Answer:
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
