# A force of 200 N is applied to the long arm of the 60 cm long arm. What weight is suspended

A force of 200 N is applied to the long arm of the 60 cm long arm. What weight is suspended from the short arm of the arm 20 cm long if the arm is in equilibrium.

Data: l1 (length of the long arm of the arm) = 60 cm; F1 (force applied to the long arm) = 200 N; l2 (short shoulder length) = 20cm; the lever taken is in balance.

Reference values: g (acceleration due to gravity) ≈ 10 m / s2.

The mass of the load suspended from the short arm of the lever is determined from the equality: F1 * l1 = M (moment of force) = F2 * l2 = m * g * l2, whence m = F1 * l1 / (g * l2).

Calculation: m = 200 * 60 / (10 * 20) = 60 kg.

Answer: A load weighing 60 kilograms is suspended from the short shoulder.

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