The normal reactions of the wheel is determined as 13222.22 lb at point F and 21888.89 lb for point E and D.
Free body diagram of the plane is drawn by removing all the supports and replacing with forces. Equilibrium of moment of forces about x-axis is considered and relation between the reaction forces at point D and E is determined. Equilibrium of moment of forces about y-axis is considered and value of the reaction force at point F is determined. Moment of axis (line) is considered to eliminate an unknown force and get equation with other two variables.
After drawing the free body diagram, apply force balance about x-axis:
= ∑Mₓ = 0
= - (R₄ X 14) + (R₅ X14) + (W₂ X 6) - (W₃ X 8) = 0
= (R₅ - R₄) X 14 = (W₃ X 8) - (W₂ X 6)
= R₅ = R₄
Apply moment balance about y axis,
= ∑Mₙ = 0
= - (R₆ X 27 ) + (W₁ X 7) + (W₂ X 4) + (W₃ X 4) = 0
= R₆ = 13222.22 lb
Next, consider force equilibrium equation.
Write the force equilibrium equation along vertical axis, z-axis;
= ∑F = 0
= R₆ + R₅ + R₄ - (W₁ + W₂ + W₃) = 0
where, R represents reaction and W represents load.
= R₆ + R₅ + R₄ - (W₁ + W₂ + W₃) = 0
= R₅ = R₄ = 21888.89 lb
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