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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.

Update:

Hint: Write the equations for the man's and child's absolute velocities using the relative velocity equation 

Answer- 10.73 ft

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  • 1 month ago
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    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.

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