Question #283064

Two forces F1 and F2 act on a particle. F1 has magnitude 6N and in direction 45°, and F2 has magnitude 7N and in direction 154°. Find the magnitude and direction of their resultant

1
Expert's answer
2021-12-27T13:02:37-0500

The resultant force

F=F1+F2{\bf F}={\bf F}_1+{\bf F}_2

In projections

Fx=F1x+F2x=F1cosθ1+F2cosθ2Fy=F1y+F2y=F1sinθ1+F2sinθ2F_x=F_{1x}+F_{2x}=F_1\cos \theta_1+F_2\cos\theta_2\\ F_y=F_{1y}+F_{2y}=F_1\sin \theta_1+F_2\sin\theta_2

Fx=6cos45+7cos154=2.0Fy=6sin45+7sin154=7.3F_x=6\cos 45^\circ+7\cos154^\circ=-2.0\\ F_y=6\sin 45^\circ+7\sin154^\circ=7.3

The magnitude of resultant force

F=Fx2+Fy2=(2.0)2+7.32=7.6F=\sqrt{F_x^2+F_y^2}=\sqrt{(-2.0)^2+7.3^2}=7.6

The direction of resultant force

θ=tan1FyFx=tan17.32.0=105\theta=\tan^{-1}\frac{F_y}{F_x}=\tan^{-1}\frac{7.3}{-2.0}=105^\circ


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