Question

Due Date: 02.04.2020 STATICS (CE102) Department of Civil Engineering Lecturer : Asst. Prof. Umit N. ARIBAS Question 1: The lift sling is used to hoist a container having a mass of 500 kg. Determine the force in each of the cables AB and AC as a function of e. If the maximum tension allowed in each cable in 5 kN, determine the shortest length of cables AB and AC that can be used for the lin. The center of gravity of the container is located at G.

Answer 1

page-1 1.6 TC VW - 5ookg = 5KN This can be solved balancing i & Fverticalo 8 F = 5KN Now at joint A af5KN FAB COSO = FAC Coso FAB= FAC - NOW & Fy=0 tna ogo 2 FAB Sino = 5kN FAB - FAC= 2.5 Coseco. Forus as a function of o VISION

Page - 2 pant -2 ab Callowed) = 5KN 5 = 2.5 coseco sino = = 0.5 30° P LAB = Seco 15 1.5 LAB - Coso 1.732 m 1.5 Cosso" are summary is allowed max tension SKN then the length of cables (ABO AC) that are 1.732 m .. en cables (shortest can be exed P18 upvote if you are satisfied. vision