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 of the man-board system?

Answer 1

Answer:

x=1.25m

Explanation:

The Center of mass of the system is defined as the point where whole mass of the body is appeared to be  concentrated.

The center of mass of the system is given by

         x=  $\frac{m1x1+m2x2}{m1+m2}$      

where m1 is mass of man =60 kg

           m2 mass of board =20 kg

let the man be at the origin  x1 =0 , x2 =5m

by substituting in above formula

x =$\frac{(60*0)+(20*5)}{60+20}$ = $\frac{100}{80}$ =1.25 m

x=1.25m

So the center of mass of the system is at 1.25 m from man.

Answer 2

Answer:

The center of mass of the man-board system is = 1.25 m

Explanation:

Mass of a man = 60 kg

Mass of rod = 20 kg

Length of the rod = 10 m

The center of mass is given by

$x =\frac{ m_{1} x_{1} + m_{2} x_{2}}{m_{1} +m_{2} }$

$x = \frac{(60)(0)+ (20) (5)}{60+ 20}$

$x = \frac{100}{80}$

x = 1.25 m

Therefore the center of mass of the man-board system is = 1.25 m