Example1- determining the components of the specified forces inthe direction of the x and y axesnumerically+ yF1x = F1 cos a, = 400N cos 45 = 282,8NF2x = F2 cos a, = 200N cos 130 = -128,6NF3x = F3 cos oaz = 300N cos 220 = - 229,8NFax = F4 cos a= 350N cos 45 = 0 N%3DF:F1+ XF3= F1 sin a, = 400N sin 45 = 282,8 NF4%3D%3D1yF = F2 sin a, = 200N sin 130 = 153,2 N%3D2yF3y = F3 sin a, = 300N sin 220 = - 192,8 NFay = F4 sin a, = 350N sin 270 = - 350 N%3D-y

Draw the free body diagram of the force system.

Example 1- determining the components of the specified forces in the direction of the x and y axes numerically + y F1x = F1 cos a, = 400N cos 45 = 282,8N Fx = F2 cos oa, = 200N cos 130 = -128,6N F3x = F3 cos a3 = 300N cos 220 = - 229,8N Fax = F4 cos a = 350N cos 45 = 0 N %3D %3D %3D %3D 2X %3D F1 F1 %3D F3 = F1 sin a, = 400N sin 45 = 282,8 N F2y = F2 sin a, = 200N sin 130 = 153,2 N F3y = F3 sin a, = 300N sin 220 = - 192,8 N = F4 sin a, = 350N sin 270 = - 350 N %3D F1 1y %3D %3D %3D %3! 4y -y

Example 1- determining the components of the specified forces in the direction of the x and y axes numerically + y F1x = F1 cos a, = 400N cos 45 = 282,8N Fx = F2 cos oa, = 200N cos 130 = -128,6N F3x = F3 cos a3 = 300N cos 220 = - 229,8N Fax = F4 cos a = 350N cos 45 = 0 N %3D %3D %3D %3D 2X %3D F1 F1 %3D F3 = F1 sin a, = 400N sin 45 = 282,8 N F2y = F2 sin a, = 200N sin 130 = 153,2 N F3y = F3 sin a, = 300N sin 220 = - 192,8 N = F4 sin a, = 350N sin 270 = - 350 N %3D F1 1y %3D %3D %3D %3! 4y -y

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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