Torque seesaw question
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Torque seesaw question

[From: ] [author: ] [Date: 13-03-22] [Hit: ]
The answer doesnt seem right-The seesaws length is split (on either side of pivot) into the same proportions as the masses of the individuals at either end of it.However the shorter length goes with the larger mass etc.ratio of masses = 86/11 = 7.ratio of distances = 7.818 = L/S{L = longer length,L + S = 5.......
A 86kg adult sitting on one end of a seesaw with his child(11kg) on the other end. The length of the seesaw is 5.3m, where should the pivot be placed to balance the seesaw.
Distance from pivot to child dc and from pivot to adult da
I got da=0.78m and dc=4.52m
The answer doesn't seem right

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The seesaw's length is split (on either side of pivot) into the same proportions as the masses of the individuals at either end of it. However the shorter length goes with the larger mass etc.
ratio of masses = 86/11 = 7.818
ratio of distances = 7.818 = L/S {L = longer length, S = shorter length}
L + S = 5.3
L = 5.3 - S
7.818 = 5.3-S/S
7.818S = 5.3 - S
8.818S = 5.3
S = 5.3/8.818 = 0.60 m
L = 5.3 - 0.60 = 4.7 m
U can now select distance from child and distance from adult to pivot OK? :>)

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1) da + dc = 5.3 m (known length of seesaw)
2) 11dc = 86da (ΣM = 0)
Solving simultaneously,

da = 0.601 m
dc = 4.699 m
1
keywords: seesaw,Torque,question,Torque seesaw question
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