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

Š‹• ƒ’’Ž‹‡• ‡˜‡ –‘ ƒ ‹ Ž‹‡† •–”ƒ‹‰Š– Ž‹‡Ǥ ‘™‡˜‡”ǡ ‘‡ ƒ••—’–‹‘ –Šƒ– ™‡ Šƒ˜‡ ƒ†‡ ‹• –Šƒ–ǡ
–Š‡ ‡–”‹ˆ—‰ƒŽ ˆ‘” ‡ǡ –Šƒ– ‹ ”‡ƒ•‡• –Š‡ ‘”ƒŽ ”‡ƒ –‹‘ ‹• ‘– ƒ ‘—–‡† ˆ‘”Ǥ Š‹• ‹• –‘ ƒ˜‘‹†
ƒ–Š‡ƒ–‹ ƒŽ ‘’Ž‡š‹–›Ǥ Š‹• ƒ••—’–‹‘ †‘‡• ‘– •‹‰‹ϐ‹ ƒ–Ž› ƒŽ–‡” –Š‡ ”‡•—Ž– ˆ‘” —”˜‡• ™‹–Š
Žƒ”‰‡ ”ƒ†‹—• ƒ† ‘†‡”ƒ–‡ ˜‡Ž‘ ‹–›Ǥ

‘Ž—–‹‘ ȋ ‘–ǥȌǣ „› ‡‡”‰› „ƒŽƒ ‡ ȋ‡‰Ž‡ –‹‰ ‡‡”‰› Ž‘•– ‹ ‘ŽŽ‹•‹‘Ȍǡ ™‡ ‰‡–ǣ

Initial potential energy = frictional work done + final potential energy + spring work done

1
2
= mgH = μmg L + mg h + x sinθ + kx - - - - (2.12)
2
The final potential energy mg h + x sin θ , is got as follows:

Ø Initially the mass climbs a height (h).
Ø Further, as the momentum of the mass compressed the spring, the mass gains further
height, which is sinθ times the spring compression (x), as shown in [Fig 2.8]

[Fig 2.8: compression of the spring (x) due to momentum of the mass]

Simplifying equation (2.12) we get:

1
2
H = μ L + h + x sinθ + kx - - - - (2.13)
2mg
k
For ease of simplification, let us assume to be C and (H − μL − h = C )
1
2mg
Rearranging the expression (2.13) and applying the assumption, we get:

1
2
2
2
−(H + μ L + h) + x sin θ + kx = 0 = −C + x sinθ + Cx = 0 or Cx + x sinθ − C = 0
1
1
2mg

Page 14 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,