A60-kg uniform board 2.4 m long is supported by a pivot 80 cm from the left end and by a scale at the right end (see the fig. below). (a) how far from the left end should a 40-kg child sit for the scale to read zero? (b) where should the child sit if the scale is to read (i) 100 n and (ii) 300n?

a) x = 0.61 m

b) x = 1.4 m

Explanation:

Given data;

M = 60 kg,

m = 40 kg,

F considered to be a  reading

Torque of N about the given pivot point = F*(2.4 - 0.80) = 1.6 F (counterclockwise)

Torque of Mg about the given pivot Point = Mg*(1.2 - 0.80)

= 0.4*60*9.8 = 235.2 N-m (clockwise)

a) F = 100 N

note 1.6F = 160 < 235.2, so

mass m should be placed at left of pivot,

its torque = mg*(0.80 - x)

                = 40*9.8*(0.80 - x) = 392(0.80 -x)

160 + 392(0.8 - x) = 235.2

x = 0.61 m

b) N = 300 N

note 1.6F = 480 > 235.2, so

mass m should be placed at right of pivot,

its torque = mg*(x - 0.80) = 40*9.8*(x - 0.80) = 392(x -0.80)

480 = 392(x - 0.8) + 235.2

x = 1.4 m