Question

Determine the magnitude, direction, and point of application of the the resultant force.

z 20 kip 10kip A 112 kip 6 kip B х D 5 Y I ft 2 ft 4 ft

Answer 1

E fz (-20 skips -10 - 12 y y to a - 48k Kips. g 1 48 = A Resoltant force fo Magnitude: 48 kips Dinection: - Isegative z direction) for point of Application! My = (12x5) + 16+7) 102 kip-ft Clockwise Positive) Bute. Mx=lfely a ya 102 48 2.8125 ft y=2.125 st = 116 kip.ft My a (1045) + (1275) +(6x1) 116 a 22 My = | Feli 48 a 2.4167 ff 2 2.417 ft.

FR = 48 kip 9 (44) y Magnitude = 48 kips direction - (negative z-direction) Point of application (22.4167, 2.280) ff (051 ( 22 22 17 ) st.