A metre scale is pivoted at its mid point. A 0.60 N weight is suspended
from one end. How far from the other end must a 1 N weight be suspended for
the rule to balance ?
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In solving this problem, we will assume that the metre scale has no weight
Taking moments;
Anticlockwise moments = clockwise moments
0.6 x 50 = 1 x y (y is the distance of the second force from the centre)
y = 30
This means the 1N force must be suspended (50 - 30) = 20 cm from the other end.
The answer is 20 cm.
Taking moments;
Anticlockwise moments = clockwise moments
0.6 x 50 = 1 x y (y is the distance of the second force from the centre)
y = 30
This means the 1N force must be suspended (50 - 30) = 20 cm from the other end.
The answer is 20 cm.
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