Question

Chapter 03, Problem 017 rhree vectors a, b, and T, each have a magnitude of 45.0 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 25.0 (-180° 180°) and (c) the magnitude and (d) the angle of a b + ·in the range of (-180e, i 80e)? what are (e) the magnitude and (f) the angle (in the enaciean T 192 , and 313,respectively. What are (a) the magnitude and (b) the angle of the vector (relative to the tx direction in the range of range of (-1809, 180°)) of a fourth vector d such that a-+d) o?

Answer 1

a)

$\overrightarrow{a}=45Cos25\widehat{i}+45Sin25\widehat{j}$

$\overrightarrow{b}=45Cos192\widehat{i}+45Sin192\widehat{j}$

$\overrightarrow{c}=45Cos313\widehat{i}+45Sin313\widehat{j}$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(27.457)\widehat{i}+(-23.249)\widehat{j}$

Magnitude

$\left | \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right |=\sqrt{27.457^{2}+(-23.249)^{2}}=35.98 m$

b)

$\theta =tan^{-1}\left ( \frac{-23.249}{27.457} \right )=-40.26^{o}$

c)

$\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}=115.49\widehat{i}+(-4.537)\widehat{j}$

Magnitude

$\left | \overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c} \right |=\sqrt{115.49^{2}+(-4.537)^{2}}=115.58 m$

d)

direction

$\theta =tan^{-1}\left ( \frac{-4.537}{115.49} \right )=-2.25^{o}$

e)

$(\overrightarrow{a}+\overrightarrow{b})-\left ( \overrightarrow{c}+\overrightarrow{d} \right )=0$

$\overrightarrow{d}=\overrightarrow{a}+\overrightarrow{b}-\lef\overrightarrow{c} \right )$

$\overrightarrow{d}=\left ( -33.923 \right )\widehat{i}+42.573\widehat{j}$ m

Magnitude

$\left | \overrightarrow{d} \right |=\sqrt{(-33.923)^{2}+(42.573)^{2}}=54.435 m$

f)

$\theta =tan^{-1}\left ( \frac{42.573}{-33.923} \right )=180-51.45=128.55^{o}$