A 0.115-kg, 53.6-cm-long uniform bar has a small 0.045-kg mass glued to its left end and...
A 0.115-kg, 53.6-cm-long uniform bar has a small 0.045-kg mass glued to its left end and a small 0.150-kg mass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.
How far from the left end should the fulcrum be placed?
Homework Answers
bar will be balanced at the centre of gravity (centre of mass) of bar plus mass system.
for centre of mass:
Xcm = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)
suppose distace of Xcm from left end is d then
d = [ (53.6/2 * 0.115) + (0.045 * 0) + (53.6 * 0.150 )] / (0.115 +
0.045 + 0.150)
d = 35.88 cm from left end
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