Question

A load of lumber of weight W 25 kN is being raised by a mobile crane. The weight of boom ABC and the combined weight of the truck and driver are as shown. Determine the reaction at each of the two (a) front wheels H, (b) rear wheels K. 0.6 m 0.4 m 0.3 m 3 kN 50 kN 2.0 m2.0 m 0.9 m 0.5 m

Answer 1

Given that, Draw the free body diagram of the truck. Consider the force equilibrium equation in the y - direction sigma F_y = 0 z F_w + 2 F_k - w - 3 - 50 = 0 Here, F_H, F_K are the reaction force at each wheel and w is the weight of the substitute 25 Kn for w \r\n 2F_H + 2 F_k - w - 3 - 50 = 0 here, F_H + F_k = 39 -------------(1) consider the moment equilibrium about point k sigma M_k = 0 (w times 5.4) + (3 times 3.4) + (50 times 0.5) - (2 F_H times 2.5) = 0 substitute 25 KN for w (25 times 5.4) + (3 times 3.4) + (50 times 0.5) - (2 F_H times 2.5) = 0 2 F_H = 68.08 F_H = 34.04 KN therefore, the reaction force at wheel H is 34.04 KN substitute 34.04 KN F_H in equation 1 F_H + F_k = 39 F_k = 39 - 34.04