Moment (N m) = Force (N) × perpendicular distance from the pivot (m)

The force might not always be perpendicular to the distance
A uniform metre rule is pivoted at the 50 cm mark.
A 0.5 kg weight is suspended at the 80 cm mark, causing the rule to rotate about the pivot.
Assuming the weight of the rule is negligible, what is the turning moment about the pivot?


If not already given, drawing all the forces on an object in the diagram will help you see which ones are perpendicular to the distance from the pivot. Not all the forces will provide a turning effect and it is not unusual for a question to provide more forces than required to throw you off!
For a system to be in equilibrium, the sum of clockwise moments about a point must be equal to the sum of the anticlockwise moments (about the same point)
Diagram showing the moments acting on a balanced beam
A uniform beam of weight 40 N is 5 m long and is supported by a pivot situated 2 m from one end.When a load of weight W is hung from that end, the beam is in equilibrium as shown in the diagram.

What is the value of W?
A? ? ?10 N? ? ? ? ? ? ? ?B? ???50 N? ? ? ? ? ? ? ?C? ???25 N? ? ? ? ? ? ? ?D? ? ?30 N


Make sure that all the distances are in the same units and you’re considering the correct forces as clockwise or anticlockwise, as seen in the diagram below
轉(zhuǎn)載自savemyexams
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