A uniform meter rule of mass 150g is pivoted freely at the 0cm mark.what force applied vertically upwards at the 60cm mark is needed to maintain the rule horizontally
To maintain the rule horizontally, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
Let the force applied at the 60cm mark be F.
Clockwise moment = Force x Distance
Clockwise moment = F x 60
Anticlockwise moment = weight x Distance
Weight = mass x gravity
Weight = 0.15kg x 9.8 m/s^2
Weight = 1.47N
Anticlockwise moment = Weight x Distance
Anticlockwise moment = 1.47 x 60
Since the rule is balanced horizontally,
F x 60 = 1.47 x 60
F = 1.47N
Therefore, a force of 1.47N applied vertically upwards at the 60cm mark is needed to maintain the rule horizontally.