Question

Q.1) The crane arm is pinned at point A and has a mass of 200 kg whose weight is acting at a point 2 m to the right of point A. The mass of the block is 800 kg. If the crane is in equilibrium, find (a) the magnitude of the force exerted on the arm by the hydraulic cylinder BC, (b) the support reactions at point A.

Answer 1

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   (↓)