Same force

The trunk of 5m length and 95 kilograms has a center of gravity 2m from the thicker end. The tribe is carried by two men, one at the thicker end. At what distance does the trunk carry a man from the other end to make the same force on it?

Result

x =  1 m

Solution:

l=5 m m=95 kg y=2 m  ΣM=0  F1 y=F2 (lyx)=F2 (521)=0 F1=F2  y=lyx  x=l2 y=52 2=1 ml=5 \ \text{m} \ \\ m=95 \ \text{kg} \ \\ y=2 \ \text{m} \ \\ \ \\ Σ M=0 \ \\ \ \\ F_{1} \cdot \ y=F_{2} \cdot \ (l-y-x)=F_{2} \cdot \ (5-2-1)=0 \ \\ F_{1}=F_{2} \ \\ \ \\ y=l-y-x \ \\ \ \\ x=l-2 \cdot \ y=5-2 \cdot \ 2=1 \ \text{m}



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