Hooke's Law
I have designed the experiment to measure spring constants when the
springs are in series and in parallel. The theory is based on Hooke's
law which is:
F = kx
where F = Force, k = Constant and x = Extension [Ref. 1].
Unfortunately with the springs I have, I can only measure extension,
not compression for which Hooke's is also valid.
Prediction
Single Spring:
Hooke's law, where F = kx. I predict that I if I plot Force on the Y
axis and extension, x, on the X axis, it will be a straight line and
the gradient will be the spring constant.
Parallel Springs:
In the diagram above, T1 is the tension in spring 1, and T2 is the
tension is spring 2. So T1 + T2 has to equal the Force, F. I assume
the two springs are originally the same length and the extensions are
the same in both springs. So this means that T1 = k1x and T2 = k2x. So
by knowing this, you get the formula:
F = k1x + k2x = x (k1 +k2).
So overall, the spring constant for the two springs is k1 + k2. For
the graph, my prediction is the same as for a single spring as
mentioned earlier but with a different constant.
Springs In Series:
In this diagram, T1 is the tension in spring 1, and T2 is the tension
is spring 2. If I ignore the mass of the 2nd spring, the two tensions
have to be equal and both equal to the force. I have to find a final
expression in the form F = k (x1 + x2) because x1 + x2 is the total
extension. So from Hooke's law you can get x1 = F / k1 and x2 = F / k2.
Knowing this, gives me a new formula: F = k (F / k1 + F / k2). Here,
you can then take out the common factor, F and divide both sides by kF
which gives you a final formula: 1 / k = 1 / k1 + 1 / k2 .
Elastic strain region at small and big end of connecting rod is shown in figure no. 10. The maximum and minimum equivalent strain values are 0.00033975 and 2.1407e-10 respectively. Due to applied pressure there will be change in original dimensions of the connecting rod and hence strain developed can be
When the building begins to oscillate or sway, it sets the TMD into motion by means of the spring and, when the building is forced right, the TMD simultaneously forces it to the left.
Assuming that x2 = 20 basis, and x1 = 100 basis. We can conclude the
Our predicted points for our data are, (13, -88.57) and (-2, -29.84). These points show the
Show your work. Note that your answer will probably not be an even whole number as it is in the examples, so round to the nearest whole number.
method can be produced and a graph of the function can be made. From the graph,
contains for stresses; there is a strong caesura in the middle of the lines and
In analyzing the force associated with a certain spring, whether it is in you pen or under your truck, Hooke’s Law applies.
The optimum weight of the cold formed steel truss is the minimum weight of 48.5 kg corresponding to the central rise of 1.02 m as shown in the figure 2.
Fig. 6 © HowStuffWorks 2002. How Seatbelts Work [online]. Available at: http://static.ddmcdn.com/gif/seatbelt-spring.gif [Accessed 17th November 2012
According to Newton’s Second Law of Motion, the vector sum of the total forces in a system is equal to the product of the mass (m) and acceleration ( a ) of the system.
The second law is, “the relationship between an objects mass (m), its acceleration (a), and the applied force (f) is F= ma.” The heavier object requires more force to move an object, the same distance as light object. The equation gives us an exact relationship between Force, mass, and acceleration.
In the expression of potential energy (V) given by equation (2.03), the higher order terms can be neglected for sufficiently small amplitudes of vibration. To make coinciding with the equilibrium position, the arbitrary zero of potential must be shifted to eliminate V_0. Consequently the term (∂V/〖∂q〗_i ) becomes zero for the minimum energy in equilibrium. Therefore, the expression of V will be reduced to