physics help?

Question

A 75 kg adult sits at one end of a 11.0 m board, on the other end of which sits his 25 kg child. Where should the pivot be placed so the board (ignore its mass) is balanced?

Answer 1

x=distance from pivot to adult
y=distance from pivot to child
75x=25y
x+y=11
x:y=1:3
11/4=x
The pivot should be placed 2 3/4 metres away from the adult, 8 1/4 metres away from the child.

Answer 2

The child's mass is 1/3 the mass of the adult, so the distance from the child to the pivot has to be 3 times the distance from the adult to the pivot. Since the board is 11 m long, 1/4 of the board's length must be on the adult's side, and 3/4 must be on the child's side. Thus, the pivot must be 2.75 m from the adult, and 8.25 m from the child.

Answer 3

the answer lies in the fact that you want zero movement when it is balance, and that represents a balance of torque. So if the force of the kid pushing down is equal to the force of the adult pushing down, then the pivot point is in the correct place.

massAdult = 75kg
massKid = 25kg
momentArmAdult = mAA
momentArmKid = mAK
totalLength = 11.0 = mAA + mAK
or (mAK = 11.0 - mAA)

To make torques equal

massKid*mAK = massAdult*mAA

do a little subsitution for mAK

massKid*(11.0-mAA) = massAdult*mAA

substitue the weight massKid and massAdult

25 * (11.0 -mAA) = 75 * mAA

275.0 - 25*mAA = 75*mAA

275.0 = 100*mAA

mAA = 2.75
mAK = 11 - 2.75 = 8.25

Basically since the adult is 3 times the size of the kid, alltogether they are 4 units, the pivot point is 3/4 of the way from they kid to the adult.

Answer 4

3.67 meters from the 75kg