Question

A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass (or center of gravity) of the man-board system?

Answer 1

= 1.25 m

Step-by-step explanation:

mass of man (Mm) = 60 kg

mass of board (Mb) = 20 kg

length of board (L) = 10 m

Firstly lets take the center of the board as our origin, this means that the center of the board will be x = 0 and the man will be standing at one end of the board which will be point x = 5. Therefore:

Point of action of boards mass (P1) = 0 m

Position of the man (P2) = 5 m

center of mass =
`(Mm.P2 + Mb.P1)/(Mm + Mb)`

center of mass =
`((60x5)+(20x0))/(60 + 20)`

center of mass = 300 / 80 = 3.75

this means the total center of mass of the man-board system is at 3.75 m from the origin.

The distance of the center of mass from the position of the man = 5 - 3.75

= 1.25 m