# Weights of 5 kg and 15 kg are suspended at the ends of the 2 m long arm.

Weights of 5 kg and 15 kg are suspended at the ends of the 2 m long arm. How far from the middle of the lever should the support be placed to keep the lever in balance?

Task data: L (total arm length) = 2 m; mg1 (mass of the first load) = 5 kg; mg2 (mass of the second load) = 15 kg.

Equilibrium condition of the specified lever: Ft1 * l1 = Ft2 * l2; mg1 * l1 = mg2 * l2 and mg1 * (L – l2) = mg2 * l2.

Substitute the variables: 5 * (2 – l2) = 15 * l2.

10 – 5l2 = 15l2

10 = 20l2 and l2 (smaller shoulder) = 10/20 = 0.5 m.

Larger shoulder: l1 = L – L2 = 2 – 0.5 = 1.5 m.

Answer: The support should be at a distance of 0.5 m from a load weighing 15 kg (at a distance of 1.5 m from a load weighing 5 kg).

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