Forces of 120 N and 90 N act on the arms of the lever. The distance from the fulcrum to the action
| March 13, 2021 | education
Forces of 120 N and 90 N act on the arms of the lever. The distance from the fulcrum to the action of the greater force is 30 cm. Determine the arm of the lower force.
F1 = 120 N.
F2 = 90 N.
L1 = 30cm = 0.3m.
L2 -?
When the lever is in equilibrium, the moments of forces that act from opposite sides of the lever are equal to each other: M1 = M2.
The moment of force M is called the product of the applied force F to the smallest distance from the application of force in the axis of rotation of the lever L: M = F * L.
F1 * L1 = F2 * L2.
The shoulder L2, on which the smaller force F2 acts, will be determined by the formula: L2 = F1 * L1 / F2.
L2 = 120 N * 0.3 m / 90 N = 0.4 m.
Answer: the shoulder of the lesser force is L2 = 0.4 m.
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