The length of one arm of the lever is 50 cm, the other is 10 cm. A force of 400 N acts on the larger arm.

| March 28, 2021 | education

The length of one arm of the lever is 50 cm, the other is 10 cm. A force of 400 N acts on the larger arm. What force must be applied to the smaller arm in order for the lever to be in balance?

L1 = 50 cm = 0.5 m.

L2 = 10 cm = 0.1 m.

F1 = 400 N.

F2 -?

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, which is called the lever arm: M = F * L.

F1 * L1 = F2 * L2.

The force that should act on the smaller shoulder is determined by the formula: F2 = F1 * L1 / L2.

F2 = 400 N * 0.5 m / 0.1 m = 2000 N.

Answer: in order for the lever to be in balance, a force F2 = 2000 N must be applied to the smaller arm.



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