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posted by  pbodie on 7/16/2008 3:05:02 PM  |  status: Live  

prob. 13.65

Course Textbook Chapter Problem
Classical Mechanics Vector Mechanics for Engineers: Dynamics (8th) by Johnston, Beer, Clausen 13 65P
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posted by ClayL on 7/16/2008 5:21:56 PM  |  status: Live
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This is a conservation of energy problem:
Position 1 is at the top of the incline, and position 2 is when the spring has a maximum deformation
k = 1500 lb/ft
Where T1 + V1 = T2 + V2
So
at position (1):
T1 + (1/2)*m*V12 = (1/2)(200/32.2)(64) = 198.76 ft*lb
so V1 = Vg1 + Ve1 = mgz1 = (1/2)kx12 (the datum at point 2)
 = 200(25-x)*sin20 + (1/2)(1500)(.52)
 x = deformation of the spring
V1 = 1710.1 + 68.404x + 187.5

At position (2):
T2 = 0
V2 = Vg2 + Ve2 = (1/2)kx22 = (1/2)(1500)(.5+x)2

Now substituting this back into 1:
198.78 + 1710.1 + 68.404x + 187.5 = 750(.5+x)2
Solve for x using the quadratic formula:
x = -2.11 or +1.2044
so x = 1.204 ft = 14.45 inches
Engineering ain't easy...
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