Answer to Question #171449 in Classical Mechanics for Jethro Torio

Question

Two vectors, 300 and 400, have a common origin and are 60° apart. Find the magnitude and the direction of the resultant using a.) The cosine law and b.) The i and j unit vectors

Answer 1

a) Let $a=300$, $b=400$ and angle between them $\beta=180^{\circ}-60^{\circ}=120^{\circ}$. Then, according to the cosine law, we get:

$$R^2=a^2+b^2-2abcos\beta,$$ $$R=\sqrt{a^2+b^2-2abcos\beta},$$ $$R=\sqrt{(300)^2+(400)^2-2\cdot300\cdot400\cdot cos120^{\circ}}=608.3$$

We can find the direction as follows:

$$b^2=a^2+R^2-2aRcos\alpha,$$ $$\alpha=cos^{-1}{\dfrac{a^2+R^2-b^2}{2aR}},$$ $$\alpha=cos^{-1}{\dfrac{(300)^2+(608.3)^2-(400)^2}{2\cdot300\cdot608.3}},$$ $$\alpha=34.7^{\circ}.$$ $$\theta=60^{\circ}-\alpha=60^{\circ}-34.7^{\circ}=25.3^{\circ}.$$

b) Let's find $x$ and $y$ components of the resultant:

$$R_x=400i+(300\cdot cos60^{\circ})i=550i,$$ $$R_y=0j+(300\cdot sin60^{\circ})j=260j.$$

Then, we can write the resultant in unit vector notation as follows:

$$R=550i+260j.$$

We can find the magnitude of the resultant from the Pythagorean theorem:

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(550)^2+(260)^2}=608.3$$

We can find the direction from the geometry:

$$\theta=cos^{-1}(\dfrac{R_x}{R})=cos^{-1}(\dfrac{550}{608.3})=25.3^{\circ}.$$