Find the magnitude and direction of the resultant of the following forces impressed upon a particle, 3?2 kgf due NE, 6?2 kgf due SE, ?2 kgf due NW.

F1 = 3?2 kgf due NE 

Writing it into x and y components

F1 = 3?2 kgf (cos45 i + sin45j)

F1 = 3i + 3j kgf

 

Similarly, F2 = 6?2 kgf due SE 

F2 = 6?2 kgf (cos45 i - sin45j)

F2 = 6i - 6j kgf

 

F3 = ?2 kgf due NW

F3 = ?2 kgf (-cos45 i + sin45j)

F3 = -i + j kgf

 

hence, Fres = F1+F2+F3 

Fres = 3i + 3j + 6i - 6j -i + j  = 8i -2j 

 

Hence, magnitude of Fres = sqrt(8^2 + 2^2) = sqrt(68)

direction = tan^-1 (-2/8) = tan^-1(-1/4)