Question

A 60 kg rock climber has positioned himself as shown in the figure (with his hands on one side of a gap and his feet on the other side). Both friction and normal forces act on his hands and feet. The width of the gap, w, is 0.24 m and the climber's center of mass is a horizontal distance d 0.54 m from the gap. The coefficient of static friction = 0.48 between his hands and the rock and 2 1.17 between his boots and the rock. For the minimum pull by the hands and push by the feet that will keep the climber stationary, what must the vertical distance h between hands and feet be? Your answer should be in N, but enter only the numerical part in the box. h com

Answer 1

the net Under laudibrium Condhon teque is zero Errero The net falle along nanil Efn=0 & Fy=0 mg MAN+MON N = mg 60 (9.8) CH+48 0.48+1-17 N = 356.36N mg (d+w)- NelW - Nh=0 Nh- mold+w) - NUH W. = 60 (9.8) (0.54+0.24) - 356.36 (0.48) (0-24) Nh= 417.5869 17.5869 = 1.17 m 356.36 n = 11:17]