
calculating the gradient of the graph for each material by using the formula :-
(y2 - y1)
By using the
formula we get :-
Spring constant, k
(y1) = 1.5583 N/m
Spring constant, k
(y2) = 2.0583 N/m
Table2:
x(N)
z(mm)
1.00
2.375
2.00
9.375
3.00
28.375
4.00
65.375
5.00
126.375
6.00
217.375
7.00
344.375
8.00
513.375
9.00
730.375
From table 2 we can see that x is the weight or
load(N) that is applied on the material z and z is deformation of the material
z and a graph2 of x against z is plotted using the equation z = (x^3) +1.375
Based on the graph 2 material z against x
the weight or load(N) we can see that the weight or load(N) ,x is not
directly proportional to the material z, z because it
does not obey the Hooke's law which states that "the
extension is proportional to the force". We can also see that material z
is out of the elastic region and it's in the plastic deformation region where
the material won't return to its original shape and in the long term when
more load is applied and therefore the material will reach
its ultimate tensile strength and the material will break.
Possible Errors
When looking at graph1 we can see that not
all the coordinates are touching the best fit line, this is because of human
error where the reading of extension isn't accurate because when
taking the reading it is not exactly 90 degree to the ruler. Another possible
error is when doing the calculation to find the load in Newton when using
the equation F = mg.
Table1:
x(N)
|
y1(mm)
|
y2(mm)
|
1.00
|
3.00
|
2.2583
|
2.00
|
4.50
|
4.3166
|
3.00
|
6.00
|
6.3749
|
4.00
|
7.50
|
8.4332
|
5.00
|
9.00
|
10.4915
|
6.00
|
10.50
|
12.5498
|
7.00
|
13.00
|
14.6081
|
8.00
|
14.00
|
16.6664
|
9.00
|
15.00
|
18.7247
|
From the graph, we
can also get the spring constant, k for both the materials y1 and y2 by
calculating the gradient of the graph for each material by using the formula :-
(y2 - y1)
(x2 - x1)
By using the
formula we get :-
Spring constant, k
(y1) = 1.5583 N/m
Spring constant, k
(y2) = 2.0583 N/m
We can get the
intersection point where both of the material meet. Just by looking at the
graph the estimated value of intersection is(2.40 , 5.00). The real value of intersection
is gotten using two of the formula, y1 = 1.5583x + 1.375 and y2 = 2.0583 + 0.2
which is (2.35 , 5.04). From the value we got from the calculation above, we
now know that the spring constant of y2 is higher than y1 therefore material y2
needs more force to deform the material compared to material y1.
The meaning of the
intersection point(2.35 , 5.04) means that both of the material deforms to
5.04mm when a force of 2.35 Newton is applied to the material.
Table2:
x(N)
|
z(mm)
|
1.00
|
2.375
|
2.00
|
9.375
|
3.00
|
28.375
|
4.00
|
65.375
|
5.00
|
126.375
|
6.00
|
217.375
|
7.00
|
344.375
|
8.00
|
513.375
|
9.00
|
730.375
|
From table 2 we can see that x is the weight or
load(N) that is applied on the material z and z is deformation of the material
z and a graph2 of x against z is plotted using the equation z = (x^3) +1.375
Based on the graph 2 material z against x
the weight or load(N) we can see that the weight or load(N) ,x is not
directly proportional to the material z, z because it
does not obey the Hooke's law which states that "the
extension is proportional to the force". We can also see that material z
is out of the elastic region and it's in the plastic deformation region where
the material won't return to its original shape and in the long term when
more load is applied and therefore the material will reach
its ultimate tensile strength and the material will break.
Possible Errors
When looking at graph1 we can see that not
all the coordinates are touching the best fit line, this is because of human
error where the reading of extension isn't accurate because when
taking the reading it is not exactly 90 degree to the ruler. Another possible
error is when doing the calculation to find the load in Newton when using
the equation F = mg.
