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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.

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Chrisnando

Answer:

Question is incomplete, missing image is in attachment.

a) [tex] F_H = 28.6 KN [/tex]

b) |A| = 20.3 KN

Explanation:

let's use the equilibrium equation

[tex] EF_x = A_x + F_H CosΦ = 0 [/tex]

[tex] EF_y = A_y + F_H CosΦ =0 [/tex]

[tex] -(800Kg)(9.81m/s^2) [/tex]

[tex] EM_A = F_H sinΦ(3m) - F_H cosΦ (1.4m) [/tex]

[tex] -(200kg)(9.81m/s^2)(2m) [/tex]

[tex] -(800kg)(9.81m/s^2)(7m) =0 [/tex]

[tex] Therefore A_x = -12.8KN [/tex]

[tex] A_y = -15.8KN [/tex]

[tex] F_H = 28.6KN [/tex]

The magnitude of the force exerted on the arm by the hydraulic cylinder BC = 28.6KN

b) We now have

[tex] A_x = -12.8KN A_y = -15.8KN FH = 28.6KN [/tex]

[tex] |A| = (sqrt(-12.8KN^2)+(-15.8KN^2)) = 20.3 [/tex]

Therefore |A| = 20.3KN

Note 'E'represents summation

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