A uniform metre rule of length 100 cm is balanced at pivot point (at 70 cm mark )50 gf which is balance at 100cmmark. Find the moment of force about the pivot point.
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Answer:
s the scale is uniform, so the weight will be balanced at mid point i.e. 50 cm of the scale.
Let, M be the mass of the meter scale.
Using principle of moments, for equilibrium, sum of moment of force on left side of knife edge is equal to sum of moment of force on right side.
So, (70−50)Mg=(94−70)0.05g (as moment of force = force × perpendicular distance from fulcrum)
or M=0.06kg
Explanation:
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