A board weighing 28 kg and a length of 1.2 m lies on two supports, the left one at a distance of 30 cm

A board weighing 28 kg and a length of 1.2 m lies on two supports, the left one at a distance of 30 cm from the board, the right one 15 cm.What force must be applied to bring the board a) to the left side b) to the right

A board with a mass of m = 28 kg and a length of L = 1.2 m lies on two supports. From the side of the ground, the gravity force acts on the board Ftyach = m • g, Ftyach = 28 kg • 9.8 N / kg; Ftyazh = 274.4 N. It is attached to the center of the board.
one). To determine what force F1 must be applied to lift the board by the right edge, we will use the Archimedes rule for the lever: F1 • L1 = F2 • L2; we get: F1 = (F2 • L2): L1. In this case, the lever will rotate around the left point. Since the left fulcrum is located at a distance of 30 cm = 0.3 m from the edge of the board, the shoulder of the applied force will be L1 = 1.2 m – 0.3 m; L1 = 0.9 m.
The gravity of the board F2 = Ftyazh = 274.4 N will have a shoulder L2 = L / 2 – 0.3 m; L2 = 0.6 m – 0.3 m; L2 = 0.3 m. The applied force will be: F1 = (274.4 N * 0.9 m): 0.3 m; F1 = 823.2 N.
2). To determine what force F1 must be applied in order to lift the board by the left edge, we will carry out similar reasoning. L1 = 1.2 m – 0.15 m; L1 = 1.05 m, since the right fulcrum is 15 cm = 0.15 m from the edge of the board. The gravity of the board will have a shoulder L2 = L / 2 – 0.15 m; L2 = 0.6 m – 0.15 m; L2 = 0.45 m. The applied force will be: F1 = (274.4 N * 1.05 m): 0.45 m; F1 ≈ 640.3 N.

Answer: to lift the board a) by the left edge, the applied force will be ≈ 640.3 N; b) for the right edge, the applied force will be 823.2 N.