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 = 35.0o .
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 45o line in the first or third quadrants the two components will be exactly equal. If the vector lies along the 45o 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
P1 = (2.00, - 4.00) mP1's distance from the origin, or its radius r1, is
r1 = SQRT [ (2.00)2 + ( - 4.00 )2 ] m = SQRT [ 4 + 16 ] m = SQRT [ 20 ] mr1 = 4.47 m
tan [1 ] = opp/adj = y1 / x1 = ( - 4) / 2 = - 2
1 = - 63.4o
The cartesian coordinates (r, ) for point P1, are
P1 = (4.47 m, - 63.4o)Now, the same thing for point P2,
P2 = (- 3.00, 3.00) mP2's distance from the origin, or its radius r2, is
r2 = SQRT [ ( - 3.00)2 + ( 3.00 )2 ] m = SQRT [ 9 + 9 ] m = SQRT [ 18 ] mr2 = 4.24 m
tan [2 ] = opp/adj = y2 / x2 = 3 / ( - 3) = - 1
2 = 135o
The cartesian coordinates (r, ) for point P2, are
P2 = (4.24 m, 135o)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 - 45o. An angle of - 45o does, indeed, have a tangent of - 1. A point located at ( + 3, - 3) is located at an angle of - 45o (measured from the + x-axis). But our point, P2, is located at ( - 3, + 3). So, from a diagram, we conclude that it is located at an angle of 135o.
x = r cos = (100 m) cos 30o = (100 m) ( 0.866)x = 86.6 m
y = r sin = - (100 m) sin 30o = - (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 = F1 + F2 After a good diagram most vector addition problems begin with finding the components of the vectors.
F1x = F1 cos 60o = (120 N) ( 0.50) = 60 N F1y = F1 sin 60o = (120 N) ( 0.866) = 104 N
F1 = 60 N i + 104 N j
F2x = - F2 cos 75o = - (80 N) ( 0.260) = - 20.8 N
F2y = F2 sin 75o = (80 N) ( 0.966) = 77.3 N
F2 = - 20.8 N i + 77.3 N j
R = F1 + F2
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 [ Rx2 + Ry2 ] R = SQRT [ 39.22 + 181.32 ] N R = 186.5 N
Notice from the diagram that we are now measuring the angle from the positive x-axis; therefore,
tan = opp/adj = Ry / Rx = 181.3 / 39.2 = 4.65 = 78o
Go 75 paces at 240o,
turn to 135o and walk 125 paces,
then travel 100 paces at 160o.
Determine the resultant displacement from the starting point.
Each piece of these directions is a displacement vectorA: Go 75 paces at 240o Ax = A cos = (75 paces) cos 240o = (75 paces) ( - 0.5) = - 37.5 pacesAy = A sin = (75 paces) sin 240o = (75 paces) ( - 0.866) = - 64.95 paces
That is,
A = - 37.5 i - 64.95 jB: turn to 135o and walk 125 paces
Bx = B cos = (125 paces) cos 135o = (125 paces) ( - 0.707) = - 88.39 pacesBy = B sin = (125 paces) sin 135o = (125 paces) (0.707) = 88.39 paces
That is,
B = - 88..39 i + 88.39 jC: travel 100 paces at 160o
Cx = C cos = (100 paces) cos 160o = (100 paces) ( - 0.940) = - 93.97 pacesCy = C sin = (100 paces) sin 160o = (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
Rx = Ax + Bx + Cxand
Ry = Ay + By + CyUsing numerical values for these, we have
Rx = Ax + Bx + CxRx = ( - 37.50 - 88.39 - 93.97 ) paces
Rx = - 219.86 paces
and
Ry = Ay + By + CyRy = ( - 64.95 + 88.39 + 34.20 ) paces
Ry = 57.64 paces
So we expect the buried treasure to be located at
(X, Y) = (Rx, Ry) = ( - 219.86, 57.64 ) pacesOr, we can find this displacement in polar coordinates,
R = SQRT [ X2 + Y2] = SQRT [ ( - 219.86 )2 + (57.64)2 ] pacesR = 227.29 paces
tan = opp / adj = Y / X = 57 / ( - 220) = - 0.26
= 165.5o
So we can state this resultant as
R = ( R, ) = (227.3 paces, 165.5o )
(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 30.0o 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 30o i + (350 km) cos 30o 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 [ Rx2 + Ry2 ] R = SQRT [ 1252 + 4532 ] km
R = 470 km
tan = opp/adj = Rx / Ry = 125 / 453 = 0.276
= 15o
R = ( 470 km, 15o )
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 unit-vector notation) and (b) the magnitude and direction of the resultant vector (ie, in polar coordinates)
First, resolve the three vectors into their x- and y-components.
Bx = B cos 45o Bx = (40)(0.707) Bx = 28.28 Cx = C cos 45o Cx = (30)(0.707) Cx = 21.21 Rx = 0 + 28.28 + 21.21 Rx = 49.49 |
By = B sin 45o By = (40)(0.707) By = 28.28 Cy = - C sin 45o Cy = - (30)(0.707) Cy = - 21.21
Ry = 20.0 + 28.28 - 21.21 Ry = 27.07 |
R = SQRT [ Rx2 + Ry2 ]
R = SQRT [ (49.49)2 + (27.07)2 ]
R = 56.4
tan = opp/adj = Ry/Rx
tan = 27.07/49.49 = 0.547
= 28.7o
[Remember: "Displacement" is a vector so the answer is a magnitude and a direction. ]
We may as well label the vectors D11, D2, D3, and D4:
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D3x = - (150 m) cos 30o
D3x = - (150 m) (0.866) D3x = - 130 m
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D3y = - (150 m) sin 30o
D3x = - (150 m) (0.500) D3x = - 75 m
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D4x = - (200 m) cos 60o
D4x = - (200 m) (0.500) D4x = - 100 m
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D4y = - (200 m) sin 60o
D4y = - (200 m) (0.866) D4y = - 173.2 m
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Rx = - 130 m |
Ry = - 201.8 |
R = SQRT [ Rx2 + Ry2 ]
R = SQRT [ (130)2 + (201.8)2 ] m
R = 240 m
tan = Ry/Rx = - 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, 57.2o 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 R1. What is the magnitude of this vector?
(b) Obtain a vector expression for the body diagonal vector R2.
Note that R1, c k, and R2 make a right triangle, and prove that the magnitude of R2 is SQRT( a2 + b2 + c2 ).
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R1 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,
R2 is the hypotenuse of a right triangle in the plane containing R1 and c k (or the z-axis) -- or the diagonal of the rectangle in that plane. The sides are R1 (along R1) and c (along the z-axis). 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 35o = (2.50 m) ( 0.819)x = 2.05 m
y = r sin = (2.50 m) sin 35o = (2.50 m) (0.574)
y = 1.43 m
( x, y ) = (2.05 m, 1.43 m)
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