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.

Asked by  | 20th Jun, 2013, 02:44: PM

Expert Answer:

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)

Answered by  | 23rd Jun, 2013, 06:15: AM

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