A 500 cm meter rule is pivoted at its middle point. If weight of 2 N is hanged from the 20 cm point, Calculate the amount of weight required to be applied at the 80 cm mark to keep it in a balanced position
Answers
According to the principle of moments, when an object is in rotational equilibrium, then
Total anticlockwise moments = Total clockwise moments
Total anticlockwise moments:
Length of lever arm = (50 – 30) = 20 cm
= 0.20 m
Since the length of the lever arm is the distance from its mid-point, where its balanced force applied = 10 N
Anticlockwise moment = Lever arm x Force applied
= 0.20 x 10 = 2 Nm
Clockwise moment: Length of lever arm = (60 – 50)
= 10 cm
= 0.10 m.
Since the length of the lever arm is the distance from the mid-point, about which balanced Force applied = 20 N
Clockwise moment = lever arm x force applied
= 0.10 × 20 = 2 Nm
Therefore,
Since the total anti-clockwise moment = total clockwise moment = 2 Nm, according to the principle of moments, it is in rotational equilibrium ie, the meter rule remains balanced about its pivot.