Homework
Ch 3, Vectors
Ch 3; 2, 20, 37, 44, 50, 51, 57, 61
Questions 3, 5, 6, 7, 8
Additional problems from Serway's fourth edition
(4 ed) 3.1 A point is located in a polar coordinate system by the coordinates r = 2.50 m and = 35.0^{o} .
Find the cartesian coordinates of this point, assuming the two coordinate systems have the same origin.
Conceptual Questions
Q3.3 The magnitudes of two vectors A and B are A = 5 units and B = 2 units. Find the largest and smallest values possible for the resultant vector R = A + B.
If vectors A and B point in the same direction, the magnitude of R is 7 units.
If vectors A and B point in the opposite direction, the magnitude of R is 3 units.
Q3.5 If the component of vector A along the direction of vector B is zero, what can you conclude about these two vectors.
The two vectors are perpendicular (it can also be said they are orthogonal).
Q3.6 Can the magnitude of a vector have a negative value?
No, a magnitude is always positive or zero.
Q3.7 Which of the following are vectors and which are not:
force > vector
temperature > scalar
volume > scalar
rating of a television show > scalar
height > vector (a well would have a negative height)
velocity > vector
age > scalar
Q3.8 Under what circumstances would a nonzero vector lying in the xy plane ever have components that are equal in magnitude?
If the vector lies along the 45^{o} line in the first or third quadrants the two components will be exactly equal. If the vector lies along the 45^{o} line in the second or fourth quadrants the two components will be equal in magnitude.
Problems from the current (5th) edition of Serway and Beichner.
3.2 Two points in the xy plane have cartesian coordinates (2.00,  4.00) m and (  3.00, 3.00) m.
Determine
(a) the distance between these points and
We can find the distance between the two points from the Pythagorean Theorem, distance = d = SQRT [ (x)^{2} + (y)^{2} ]d = SQRT [ (  3.00  2.00 )^{2} + ( 3.00  (  4.00) )^{2} ] m
d = SQRT [ (  5 ) ^{2} + ( 7.00) ^{2} ] m
d = SQRT [ 25.00 + 49.00 ] m
d = SQRT [ 74.00 ] m
d = 8.60 m
(b) their polar coordinates
P_{1} = (2.00,  4.00) mP_{1}'s distance from the origin, or its radius r_{1}, is
r_{1} = SQRT [ (2.00)^{2} + (  4.00 )^{2} ] m = SQRT [ 4 + 16 ] m = SQRT [ 20 ] mr_{1} = 4.47 m
tan [_{1} ] = ^{opp}/_{adj} = y_{1} / x_{1} = (  4) / 2 =  2
_{1} =  63.4^{o}
The cartesian coordinates (r, ) for point P_{1}, are
P_{1} = (4.47 m,  63.4^{o})Now, the same thing for point P_{2},
P_{2} = ( 3.00, 3.00) mP_{2}'s distance from the origin, or its radius r_{2}, is
r_{2} = SQRT [ (  3.00)^{2} + ( 3.00 )^{2} ] m = SQRT [ 9 + 9 ] m = SQRT [ 18 ] mr_{2} = 4.24 m
tan [_{2} ] = ^{opp}/_{adj} = y_{2} / x_{2} = 3 / (  3) =  1
_{2} = 135^{o}
The cartesian coordinates (r, ) for point P_{2}, are
P_{2} = (4.24 m, 135^{o})NOTE! Always use caution with the inverse tangent function (and all other inverse trig functions). When you tell your calculator that you want the inverse tangent of (  1) it will probably tell you the angle is  45^{o}. An angle of  45^{o} does, indeed, have a tangent of  1. A point located at ( + 3,  3) is located at an angle of  45^{o} (measured from the + xaxis). But our point, P_{2}, is located at (  3, + 3). So, from a diagram, we conclude that it is located at an angle of 135^{o}.
x = r cos = (100 m) cos 30^{o} = (100 m) ( 0.866)x = 86.6 m
y = r sin =  (100 m) sin 30^{o} =  (100 m) (0.500)
y =  50.0 m
( x, y ) = (86.6 m,  50.0 m)
Find
(a) the single force that is equivalent to the two forces shown, and
(b) the force that a third person would have to exert on the mule to make the resultant force equal to zero.
We want the resultant R,R = F_{1} + F_{2} After a good diagram most vector addition problems begin with finding the components of the vectors.
F_{1x} = F_{1} cos 60^{o} = (120 N) ( 0.50) = 60 N F_{1y} = F_{1} sin 60^{o} = (120 N) ( 0.866) = 104 N
F_{1} = 60 N i + 104 N j
F_{2x} =  F_{2} cos 75^{o} =  (80 N) ( 0.260) =  20.8 N
F_{2y} = F_{2} sin 75^{o} = (80 N) ( 0.966) = 77.3 N
F_{2} =  20.8 N i + 77.3 N j
R = F_{1} + F_{2}
R = (60 N i + 104 N j) + (20.8 N i + 77.3 N j)
R = ( 60  20.8 ) N i + ( 104 + 77.3 ) N j
R = 39.2 N i + 181.3 N j
As before, we now need to find the magnitude of the resultant and its direction,
R = SQRT [ R_{x}^{2} + R_{y}^{2} ] R = SQRT [ 39.2^{2} + 181.3^{2} ] N R = 186.5 N
Notice from the diagram that we are now measuring the angle from the positive xaxis; therefore,
tan = ^{opp}/_{adj} = R_{y} / R_{x} = 181.3 / 39.2 = 4.65 = 78^{o}
Go 75 paces at 240^{o},
turn to 135^{o} and walk 125 paces,
then travel 100 paces at 160^{o}.
Determine the resultant displacement from the starting point.
Each piece of these directions is a displacement vectorA: Go 75 paces at 240^{o}^{} A_{x} = A cos = (75 paces) cos 240^{o} = (75 paces) (  0.5) =  37.5 pacesA_{y} = A sin = (75 paces) sin 240^{o} = (75 paces) (  0.866) =  64.95 paces
That is,
A =  37.5 i  64.95 jB: turn to 135^{o} and walk 125 paces
B_{x} = B cos = (125 paces) cos 135^{o} = (125 paces) (  0.707) =  88.39 pacesB_{y} = B sin = (125 paces) sin 135^{o} = (125 paces) (0.707) = 88.39 paces
That is,
B =  88..39 i + 88.39 jC: travel 100 paces at 160^{o}
^{} C_{x} = C cos = (100 paces) cos 160^{o} = (100 paces) (  0.940) =  93.97 pacesC_{y} = C sin = (100 paces) sin 160^{o} = (100 paces) (0.342) = 34.20 paces
That is,
C =  93.97 i + 34.2 jNow we add these displacement vectors to find the resultant, R
R = A + B + CRemember, tho', that vector notation or vector addition is really elegant shorthand for the two scalar equations
R_{x} = A_{x} + B_{x} + C_{x}and
R_{y} = A_{y} + B_{y} + C_{y}Using numerical values for these, we have
R_{x} = A_{x} + B_{x} + C_{x}R_{x} = (  37.50  88.39  93.97 ) paces
R_{x} =  219.86 paces
and
R_{y} = A_{y} + B_{y} + C_{y}R_{y} = (  64.95 + 88.39 + 34.20 ) paces
R_{y} = 57.64 paces
So we expect the buried treasure to be located at
(X, Y) = (R_{x}, R_{y}) = (  219.86, 57.64 ) pacesOr, we can find this displacement in polar coordinates,
R = SQRT [ X^{2} + Y^{2}] = SQRT [ (  219.86 )^{2} + (57.64)^{2} ] pacesR = 227.29 paces
tan = opp / adj = Y / X = 57 / (  220) =  0.26
= 165.5^{o}
So we can state this resultant as
R = ( R, ) = (227.3 paces, 165.5^{o} )
(a) The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot travel in this direct flight?
(b) How far will the pilot travel in this direct flight?
We can describe each leg of this airplane's path as a vector:The airplane flies 300 km eastthen 350 km 30.0^{o} west of north
and then 150 km north
Now we can add those vectors to find the resultant R,
To carry out this vector addition, we can write vectors A, B, and C in component form. Remember, this time we are given, and will find, angles measured from North (or y). Be careful as you use the trig functions.
A = 300 km i + 0 j B =  (350 km) sin 30^{o }i + (350 km) cos 30^{o }j
B =  (350 km) (0.500) i + (350 km) (0.866) j
B =  175 km i + 303 km j
C = 0 i + 150 km j
R = A + B + C
R = (300 km i + 0 j) + (  175 km i + 303 km j) + (0 i + 150 km j)
R = (300  175 + 0 ) km i + ( 0 + 303 + 150 ) km j
R = 125 km i + 453 km j
Now we want to write this resultant in polar coordinates, finding its length and its direction.
R = SQRT [ R_{x}^{2} + R_{y}^{2} ] R = SQRT [ 125^{2} + 453^{2} ] km
R = 470 km
tan = ^{opp}/_{adj} = R_{x} / R_{y} = 125 / 453 = 0.276
= 15^{o}
^{}
R = ( 470 km, 15^{o} )
3.51 Three vectors are oriented as shown in Figure P3.51, where A = A = 20.0 units, B = B = 40.0 units, and C = c = 30.0 units.
Find (a) the x and y components of the resultant vector (expressed in unitvector notation) and (b) the magnitude and direction of the resultant vector (ie, in polar coordinates)
First, resolve the three vectors into their x and ycomponents.
B_{x} = B cos 45^{o} B_{x} = (40)(0.707) B_{x} = 28.28 Cx = C cos 45^{o} Cx = (30)(0.707) Cx = 21.21 R_{x} = 0 + 28.28 + 21.21 Rx = 49.49 
B_{y} = B sin 45^{o} B_{y} = (40)(0.707) B_{y} = 28.28 Cy =  C sin 45^{o} Cy =  (30)(0.707) Cy =  21.21
Ry = 20.0 + 28.28  21.21 Ry = 27.07 
R = SQRT [ R_{x}^{2} + R_{y}^{2} ]
R = SQRT [ (49.49)^{2} + (27.07)^{2} ]
R = 56.4
tan = ^{opp}/_{adj} = R_{y}/R_{x}
tan = 27.07/49.49 = 0.547
= 28.7^{o}
[Remember: "Displacement" is a vector so the answer is a magnitude and a direction. ]
We may as well label the vectors D1_{1}, D_{2}, D_{3}, and D_{4}:







D_{3x} =  (150 m) cos 30^{o}
D_{3x} =  (150 m) (0.866) D_{3x} =  130 m

D_{3y} =  (150 m) sin 30^{o}
D_{3x} =  (150 m) (0.500) D_{3x} =  75 m


D_{4x} =  (200 m) cos 60^{o}
D_{4x} =  (200 m) (0.500) D_{4x} =  100 m

D_{4y} =  (200 m) sin 60^{o}
D_{4y} =  (200 m) (0.866) D_{4y} =  173.2 m


R_{x} =  130 m 
R_{y} =  201.8 
R = SQRT [ R_{x}^{2} + R_{y}^{2} ]
R = SQRT [ (130)^{2} + (201.8)^{2} ] m
R = 240 m
tan = R_{y}/R_{x} =  201.8 /( 130) = 1.55
According to my calculator, this means
Is that correct?
That depends. Be careful here! The angle (or direction) is, indeed, 57.2^{o} as indicated in the diagram. Normally, tho', we would consider angles counterclockwise as positive, so we would write this as
Never blindly write down an answer. Always be sure you understand what it means. This is very important!
3.61 A rectangular parallelepiped has dimensions a, b, and c, as in Figure P3.61.
(a) Obtain a vecor expression for the face diagonal R_{1}. What is the magnitude of this vector?
(b) Obtain a vector expression for the body diagonal vector R_{2}.
Note that R_{1}, c k, and R_{2} make a right triangle, and prove that the magnitude of R_{2} is SQRT( a^{2} + b^{2} + c^{2} ).

R_{1} is the hypotenuse of a right triangle in the xy plane  or the diagonal of the rectangle in the xy plane. The sides are a (along x) and b (along y). Therefore,
R_{2} is the hypotenuse of a right triangle in the plane containing R_{1} and c k (or the zaxis)  or the diagonal of the rectangle in that plane. The sides are R_{1} (along R_{1}) and c (along the zaxis). Therefore,
Solutions to the additional problems from Serway's fourth edition
Find the cartesian coordinates of this point, assuming the two coordinate systems have the same origin.
x = r cos = (2.50 m) cos 35^{o} = (2.50 m) ( 0.819)x = 2.05 m
y = r sin = (2.50 m) sin 35^{o} = (2.50 m) (0.574)
y = 1.43 m
( x, y ) = (2.05 m, 1.43 m)
 Hmwk, Ch 2  Homework Assignment Page  PHY 1350's Home Page  Hmwk, Ch 4 