Question

Q2. The motion of the bucket of the front-end loader shown is controlled by two arms and a linkage that are pin-connected at D. The arms are located symmetrically with respect to the central, vertical, and longitudinal plane of the loader; one arm AFJ and its control cylinder EF are shown. The linkages GH, HDB and its control cylinder BC are located in the plane of symmetry. For the position and loading shown, determine the force exerted (a) by cylinder BC, (b) by cylinder EF. 20 in. 28 in 12 in C 112 in. E 12 in 20 in. 18 in 24 in. 4500 lb 22 in. 10 in

Answer 1

Given that: Let F_BC = and F_EF be the forces in the cylinders BC and EF. A) free body diagram of the bucket Taking moment about J Sigma M_j = 0 4500 (20) = T_GH (2.2) = 0 9000 - 22FGH - C 22 Fgh = 90000 F_GH = 4090.91 IB free body diagram to the member BH Taking moment about D Sigma M_D = 0 F_BC = (20) - F_GH (24) = 0 F_BC (20) - 4090.91 (24) = 0 20 F_BC = 98181.84 F_BC = 4909 .09 Ib therefore the compressive force acting in the members BC is F_BC = 4.91 kpps b) Free body diagram at the mechanism with bucket Now find theta tan theta = 65/1 8 theta = tan^-1 (65/18) = 74.5 degree taking moment about A Sigma M_a = 0 4500 (123) - F_BC (12) - (2 F_EF sin 74.5 degree)(24) = 0