Relative velocity for man and child on platform ?
A 180 lb man A and a 40 lb child C are at the opposite ends of a
250 lb floating platform P with a length Lfp D 15 ft. The man,
child, and platform are initially at rest at a distance ı D 1 ft from a
mooring dock. The child and the man move toward each other with
the same speed v0 relative to the platform. If the drag force due to
the water is negligible, determine the distance d from the mooring
dock where the child and man will meet.
Hint: Write the equations for the man's and child's absolute velocities using the relative velocity equation
Answer- 10.73 ft
- Andrew SmithLv 71 month agoFavorite Answer
As they are moving at the same speed relative to the dock they must meet at the middle of that dock. The position of the centre of mass of the system cannot shift. Initially the MOMENT about the end of the dock is (40d + 180 L+d + 250*(L/2)+d) At the end the moment is (40+180+250) *x
-> x = (40*1+180*16+250*8.5) / (40+180+250) = 10.7 feet. ie they meet at the position of the CENTRE OF MASS.