Problem Number 2:The Force system shown has a resultant of 320 lbs pointing down along the Y-axis. Compute the values of F and e required to give the resultant.S=420 LbsSy420 IbsSxSxSxe = ?T = 900 Lbs25°F = ?R=320 Ibs こdcy-0 sinoFes -t oos 25Fye-Fsin25-320Ex= 900-4200os 0-Fcos25 =0eye 420 sino Fsims-320900 -420 Cos e f0s 25%3Dcos 25S32> ノ- SocySinas) = -320%3DCOL 25(szusX- レS 81900-920 Cos0) (tan 25s) = -320Etoo) (tan 25) +420) (tan 25)(o) 320426 sine 1420ltanes) (ase)= (900 fan 25) - 30%3D900 fan25-320Sino + Han 25Y Coso) =420七と2 -0 esa0 ク

Rx = 0 -F cos 25° - 420 cos θ + 900 = 0 -0.906 F - 420 cosθ + 900 =0 -0.906 F + 900 =420 cosθ…

Answered: The Force system shown has a resultant…

The Force system shown has a resultant of 320 lbs pointing down along the Y- axis. Compute the values of F and e required to give the resultant.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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