Question

) A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. (a) Where should the pivot be placed so that the board is balanced, ignoring the board's mass? (b) Find the pivot point if the board is uniform and has a mass of 15 kg.

Answer 1

Answer with Explanation:

We are given that

Mass of adult,M=75 kg

Length of board=l=9 m

Mass of child=m=25 kg

a.Let pivot placed at point O so that the board is  balanced ignoring the board's mass

Moment about the point O

$75x=25(9-x)$

$75x=225-25x$

$75x+25x=225$

$100x=225$

$x=\frac{225}{100}=2.25 m$

Hence, the pivot placed at 2.25 m from the adult.

b.Mass of board=m''=15 kg

Center of board=$\frac{9}{2}=4.5 m$

Moment about point O

$75x=15(4.5-x)+25(4.5+(4.5-x))$

$75x=67.5-15x+225-25x$

$75x+25x+15=67.5+225$

$115x=292.5$

$x=\frac{292.5}{115}=2.5 m$

Hence, the pivot point is 2.5 m from the adult.