Asked by gian
                Two forces of 3N and 2N are acting at an angle of 45 to each other. Find the magnitude and direction of the resultant force?
            
            
        Answers
                    Answered by
            Damon
            
    3 in x direction
2 /sqrt 2 in x and 2/sqrt2 in y
x force = 3 + 2/sqrt2
y force = 2/sqrt 2
}F} = sqrt [ (3 + 2/sqrt2)^2 + 9 ]
tan angle above x axis = [ (2/sqt2)/(3 +2/sqrt2)]
    
2 /sqrt 2 in x and 2/sqrt2 in y
x force = 3 + 2/sqrt2
y force = 2/sqrt 2
}F} = sqrt [ (3 + 2/sqrt2)^2 + 9 ]
tan angle above x axis = [ (2/sqt2)/(3 +2/sqrt2)]
                    Answered by
            Reiny
            
    or
complete the parallelogram
r^2 = 3^2 + 2^2 - 2(3)(2)cos135°
= 13 - 12(-1/√2)
r = appr 4.6 N
or
direction vector = (3,0) + (2cos45, 2sin45)
= (3,0) + (1.414, 1.414) = (4.414 , 1.414)
magnitude = √(4.414^2 + 1.414^2) = appr 4.6
    
complete the parallelogram
r^2 = 3^2 + 2^2 - 2(3)(2)cos135°
= 13 - 12(-1/√2)
r = appr 4.6 N
or
direction vector = (3,0) + (2cos45, 2sin45)
= (3,0) + (1.414, 1.414) = (4.414 , 1.414)
magnitude = √(4.414^2 + 1.414^2) = appr 4.6
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