vikas askiitian expert
Last Activity: 13 Years ago
accleration = 4
it makes 160o with +ve x axis so 20o from -ve x axis ...
now resolve accleration into two rectangular components ,
a = 4cos20o (-i) + 4sin20o (j)
net force = ma = 4cos20(-i) + 4sin20(j)
(sin20o similar to sin18o = 0.3 &
cos20o is similar to cos18o = 0.67 )
Fnet = -2.7(i) + 1.2 (j) ...........1
force on particle (Fnet) = F1 + F2
F1 = 2.5(i) + 4.6 (j) so
F2 = -5.2(i) - 3.4 (j) ............2 ans a)
magnitude = (sum of squares of cofficient of x comp & y comp)1/2 = 6.2 (ansc)
unit vector = vector/magnitude
unit vector = -(5.2i + 3.4j)/6.2 = -0.83(i) - 0.55(j) (ans b)
approve if u like my ans