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Answer:
Ra = 2.24 kN
Rb = 3.34 kN
Explanation:
Draw a free body diagram. There are 4 forces on the board:
Weight of the board mg pulling down at the center,
Weight of the diver Mg pulling down at the end,
Reaction Ra pushing up at A,
and reaction Rb pushing up at B.
The total length of the board is 1.2 m + 3.9 m = 5.1 m.
Half of the length is 5.1 m / 2 = 2.55 m.
(part 1) Sum the moments about point B (take clockwise to be positive):
∑τ = Iα
Ra (1.2 m) + mg (2.55 m − 1.2 m) + Mg (3.9 m) = 0
Ra (1.2 m) + mg (1.35 m) + Mg (3.9 m) = 0
1.2 Ra + g (1.35 m + 3.9 M) = 0
Ra = -g (1.35 m + 3.9 M) / 1.2
Ra = -9.81 m/s² (1.35 × 30 kg + 3.9 × 65 kg) / 1.2
Ra = -2244 N
Ra = -2.24 kN
Notice the sign is negative, which means Ra is actually pulling down, not pushing up.
(part 2) Sum the moments about point A (take clockwise to be positive):
∑τ = Iα
-Rb (1.2 m) + mg (2.55 m) + Mg (5.1 m) = 0
-1.2 Rb + g (2.55 m + 5.1 M) = 0
Rb = g (2.55 m + 5.1 M) / 1.2
Rb = 9.81 m/s² (2.55 × 30 kg + 5.1 × 65 kg) / 1.2
Rb = 3335 N
Rb = 3.34 kN
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