A uniform metre rule of Mass 120g is pivoted at 60cm Mark. At what point on metre rule should a mass of 50g be suspended for it to balance horizontally
the 120 g mass acts at the center of the metre rule (50 cm mark)
... 10 cm from the pivot
the 50 g mass needs to be on the other side of the pivot (> 60 cm mark)
120 g * 10 cm = 50 g * ? cm
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Physics
A metre rule is balanced by masses of 24g and 16g suspended from its ends .find the position of the pivot??
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A uniform m metre rule of mass 110g is balance at xcm marks when a mass of 60g is lang at 10cm marks calculate. 1 the balance point x 2 the reaction at turning point
To find the point on the meter rule where a 50g mass should be suspended for it to balance horizontally, we can use the principle of moments.
The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments about any point.
In this case, we can choose the pivot point at the 60cm mark as our reference point.
Let's assume the distance from the pivot point to the 50g mass is x cm.
The moment generated by the 120g mass at the 60cm mark is (120g) * (60cm) = 7200 g.cm clockwise.
The moment generated by the 50g mass at the x cm mark is (50g) * (x cm) anticlockwise.
For the meter rule to balance horizontally, these two moments should be equal. Therefore, we have the equation:
7200 g.cm = 50g * x cm
Simplifying the equation:
7200 = 50 * x
x = 7200 / 50
x = 144 cm
Hence, the 50g mass should be suspended at the 144cm mark on the meter rule to balance it horizontally.