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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...Asked by Samson Opiyo
A uniform meter rule of mass 120 g is pivoted at the 60 cm mark. At what point on the meter rule should a mass of 50 g be suspended for it to balance horizontally?
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Answered by
Samson Opiyo
am confused
Answered by
R_scott
the center-of-mass of the rule is the 50 cm mark
... 10 cm from the pivot
the suspended mass is on the other side of the pivot
the moments are equal ... 120 g * 10 cm = 50 g * d
... 10 cm from the pivot
the suspended mass is on the other side of the pivot
the moments are equal ... 120 g * 10 cm = 50 g * d
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