SOLUTION
a) $\vec{R}=\vec{A}+\vec{B}+\vec{C}=$?
We already found the $x$- and $y$-components of these vectors in Exercise 1.31,
$A_x=0 \ \ \ \ A_y= -8.00 \ m$
$B_x=7.50 \ m \ \ \ \ B_y= 13.0 \ m$
$C_x=-10.9 \ m \ \ \ \ C_y= -5.07 \ m$
Then the components of the resultant $\vec{R}$ are,$R_x=A_x+B_x+C_x=-3.40 \ m$
$R_y=A_y+B_y+C_y=-0.070 \ m$
Using the Pythagorean theorem and trigonometry to solve for the magnitude and direction,
$R=\sqrt{(R_x)^2+(R_y)^2}=\sqrt{(-3.4 \ m)^2+(-0.070 \ m)^2}=3.4 \ m$
$tan \ \theta=\frac{R_y}{R_x}$
$\theta=1.18^o$
The direction is $1.18^o$ measured clockwise from the -$x$-axis or $1.18^o$ south of west. Without a diagram we can still identify which quadrant is the angle by looking at the components of a vector. Since $\vec{R}$ has negative $x$- and $y$-components then it is in the third quadrant of an $xy$-plane. |
| Vector diagram of (a) |
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| Vector diagram of (b) |
b) $\vec{S}=\vec{C}-\vec{A}-\vec{B}=$?
The components of the resultant $\vec{S}$ are,
$S_x=C_x-A_x-B_x=-10.9\ m -0-7.50 \ m=-18.4 \ m$
$S_y=C_y-A_y-B_y=-5.07 \ m -(-8.00 \ m)-13.0 \ m= -10.07 \ m$
Using the Pythagorean theorem and trigonometry to solve for the magnitude and direction,
$S=\sqrt{(S_x)^2+(S_y)^2}=\sqrt{(-18.4 \ m)^2+(-10.07 \ m)^2}=21.0 \ m$
$tan \ \theta=\frac{S_y}{S_x}$
$\theta=28.7^o$
The direction is $28.7^o$ measured clockwise from the -$x$-axis or $28.7^o$ south of west.
