Question

I need help with how to solve these simultaneous equations. Could someone please show me the steps to get to the answer.

Sum of forces in the x direction is zero i.e. FCD COs 33.69"-FAD COS 33.69"-FDE COS 33.690-0 54.08xcos 33.69 - FAD coS 33.69 F cos 33.69... .(5) Sum of forces in the y direction is zero i.e FRD +Fcn sin 33.69" + FAD sin 33.69 - FDE sin 33.690 -20-54.08x sin 33.69" + FD sin 33.69-Fg sin 33.690 (6) By solving the equations (5) and (6), we get FAD 18.03 kN and FDE72 kN

Answer 1

we people generally use scientific calculator for calculation of equation.

-54.08xCos33.69^o-F_ADCos33.69^o-F_DECos33.69^o =0 ------(5)

-20-54.08xSin33.69^o+F_ADSin33.69^o-F_DESin33.69^o =0 --------(6)

divide whole term in equation (5) by Cos33.69^o

-54.08-F_AD-F_DE =0

or,F_AD+F_DE = -54.08 -----(5A)

divide whole term in equation (6) by Sin33.69^o

-20/(Sin33.69^o)-54.08+F_AD-F_DE = 0  

Sin33.69^o = 0.5547 ( you have to use scientific calculator for this,there is no other way)

-20/0.5547-54.08+F_AD-F_DE = 0

F_AD-F_DE = 90.1355 ------(6A)

now you can use scientific calculator for solving (5A) and (6A) or,

add (5A) and (6A) F_AD+F_DE +F_AD-F_DE= -54.08+ 90.1355

2F_AD =36.0555

  F_AD = 18.02775KN = 18.03KN

now put this value of F_AD in equation (5A) or (6A),let's put in (5A)

18.03+F_DE = -54.08

F_{DE = -72.11KN}