Hooke's Law
Introduction
Introduction
Hooke's law, law of elasticity discovered
by the English scientist Robert Hooke in 1660, which states that, for
relatively small deformation of an object, the displacement or size
of the deformation is directly proportional to the deforming force or load.
Under these conditions the object returns to its original shape and size upon
removal of the load. Elastic behavior of solids according to Hooke’s law
can be explained by the fact that small displacements of their constituent
molecules, atoms and ions from normal positions is also proportional to
the force that causes the displacement.
Figure A
Figure B
An object only obeys
Hooke's Law when its elastic limit is not exceeded. When elastic limit of the
spring is exceeded, the graph will not remain in straight line. The graph will
be curved (as shown in Figure B).The spring is said to undergo plastic deformation
causes the spring to irreversibly change its shape when the elastic limit of a
spring is exceeded.
Experiment
Aim
To study the relationship between the extension and the force applied to an elastic material.
Apparatus
3 different types of elastic materials, ring stand and clamps, meter rule, and slotted weight
Procedure
1. The original length
of Material 1 is recorded.
2. The apparatus is set
up by hanging up Material 1 on the retort stand and clamp.
3. Slotted weight of 1N
is hung at the end of Material 1.
4. The final length of
Material 1 is recorded.
5. The extension of
Material 1 is calculated by the difference between the original length and the
final length of Material 1.
6. The experiment is
repeated using slotted weight of 2N, 3N, 4N, 5N, 6N, 7N, 8N and 9N.
7. The data of Material
2 and 3 is calculated using the formula given.
8. The data is tabulated
in a table.
9. Graph of extension
against force applied for all 3 materials is plotted.
Diagram 1.1
Tabulation of data
The result of
Material 1 and Material 2 are tabulated in Table 1 and Material 3 is tabulated
in table 2.
Given that y1=ax+b and y2=(a+0.5)x+c where c=0.2
Force,x/N
|
Extension of
Material 1,y1/mm
|
Extension of
Material 2, 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
|
Table 1
Given that z= x 3+ b
Force,x/N
|
Extension of
Material 3,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
|
Table
2
Discussion
A
graph of Extension, y/mm against Force, x/N of Material 1&2 is obtained.
Figure 1
-From
the graph 1, we can get the values of a and b with the equation of y1= 1.5583x
+ 1.375 which mean that a=1.5583 and b=1.375.
-We
can solve the equations from the graph using simultaneous equation to find the
intersection point (x,y).
y1= 1.5583x+ 1.375 – (1)
y2= 2.0583x+ 0.2 – (2)
By
putting y2 into equation (1), we get
2.0583x+ 0.2 = 1.5583x+ 1.375
2.0583x- 1.5583x = 1.375- 0.2
0.5x= 1.175
x= 2.35
By
putting x= 2.35 into equation (2), we get
y= 2.0583(2.35) + 0.2
y=5.037005
Thus,
the intersection point of y1 and y2 is at (2.35, 5.037005).
