Given:
A x = 7.0 , A y = 0 A_x=7.0,\quad A_y=0 A x = 7.0 , A y = 0
B x = 6 cos 3 0 ∘ = 5.2 , B y = 6 sin 3 0 ∘ = 3.0 B_x=6\cos 30^{\circ}=5.2,\quad B_y=6\sin 30^{\circ}=3.0 B x = 6 cos 3 0 ∘ = 5.2 , B y = 6 sin 3 0 ∘ = 3.0
C x = − 4 cos 4 5 ∘ = − 2.8 , C y = 4 sin 4 5 ∘ = 2.8 C_x=-4\cos 45^{\circ}=-2.8,\quad C_y=4\sin 45^{\circ}=2.8 C x = − 4 cos 4 5 ∘ = − 2.8 , C y = 4 sin 4 5 ∘ = 2.8
The resultant vector have compotents
D x = A x + B x + C x = 7.0 + 5.2 − 2.8 = 9.4 D y = A y + B y + C y = 0 + 3.0 + 2.8 = 5.8 D_x=A_x+B_x+C_x=7.0+5.2-2.8=9.4\\
D_y=A_y+B_y+C_y=0+3.0+2.8=5.8 D x = A x + B x + C x = 7.0 + 5.2 − 2.8 = 9.4 D y = A y + B y + C y = 0 + 3.0 + 2.8 = 5.8 The magnitude
R = R x 2 + R y 2 = 9. 5 2 + 5. 8 2 = 11 R=\sqrt{R_x^2+R_y^2}=\sqrt{9.5^2+5.8^2}=11 R = R x 2 + R y 2 = 9. 5 2 + 5. 8 2 = 11 Angle
θ = tan − 1 R y R x = tan − 1 5.8 9.4 = 3 2 ∘ N o f E \theta=\tan^{-1}\frac{R_y}{R_x}=\tan^{-1}\frac{5.8}{9.4}=32^{\circ}\quad \rm N\:of\:E θ = tan − 1 R x R y = tan − 1 9.4 5.8 = 3 2 ∘ N of E
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