Asked by Josh
A brother (mass 20 kg) and his big sister (mass 32 kg) want to play on a teeter-totter. The beam is 7 m long and has a mass of 41 kg. Suppose the brother sits at the end of his side, 3.5 m away from the pivot point. How far away from the pivot point does the sister need to sit for the system to be in rotational equilibrium?
Answers
Answered by
Mila
Use the center of mass fomula:
Xcm = (M1X1 + M2X2 + ... + MnXn)/(M1 + M2 + ... + Mn)
Don't forget the mass of the teeter-totter and its location at its geometric center (M1 = 41 kg, X1 = 3.5 since the fulcrum is in the middle of the beam).
Set the location of the sister at the origin (X = 0) and the brother at the opposite end of the beam (X = 7m).
Xcm = (20 x 7 + 32 x 0 + 41 x 3.5)/(20 + 32 + 41) = 283.5/93 = 3.05
The sister is 3.05 m from the pivot point.
Xcm = (M1X1 + M2X2 + ... + MnXn)/(M1 + M2 + ... + Mn)
Don't forget the mass of the teeter-totter and its location at its geometric center (M1 = 41 kg, X1 = 3.5 since the fulcrum is in the middle of the beam).
Set the location of the sister at the origin (X = 0) and the brother at the opposite end of the beam (X = 7m).
Xcm = (20 x 7 + 32 x 0 + 41 x 3.5)/(20 + 32 + 41) = 283.5/93 = 3.05
The sister is 3.05 m from the pivot point.
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