Question

M c ㅢ1 points CJ 101P048 GO + keepers use a rope to apply a force FA, as part a of the drawing shows. By itself, however, force F A baby elephant is stuck in a mud hole. To help pull it out, game is insufficient. Therefore, two additional forces Fo and Fc are applied, as in part b of the drawing. Each of these adstional forces has the same magnitude magnitude of the resultant force acting on the elephant in part b of the drawing is k times larger than that in part a The . Find the ratio F 23.0.) GO Tutorial Additional Materials L3 Section 1.8 回

Answer 1

Given body elephant stuck in mud hole to pull it out force F^rightarrow_A applied two additional force F�rightarrow_B and F^rightarrow_C applied to find ratio F/F_A when k = 1 - 60, theta = 23 degree now force F_B and F_C are equal the horizontal component of force F_B = F_C = F cos theta net force applied on the elephant is given F_net K times F_A F_net = K F_A F_net = 1 - 60 F_A .(2) form (1) & (2) 1 - 60 F_A = F_A + 2 F cos theta 1 - 60 F_A - F_A = 2 cos theta 0.60 F_A = 2 cos theta F/F_A = 0.60/2 cos theta \r\n F/FA = 0.60/2 cos 23 degree = 0.325 therefore the ratio F/F_A = 0.325