College Physics by Openstax Chapter 3 Problem 9

Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

Figure 3.24

Solution:

So, we are given the two vectors shown below.

Vectors A and B

If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:

Figure 3.9B: Vectors A and B added graphically

The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below.

Solve for the value of the angle 𝛼 by geometry.

α=66∘+(180∘−112∘)=134∘\alpha = 66^\circ +\left( 180^\circ-112^\circ \right) = 134^\circ

Solve for the magnitude of the resultant using cosine law.

R2=A2+B2−2ABcos⁡αR=A2+B2−2ABcos⁡αR=(27.5 m)2+(30.0 m)2−2(27.5 m)(30.0 m)cos⁡134∘R=52.9380 mR=52.9 m  (Answer)\begin{align*} R^2 & = A^2+B^2-2AB\cos \alpha \\ R & = \sqrt{A^2+B^2-2AB\cos \alpha} \\ R & = \sqrt{\left( 27.5 \ \text{m} \right)^2+\left( 30.0 \ \text{m} \right)^2-2\left( 27.5\ \text{m} \right)\left( 30.0\ \text{m} \right) \cos 134^\circ} \\ R & =52.9380 \ \text{m} \\ R & = 52.9 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Solve for 𝛽 using sine law.

sin⁡βB=sin⁡αRβ=sin⁡−1(Bsin⁡αR)β=sin⁡−1(30.0 msin⁡134∘52.9380 m)β=24.0573∘\begin{align*} \frac{\sin \beta}{B} & = \frac{\sin \alpha}{R} \\ \beta & = \sin ^{-1} \left( \frac{B \sin \alpha }{R} \right) \\ \beta & = \sin ^{-1} \left( \frac{30.0\ \text{m} \sin 134^\circ}{52.9380 \ \text{m}} \right) \\ \beta & = 24.0573^\circ \end{align*}

Finally, solve for 𝜃.

θ=66∘+24.0573∘=90.1∘  (Answer)\theta = 66^\circ+24.0573^\circ = 90.1^\circ \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

The result is in conformity with that in figure 3.24 shown on the question shown above.

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