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find the magnitude of resultant of the following 3 forces acting on a particle.vecF1=20N in eastward direction,vecF2=20N due north east and F3=20N in southward direction

ISHIKA , 7 Years ago
Grade 11
anser 1 Answers
Soumya Ranjan Mohanty

Last Activity: 7 Years ago

Hello,
I posted this answer again here as the question was posted twice.
you can solve this problem by plotting the free body diagram of the particle.
So as there are 3 forces acting on the particle i.e F1=20 N towards east, F2=20 N towards north east and F3=20 N towards south, so first lets find the resultant of F1 and F3 and lets say it F4.
SO F4=sqrt(F1^2+F3^2+2cos 90) = sqrt(20^2+20^2+0)=sqrt(800)=20*sqrt(2) and direction of F4=tan^-1 (20/20) =45 degrees between east and south i.e south east.
 
now les find resultant of F2 and F4. Direction between F2 and F4 is 90 degrees.
SO resultant=sqrt(F2^2+F4^2+2cos 90) = sqrt(400+800+0)=sqrt(1200)=20*sqrt(3)=20*1.732=34.64 N
 
THerefore, resultant of forces acting on the particle is 34.64 N.

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