The length of the horizontal arm with weights 2.5 and 4 N, at the ends is 52 cm. Find the shoulders of the gravity forces

The length of the horizontal arm with weights 2.5 and 4 N, at the ends is 52 cm. Find the shoulders of the gravity forces of the weights and the force of pressure of the lever on the support. Disregard the weight of the lever itself.

L = 52 cm = 0.52 m.

F1 = 2.5 N.

F2 = 4 N.

L1 -?

L2 -?

F -?

The force of the lever pressure on the support F will be the sum: F = F1 + F2.

F = 2.5 N + 4 N = 6.5 N.

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.

L1 + L2 = L.

L1 = L – L2.

F1 * L1 = F2 * L2.

F1 * (L – L2) = F2 * L2.

F1 * L – F1 * L2 = F2 * L2.

L2 = F1 * L / (F1 + F2).

L2 = 2.5 N * 0.52 m / (2.5 N + 4 N) = 0.2 m.

L1 = 0.52 m – 0.2 m = 0.32 m.

Answer: L1 = 0.32 m, L2 = 0.2 m, F = 6.5 N.