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.

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   (↓)

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.