Answer:
a) Fbc = 22692.23 N
b) Ax = 10148.276 N (→)
Ay = 10486.552 N (↓)
Explanation:
Given
Mass of the crane arm: M = 200 Kg
rM = 2.00 m
Mass of the block: m = 800 Kg
rm = 7.00 m
Fbc = ?
rbcx = (1.80 + 1.20) m = 3.00 m
rbcy = (2.40 - 1.00) m = 1.40 m
a) We apply rotational equilibrium around point A as follows
∑τ = 0 (Counterclockwise is the positive rotation direction)
rbcx*Fbcx + rbcy*Fbcy - rM*WM - rm*Wm = 0 (I)
From the pic we have the Right triangle where
BC² = 1.2²+2.4² = 7.2 ⇒ BC = √7.2 = 2.683
then
Cos ∅ = 1.2/2.683 = 0.447
Sin ∅ = 2.4/2.683 = 0.894
If
Fbcx = Fbc*Cos ∅
⇒ Fbcx = 0.447*Fbc
Fbcy = Fbc*Sin ∅
⇒ Fbcy = 0.894*Fbc
Now, we obtain
3.00*(0.447*Fbc) + 1.40*(0.894*Fbc) - 2.00*(200*9.81) - 7.00*(800*9.81) = 0
⇒ 1.342*Fbc + 1.252*Fbc - 3924 - 54936 = 0
⇒ 2.594*Fbc = 58860
⇒ Fbc = 22692.23 N
b) We apply
∑Fx = 0 (+→)
⇒ Ax - Fbcx = 0
⇒ Ax = Fbcx = 0.447*Fbc = 0.447*(22692.23 N)
⇒ Ax = 10148.276 N (→)
∑Fy = 0 (+→)
⇒ Ay + Fbcy - WM - Wm = 0
⇒ Ay = - Fbcy + WM + Wm
⇒ Ay = - 0.894*Fbc + m*g + M*g
⇒ Ay = - 0.894*(22692.23 N) + (800 Kg*9.81 m/s²) + (200 kg*9.81 m/s²)
⇒ Ay = 10486.552 N (↓)
