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problem 7: (9% of assignment value) consider the three vectors shown in…

Question

problem 7: (9% of assignment value)
consider the three vectors shown in the figure. they have magnitudes |\vec{a}| = 31, |\vec{b}| = 7.5, and |\vec{c}| = 33.9, and the labeled angles are \theta_a = 40°, \theta_b = 20°, and \theta_c = 15°. note that the figure shows the definitions of the angles, but the arrows in the figure may not be to scale.

  • part (a)

in what quadrant is the vector \vec{a}+\vec{b}+\vec{c}? note: the quadrants are numbered countered clockwise beginning with the upper right quadrant as q1, the upper left as q2, the lower left as q3, and the lower right as q4.
q3 x incorrect!
feedback: is available.
part (b)
what is the magnitude of the vector \vec{r}=\vec{a}+\vec{b}+\vec{c}?
r = 12.04
sin() cos() tan() \pi ( ) 7 8 9 home
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atan() acotan() ln() / * 1 2 3 -
log10() exp() fact() + - 0 end
degrees radians \sqrt{} backspace del clear
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part (c)

Explanation:

Step1: Find x - components of vectors

The x - component of a vector $\vec{V}$ with magnitude $|\vec{V}|$ and angle $\theta$ with the positive x - axis is $V_x=|\vec{V}|\cos\theta$.
$A_x = |\vec{A}|\cos\theta_A=31\cos40^{\circ}\approx31\times0.766 = 23.746$
$B_x=-|\vec{B}|\cos\theta_B=- 7.5\cos20^{\circ}\approx - 7.5\times0.9397=-7.04775$
$C_x=-|\vec{C}|\sin\theta_C=-33.9\sin15^{\circ}\approx-33.9\times0.2588=-8.77332$
$R_x=A_x + B_x+C_x=23.746-7.04775 - 8.77332=7.92493$

Step2: Find y - components of vectors

The y - component of a vector $\vec{V}$ with magnitude $|\vec{V}|$ and angle $\theta$ with the positive x - axis is $V_y = |\vec{V}|\sin\theta$.
$A_y=|\vec{A}|\sin\theta_A=31\sin40^{\circ}\approx31\times0.6428 = 19.9268$
$B_y = |\vec{B}|\sin\theta_B=7.5\sin20^{\circ}\approx7.5\times0.342 = 2.565$
$C_y=-|\vec{C}|\cos\theta_C=-33.9\cos15^{\circ}\approx-33.9\times0.9659=-32.74301$
$R_y=A_y + B_y + C_y=19.9268+2.565-32.74301=-10.25121$

Step3: Determine the quadrant

Since $R_x>0$ and $R_y < 0$, the vector $\vec{R}=\vec{A}+\vec{B}+\vec{C}$ is in the fourth - quadrant (Q4).

Step4: Calculate the magnitude of $\vec{R}$

The magnitude of a vector $\vec{R}=(R_x,R_y)$ is given by $|\vec{R}|=\sqrt{R_x^{2}+R_y^{2}}$.
$|\vec{R}|=\sqrt{(7.92493)^{2}+(-10.25121)^{2}}=\sqrt{62.8044 + 105.0879}=\sqrt{167.8923}\approx12.96$

Answer:

Part (a): Q4
Part (b): $12.96$