Ch 4. Particle Momentum and Impulse Multimedia Engineering Dynamics Impulse & Momentum Consv. Linear Momentum Impact AngularMomentum MassFlow
 Chapter - Particle - 1. General Motion 2. Force & Accel. 3. Energy 4. Momentum - Rigid Body - 5. General Motion 6. Force & Accel. 7. Energy 8. Momentum 9. 3-D Motion 10. Vibrations Appendix Basic Math Units Basic Equations Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll

 DYNAMICS- CASE STUDY SOLUTION Problem Coordinate System Begin with a diagram of the car and barge in their initial state. Place the origin of the rectangular coordinate system at the initial position of the car. The velocity of the barge when the car reaches the end of the barge is      vB = -0.885i m/s Because the horizontal force of the water on the barge is neglected, the total linear momentum in the horizontal direction is conserved. Velocity Velocity of Barge and Car as Car Leaves Barge Since the car and barge are initially stationary, the total linear momentum is initially zero.      mCvC + mBvB = constant = 0 The velocity of the car at the end of the barge can now be solved as      vC = -mB/mC vB           = -(2,500/500)(-0.885)i m/s           = 4.425i m/s Position Position of Barge and Car after motion starts Barge and Car Position when Car Leaves Barge The combined CG of the car and the barge is initially stationary, so it must remain stationary. Knowing the position of the combined CG, the position of the car and the barge relative to their original positions can be determined. The diagram at the left shows the position after motion starts. The initial combined rCG is equal to zero since that is the location of the coordinate system.            0 = mC xC - mB xB There are still two unknowns, xC and XB. However, when the car leaves the barge, the combined distance xB and xC must be half the barge distance,      xC + xB = 0.5 L Combining with the previous equation gives,      0 = mC xC - mB (0.5L - xC)      0 = (500 + 2,500)xC - 2,500 [0.5 (42)]      xc = 17.5 m The distance from the dock is      xdock = 21 - 17.5      xdock = 3.5 m

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