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Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

Solving the above quadratic equation, we get:

2
− sinθ ± sin θ + 4CC 1
x =
2C
For the spring compression (x) to be positive, ‘C ’ must be greater than zero, that is:
1
H − μL − h > 0 , therefore, we get: H > μL + h
The spring compression is now given by the positive root, as follows:

2
sin θ + 4CC − sinθ
x = 1
2C

2.5.3 The 3 Equation of motion

Let us consider the energy and work relation between two positions of a particle in motion. The
work done on the particle is the change in its kinetic energy, assuming that, the potential energy
change is zero.

[Fig 2.9: a particle in uniform straight-line motion]

1 1
2
2
Work done = final kinetic energy − initial kinetic energy = F ∗ s = mv − mU
2 2
1 1
= (ma) ∗ s = mv − mU = 2(a ∗ s) = v − U = v = U + 2as - - - - (2.14)
2
2
2
2
2
2
2 2
Another method to derive equation (2.14) is as follows:
For a uniform straight line motion: Distance travelled = average speed ∗ time
v + U
Therefore, we have: s = t - - - - (2.15)
2
change in velocity v − U
t is got by the definition of acceleration, which is: a = = - - - - (2.16)
time t

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 15 age 15
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