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Find the magnitude and direction of the resultant if forces of 10N 12N 6N act on a particle in the direction due west S30E N30E respectively?
Wiki User
∙ 14y ago
Sorry we can't sketch a pictorial of the question. 1). The 10N force is all in the -x direction. 2). The 12N force has a component in the +x direction = 12sin(30)
and a component in the -y direction = 12cos(30) 3). The 6N force has a component in the +x direction = 6sin(30)
and a component in the +y direction = 6cos(30) The horizontal components are: -10 + 12sin(30) + 6sin(30) = -10 + 18sin(30) = -10 + 9 = -1N The vertical components are: -12cos(30) + 6cos(30) = -6cos(30) = -(6/2)sqrt(3) = -3sqrt(3) = -5.19615N The (magnitude)2 of the resultant = (-1)2 + (3sqrt(3))2 = 1 + 27 = 28
The magnitude of the resultant = sqrt(28) = 5.2915N (rounded) The direction of the resultant = S [tan-1(1/5.19615)] W = S30W (just a little bit rounded)
Wiki User
∙ 14y ago
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