The Principle of Moments
Question
The diagram below shows two masses balanced on a level beam.
How far is the 10 N weight from the pivot?
The beam is balanced and so from the Principal of Moments we know that:
Total clockwise moment about the pivot = Total anticlockwise moment about the pivot.
Calculate each individual moment first:
Anticlockwise moment
The distance from the pivot which makes a right angle with the line of action of the force. from the pivot = d m.
Force F = 10 N.
Anticlockwise moment = F x d = 10 N x d m = 10d Nm.
Clockwise moment
Perpendicular distance from the pivot = 1 m.
Force F = 20 N.
Clockwise moment = F x d = 20 N x 1 m = 20 Nm.
Total clockwise moment = Total anticlockwise moment
10d = 20.
\({d}=20 Nm 梅 10 N\)
d = 2 m
The 20 N weight is 2 m from the pivot.
Question
A parent and child are at opposite sides of a playground see-saw. The parent sits 0.8 m from the pivot. The child sits 2.4 m from the pivot and weighs 250 N.
Calculate the weight of the parent if the see-saw is balanced.
The see-saw is balanced and so from the Principal of Moments we know that:
Total clockwise moment about the pivot = Total anticlockwise moment about the pivot
Anticlockwise moment
The anticlockwise moment is the child's moment = Fd.
Perpendicular distance of child from the pivot = 2.4 m.
Force F = 250 N.
Anticlockwise moment = 250 N 脳 2.4 m = 600 Nm.
Clockwise moment
The clockwise moment is the parent's moment = Fd.
Perpendicular distance of adult from the pivot = 0.8 m.
Force F = F N.
Clockwise moment = F N x 0.8 m = 0.8F Nm.
Total clockwise moment = Total anticlockwise moment.
0.8 F = 600.
F = 600 Nm 梅 0.8 m.
F = 750 N.
The parent鈥檚 weight equals 750 N.