) 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.

A) The distance at which the pivot be placed so that the board is balanced, ignoring the board's mass is; x = 2.25 m from the adult.

B) The pivot point if the board is uniform and has a mass of 15 kg is; x = 2.54 m from the adult

We are given;

Mass of adult; M1 = 75 kg

Length of board = 9 m

Mass of child; M2 = 25kg

A) If we ignore the boards mass, then let the position of the pivot from the adult be x.

Thus, taking moments about the pivot gives;

75x = 25(9 - x)

75x = 225 - 25x

75x + 25x = 225

100x = 225

x = 225/100

x = 2.25 m

B) We are told that the board is now uniform and has a mass of 15 kg.

This mass of board will be acting at the center.

Center of board is 9/2 = 4.5 m.

Now, taking moments about the previous pivot gives;

75x = 15(4.5 - x) + 25(9 - x)

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

75x = 67.5 + 225 - 40x

115x = 292.5

x = 292.5/115

x = 2.54 m

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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

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

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

Center of board=

Moment about point O

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

not too sure thinking all accelerations are pointed down