The components of resultant force
F x = F 1 cos θ 1 + F 2 cos θ 2 = 10 cos 3 7 ∘ + 8 cos 33 0 ∘ = 15 N F_x=F_1\cos\theta_1+F_2\cos\theta_2\\
=10\cos 37^{\circ}+8\cos 330^{\circ}=15\:\rm N F x = F 1 cos θ 1 + F 2 cos θ 2 = 10 cos 3 7 ∘ + 8 cos 33 0 ∘ = 15 N
F y = F 1 sin θ 1 + F 2 sin θ 2 = 10 sin 3 7 ∘ + 8 sin 33 0 ∘ = 2.0 N F_y=F_1\sin\theta_1+F_2\sin\theta_2\\
=10\sin 37^{\circ}+8\sin 330^{\circ}=2.0\:\rm N F y = F 1 sin θ 1 + F 2 sin θ 2 = 10 sin 3 7 ∘ + 8 sin 33 0 ∘ = 2.0 N The components of the acceleration
a x = F x / m = 15 / 5 = 3.0 m / s 2 a_x=F_x/m=15/5=3.0\:\rm m/s^2 a x = F x / m = 15/5 = 3.0 m/ s 2
a y = F y / m = 2.0 / 5 = 0.4 m / s 2 a_y=F_y/m=2.0/5=0.4\:\rm m/s^2 a y = F y / m = 2.0/5 = 0.4 m/ s 2 The magnitude of acceleration
a = a x 2 + a y 2 = 3. 0 2 + 0. 4 2 = 3.02 m / s 2 a=\sqrt{a_x^2+a_y^2}=\sqrt{3.0^2+0.4^2}=3.02\:\rm m/s^2 a = a x 2 + a y 2 = 3. 0 2 + 0. 4 2 = 3.02 m/ s 2 Direction:
θ = tan − 1 a y a x = tan − 1 0.4 3.0 = 7. 6 ∘ \theta=\tan^{-1}\frac{a_y}{a_x}=\tan^{-1}\frac{0.4}{3.0}=7.6^{\circ} θ = tan − 1 a x a y = tan − 1 3.0 0.4 = 7. 6 ∘
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