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two forces F1=500 N due east and F2=250 N due north have their common initial point. F2-F1 is and angle is..
let
f = f1 + f2
v= f1 - f2
there fore on adding both we get
f+v = 2 f1
and f can be calculated by vetor sum of addition = 559.07N towards north east
and therefore
v = 724.06 N towards 22.5 degree north
since both are in two diff direction so
the angle between them is 900
so l F2vector+(-F1vector) l = √f12+f22-2f1f2(cos90).......................cos90=0
by puting the value of f1 f2
we get
250√5 N and angle is
Θ=tan-1(500/250) in north of west
Θ=tan-1(2)
MAGNITUDE OF RESULTANT FORCE IS CLEARLY 250√5N AND THE ANGLE BETWEEN THE RESULTANT AND F1 IS tan-1(1/2).
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