Question #54893

On her way to visit grandmother, Red Riding Hood sat down to rest and placed her 1.2-kg basket of goodies beside her. A wolf came along, spotted the basket, and began to pull on the handle with a force of 6.4 N at an angle of 25 degrees with respect to vertical. Red was not going to let go easily, so she pulled on the handle with a force of 12 N. If the net force on the basket is straight up, at what angle was Red Riding Hood pulling?

Expert's answer

Answer on Question #54893, Physics / Mechanics | Kinematics | Dynamics

On her way to visit grandmother, Red Riding Hood sat down to rest and placed her 1.2-kg basket of goodies beside her. A wolf came along, spotted the basket, and began to pull on the handle with a force of 6.4 N at an angle of 25 degrees with respect to vertical. Red was not going to let go easily, so she pulled on the handle with a force of 12 N. If the net force on the basket is straight up, at what angle was Red Riding Hood pulling?

Solution:

If net force is straight up, then the x components of both forces must cancel each out.

Let R be the net force applied by Red Riding Hood, and W be the net force applied by the wolf such that


Rsin⁡θ=Wsin⁡φR \sin \theta = W \sin \varphi


where


W=6.4 NW = 6.4 \, \mathrm{N}φ=25∘ with vertical\varphi = 25{}^{\circ} \text{ with vertical}R=12 NR = 12 \, \mathrm{N}


Thus,


sin⁡θ=Wsin⁡φR=6.4⋅sin⁡25∘12=0.2254\sin \theta = \frac{W \sin \varphi}{R} = \frac{6.4 \cdot \sin 25{}^{\circ}}{12} = 0.2254θ=sin⁡−10.2254=13.03∘ with vertical\theta = \sin^{-1} 0.2254 = 13.03{}^{\circ} \text{ with vertical}


Answer: 13.03∘13.03{}^{\circ} with vertical

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