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jeka57
19 days ago
8

A 30.0-kg child sits on one end of a long uniform beam having 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 below the beam a distance 1.10 m from the 30.0-kg child. How long is the beam?

Physics
1 answer:
Softa [2K]19 days ago
4 0

Let "L" denote the beam's length.

According to the diagram:

AD = length of the beam = L

AC = CD = AD/2 = L/2

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

BD = AD - AB = L - 1.10

m = mass of the beam = 20 kg

m₁ = mass of the child at one end = 30 kg

m₂ = mass of the child at the opposite end = 40 kg

By applying the principle of torque equilibrium around point 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

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option D.

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The correct choice is option D.

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An object is deemed to be in equilibrium when the net moment acting on it equals zero.

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15 days ago
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Maru [2360]

Answer:900 feet

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