Question

Two children are balanced on a seesaw that has a mass of 18.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the pivot. The center of mass of the seesaw is 0.133 m from the pivot (on the side of the first child). (a) If the second child has a mass of 34.4 kg, how far (in m) is she from the pivot

Answer 1

1.28 m

Step-by-step explanation:

As shown in the diagram attached,

According to the principle of moment,

For a body at equilibrium,

Sum of clockwise moment = sum of anticlockwise moment.

Taking moment about the pivot,

W₁(1.6)+W(0.133) = W₂(x)............... Equation 1

Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.

But,

W = mg

Where g = 9.8 m/s², m = mass of the body

Therefore,

W₁ = 26×9.8 = 254.8 N,

Wₓ = 18×9.8 = 176.4 N

W₂ = 34.4×9.8 = 337.12 N

Substitute these values into equation 1

(254.8×1.6)+(176.4×0.133) = 337.12(x)

407.68+23.4612 = 337.12x

337.12x = 431.1412

x = 431.1412/337.12

x = 1.2789

x ≈ 1.28 m