7.2 Rectilinear Motion
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they provide the resistive forces that are function of the deformation. Another type of forces in this category is the field force, such as the attraction force. From the gravitational law, the attraction force will be inversely proportional to the square of the separated distance. These forces induce the acceleration that is a function of the displacement. Hence we may solve for the velocity Z s Z s Z v 2 2 f (s) ds f (s) ds → v = vo + 2 vdv = vo
so
so
as a function of displacement v = g (s) The displacement may be determined explicitly by
v=
ds dt
t=
Rs
1 ds so g(s)
Inverting the above relationship, we may have s = h (t) Following are sample problems related to the rectilinear motion where we need to understand specific hidden implications and apply previous definitions in determining the answers. Example 7.1 ([3], Prob. 2/23) Small steel balls fall from rest through the opening at A at the steady rate of 2 per second. Find the vertical separation h of two consecutive balls when the lower one has dropped 3 meters. Neglect air resistance. Solution: The accceleration for the free-falling object must be equal to the constant of gravity. Also, we can integrate for the velocity and the displacement explicitly. In other words, a=g v = vo + gt s = so + vo t + gt2 /2 Since the ball is dropped from rest at the reference level of the opening, we have the initial conditions vo = 0, so = 0 Hence s = gt2 /2 Apply this relation to the ball already dropped by 3 m, the time spent would be 3 = gt2l /2, tl = 0.782 s Chulalongkorn University
Phongsaen PITAKWATCHARA