2 Three displacement vectors of a croquet ball are shown in Figure
3 P3.36, where $|\vect{A}| = 20.0\U{units}$, $|\vect{B}| =
4 40.0\U{units}$, and $|\vect{C}| = 30.0\U{units}$. Find \Part{a} the
5 resultant in unit-vector notation and \Part{b} the magnitude and
6 direction of the resultant displacement.
14 pair a = 20u * dir(90);
15 pair b = 40u * dir(45);
16 pair c = 30u * dir(-45);
18 Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$");
20 Vector B = Vector((0,0), mag=length(b), dir=degrees(b), "$\vect{B}$");
22 Vector C = Vector((0,0), mag=length(c), dir=degrees(c), "$\vect{C}$");
40 pair a = 20u * dir(90);
41 pair b = 40u * dir(45);
42 pair c = 30u * dir(-45);
45 Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$");
47 Vector B = Vector(a, mag=length(b), dir=degrees(b), "$\vect{B}$");
49 Vector C = Vector(a+b, mag=length(c), dir=degrees(c), "$\vect{C}$");
51 Vector R = Vector((0,0), mag=length(r), dir=degrees(r), "$\vect{r}$");
52 R.draw(labelOffset=-r/2);
58 The resultant displacement is
60 \vect{r} &= \vect{A} + \vect{B} + \vect{C} \\
62 + 40.0[\cos(45\dg)\ihat + \sin(45\dg)\jhat]
63 + 30.0[\cos(-45\dg)\ihat + \sin(-45\dg)\jhat]\}\U{units} \\
64 &= (49.5\ihat + 27.1\jhat)\U{units}
69 |\vect{r}| &= \sqrt{\vect{r}_x^2 + \vect{r}_y^2}
70 = \sqrt{49.5^2 + 27.1^2}\U{units}
71 = \ans{56.4\U{units}} \\
72 \theta_\vect{r} &= \arctan\p({\frac{27.1}{49.5}}) = \ans{28.7\dg}