Two men, Joe and Larry, are carrying a 24-foot-long board with its weight evenly distributed along its length.
If Joe is at one end and Larry is 6 feet from the other end, what percentage of the weight is Larry carrying?
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So you know that the weight is distributed by Larry and Joe in different ways.The center of mass is at 12f (24f/2) and Larry is applying the force at 6f from the center of mass.
1st step: Calculate the torque (T) choose the regular axis in the center of mass ∑ T = 0
TJ + TL + Tg = 0
-12J+6L=0
L=2J
2nd step: Consider the weight (G) (force that the board is applying on the man) 100% and calculate the Force as a percentage: s u m F = ma =0
J+L-G= 0
J+ L= G
J+2J=G
3J=100%
J= 33.33%
Joe is carrying 33.33% of the weight
3rd step: Substitute and find the percentage o weight that Larry is carrying: L=2J L=2*33,33% L=66.7%