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

a) cold, dry

explanation:

continental polar masses are cold and dry air masses because the cold air cannot hold much moisture. as they move into warmer air, they can trigger precipitation.

15.3 cos 65 degrees trig