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?
2 answers
am confused
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
2 answers