2019-10-20T14:09:36-04:00
THE RESULTANT OF TWO FORCES WHICH ARE OF MAGNITUDE 130N AND 90N IS 150N FIND THE ANGLE BETWEEN THE TWO FORCES AND FIND THE ANGLE BETWEEN THE 150N AND 130N FORCE
1
2019-10-23T09:42:29-0400
The law of cosines gives
15 0 2 = 9 0 2 + 13 0 2 − 2 × 90 × 130 cos ( 18 0 ∘ − α ) 150^2=90^2+130^2-2\times 90\times 130\cos(180^{\circ}-\alpha) 15 0 2 = 9 0 2 + 13 0 2 − 2 × 90 × 130 cos ( 18 0 ∘ − α ) So
cos ( 18 0 ∘ − α ) = 0.1068 \cos(180^{\circ}-\alpha)=0.1068 cos ( 18 0 ∘ − α ) = 0.1068 Therefore
α = 9 6 ∘ \alpha=96^{\circ} α = 9 6 ∘ The sine rule gives
150 sin ( 18 0 ∘ − α ) = 90 sin ( θ ) \frac{150}{\sin(180^\circ-\alpha)}=\frac{90}{\sin(\theta)} sin ( 18 0 ∘ − α ) 150 = sin ( θ ) 90 So
sin ( θ ) = 90 150 × sin ( 8 4 ∘ ) = 0.5967 \sin(\theta)=\frac{90}{150}\times\sin(84^\circ)=0.5967 sin ( θ ) = 150 90 × sin ( 8 4 ∘ ) = 0.5967 Finally
θ = 3 7 ∘ \theta=37^{\circ} θ = 3 7 ∘
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