PROBLEM 15-47 (page 246, 12th edition) The free-rolling smooth ramp weighs 120 lb. The 80 lb crate slides 15 ft down the ramp to B from rest at A. ¥ Determine the speed of the ramp when the crate reaches B. ¥ Determine the velocity of the crate when it reaches B. Express the velocity as a Cartesian vector in terms of the crate's speed and the angle the velocity makes with the horizontal. ¥ Determine the kinetic energies of the ramp and the crate when the crate reaches B. A 15 ft 5 3 4 B PROB15_047-048.jpg Copyright @ 2010 Pearson Prentice Hall, Inc.
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15-47 (page 246 12th edition) SUMMARY OF RESULTS (page 1) ¥ Data W = m g = 120 lb ! = tan"1(3/4) = 36.9° Wc= mcg= 80lb g = 32.2ft/s2 hc = 9ft · Velocities when the crate is at B VR= 8.93i ft/s Vc= 21.4(! cos51.3 i ! sin51.3 j)ft/s : ! V C/ R = 27.9(! cos 36.9" i ! sin 36.9" j )ft/s The crate slides down the ramp at angle ! = 51.3' to the horizontal. This is steeper than the plane angle ! = 36.9 because the ramp rolls to the right while the crate slides to the left and down the ramp. · Kinetic energies when the crate is at B 1 1! 120 $ m_ v2 = (8.93)2 ft'lb =149 ft'lb 2 R'R 2" 32.2 % 1 1 m v= 1! 80 $ 2 & (21.4)2 ft'lb = 571ft'lb 2H 32.2% The total kinetic energy is equal to the initial potential energy: mcghc = 80(9)ft! lb = 720fttlb