Asked by Pram
A 5 m, 10 kg seesaw is balanced by a little girl(25 kg) and her father(80 kg) at the extreme opposite ends. How far from the seesaw's center of mass must the fulcrum be placed
Answers
Answered by
MathMate
It is rare to adjust the position of the fulcrum of a seasaw to adapt to persons of different masses.
Here we assume both Dad (D) and girl (G) are seated at the extremities of the 5m seasaw, and the centre of mass (M) of the 10 kg seasaw is in the middle. We assume the fulcrum (F) is between the Dad and the middle of the seasaw.
Draw a diagram,
G---------M---F----D
We know that GM=MD=2.5m
Denote MF=x
Take moments about the fulcrum,
Mg(2.5+x)+Ms(x)-Md(2.5-x)=0
where
Mg=25 kg (mass of girl)
Ms=10 kg (mass of seasaw)
Md=80 kg (mass of Dad)
Everything in the equation above is known except for x.
Solve for x.
Here we assume both Dad (D) and girl (G) are seated at the extremities of the 5m seasaw, and the centre of mass (M) of the 10 kg seasaw is in the middle. We assume the fulcrum (F) is between the Dad and the middle of the seasaw.
Draw a diagram,
G---------M---F----D
We know that GM=MD=2.5m
Denote MF=x
Take moments about the fulcrum,
Mg(2.5+x)+Ms(x)-Md(2.5-x)=0
where
Mg=25 kg (mass of girl)
Ms=10 kg (mass of seasaw)
Md=80 kg (mass of Dad)
Everything in the equation above is known except for x.
Solve for x.
Answered by
Anonymous
what is the answer
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