High School

A 30.0-kg child sits on one end of a long uniform beam with a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The beam balances when a fulcrum is placed 1.10 m from the 30.0-kg child. How long is the beam?

Answer :

The total length of the beam is determined as 1.98 m.

Principle of moment

The length of the beam is determined by applying the principle of moment as shown below;

Take moment about the pivot;

Let the total length of the beam = x

mid point of the beam = 0.5 x

(30 x 1.1) = 20(0.5x - 1.1) + 40(x - 1.1)

33 = 10x - 22 + 40x - 44

99 = 50x

x = 99/50

x = 1.98 m

Thus, the total length of the beam is determined as 1.98 m.

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let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m