Math  /  Trigonometry

Question47. (II) An object is hanging by a string from your rearview mirror. While you are accelerating at a constant rate from rest to 28 m/s28 \mathrm{~m} / \mathrm{s} in 5.0 s , what angle θ\theta does the string make with the vertical? See Fig. 4-46.

Studdy Solution
Use trigonometry to find the angle θ\theta:
The horizontal force is ma=Tsin(θ) ma = T \sin(\theta) and the vertical force is mg=Tcos(θ) mg = T \cos(\theta) .
Taking the ratio of these forces gives:
tan(θ)=mamg=ag \tan(\theta) = \frac{ma}{mg} = \frac{a}{g}
Substitute the known values (a=5.6m/s2 a = 5.6 \, \text{m/s}^2 and g=9.8m/s2 g = 9.8 \, \text{m/s}^2 ):
tan(θ)=5.69.8 \tan(\theta) = \frac{5.6}{9.8}
Calculate θ\theta:
θ=tan1(5.69.8)29.05 \theta = \tan^{-1}\left(\frac{5.6}{9.8}\right) \approx 29.05^\circ
The angle θ\theta that the string makes with the vertical is approximately:
29.05 \boxed{29.05^\circ}

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