George has a mass of 30 kg. he sits 1.4 m from the axis of rotation of a see-saw. if susie has a mass of 27 kg, how far from the axis of rotation should she sit on the other side of the see-saw to keep it in static equilibrium? assume the center of mass of the see-saw is above the axis of rotation. g

d ₂ = 1.56 m : horizontal distance from where Susie sits to the axis of rotation.

Explanation:

M = F * d   Formula (1)

F is the magnitude of said force (N)

d: distance to the axis of rotation

The direction of the moment is determined by the direction of rotation and can be assumed positive in a counterclockwise direction and negative in the clockwise direction.

Forces on the see-saw

W: Weight: in the vertical direction and down

W₁= m₁*g =  30 kg*9.8 m/s²= 294 N

W₂= m₂*g =  27 kg*9.8 m/s²= 264.6 N    

Distances of the forces to the axis of rotation

d₁ =  1.4 m :distance from where W₁ is located to the axis of rotation

d₂ :  distance from where W is located to the axis of rotation

Problem development

For static equilibrium the sum of moments around the axis of rotation is zero:

∑M =0

W₁*d₁ - W₂*d₂ = 0

W₁*d₁ = W₂*d₂

294*1.4  = 264.6*d₂

411.6= 264.6*d₂

d₂ =( 411.6) / (264.6)

d ₂= 1.56 m

pe = weightxheight = 2000x1000=2,000,000joules

its a 5.5 n

glad that