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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 systemby the coordinates r = 2.50 m and = 35.0o .
Find the cartesian coordinates of this point, assuming the twocoordinate systems have the same origin.
Conceptual Questions
Q3.3 The magnitudes of two vectors A and B are A = 5units and B = 2 units. Find the largest and smallest valuespossible for the resultant vector R = A + B.
If vectors A and B point in the samedirection, the magnitude of R is 7 units.
If vectors A and B point in the oppositedirection, the magnitude of R is 3 units.
Q3.5 If the component of vector A along the direction of vectorB is zero, what can you conclude about these two vectors.
The two vectors areperpendicular (it can also besaid they are orthogonal).
Q3.6 Can the magnitude of a vector have a negativevalue?
No, a magnitude is always positive or zero.
Q3.7 Which of the following are vectors and which arenot:
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 inthe xy plane ever have components that are equal inmagnitude?
If the vector lies along the 45o line in the first orthird quadrants the two components will be exactly equal. If thevector lies along the 45o line in the second or fourthquadrants 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) m
P1'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 ] m r1 = 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) m
P2'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 ] m r2 = 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.
3.20 Find the horizontal and vertical components of the 100-mdisplacement of a superhero who flies from the top of a tall buildingfollowing the path shown in Figure P3.19 .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)
Turbotax 2016 deluxe torrent download for mac os x. y = - 50.0 m
( x, y ) = (86.6 m, - 50.0 m)
3.37 The helicopter view in Figure P3.37 shows two peoplepulling on a stubborn mule.Find
(a) the single force that is equivalent to the two forcesshown, and
(b) the force that a third person would have to exert onthe mule to make the resultant force equal to zero.
We want the resultant R,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 [ 39.22 + 181.32 ] NR = SQRT [ Rx2 + Ry2 ]
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
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 240oAx = A cos = (75 paces) cos 240o = (75 paces) ( - 0.5) = - 37.5 paces Ay = A sin = (75 paces) sin 240o = (75 paces) ( - 0.866) = - 64.95 paces
That is,
A = - 37.5 i - 64.95 j
B: turn to 135o and walk 125 paces
Bx = B cos = (125 paces) cos 135o = (125 paces) ( - 0.707) = - 88.39 paces By = B sin = (125 paces) sin 135o = (125 paces) (0.707) = 88.39 paces
That is,
B = - 88.39 i + 88.39 j
C: travel 100 paces at 160o
Cx = C cos = (100 paces) cos 160o = (100 paces) ( - 0.940) = - 93.97 paces Adp 7-0
Cy = C sin = (100 paces) sin 160o = (100 paces) (0.342) = 34.20 paces
That is,
C = - 93.97 i + 34.2 j
Now we add these displacement vectors to find the resultant, R
R = A + B + C
Remember, tho', that vector notation or vector addition is really elegant shorthand for the two scalar equations
Rx = Ax + Bx + Cx
7 Divided By 0
and
Ry = Ay + By + Cy
Using 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 ) paces
Or, we can find this displacement in polar coordinates,
R = SQRT [ X2 + Y2] = SQRT [ ( - 219.86 )2 + (57.64)2 ] paces R = 227.29 paces
Radium 3 0 1 – multi network radio player. 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 )3.50 An airplane starting from airport A flies 300 km east,then 350 km 30.0o west of north, and then 150 km north toarrive at airport B. There is no wind on this day.
(a) The next day, another plane flies directly from A to Bin a straight line. In what direction should the pilot travel in thisdirect flight?
(b) How far will the pilot travel in this directflight?
We can describe each leg of this airplane's path as a vector: The airplane flies 300 km east then 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 jB = - (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 + RInstall4j 7 0 11 M Ba No3 2
y2 ]R = SQRT [ 1252 + 4532 ] km
R = 470 km
tan = opp/adj = Rx / Ry = 125 / 453 = 0.276
= 15o
R = ( 470 km, 15o )
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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 (expressedin unit-vector notation) and (b) the magnitude and direction of theresultant vector (ie, in polar coordinates)
First, resolve the three vectors into their x- andy-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
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tan =opp/adj= Ry/Rx
tan =27.07/49.49 = 0.547
=28.7o
3.57 A person going for a walk follows the path shown inFigure P3.57. The total trip consists of four straight-line paths. Atthe end of the walk, what is the person's resultant displacementmeasured from the starting point?[Remember: 'Displacement' is a vector so theanswer is a magnitude and a direction. ]
We may as well label the vectors D11,D2, D3, andD4:
D3x = - (150 m) cos 30o D3x = - (150 m) (0.866) D3x = - 130 m | D3y = - (150 m) sin 30o D3x = - (150 m) (0.500) D3x = - 75 m | |
D4x = - (200 m) cos 60o D4x = - (200 m) (0.500) D4x = - 100 m | D4y = - (200 m) sin 60o D4y = - (200 m) (0.866) D4y = - 173.2 m | |
Rx = - 130 m | Ry = - 201.8 |
R = SQRT [ Rx2 +Ry2 ]
R = SQRT [ (130)2 + (201.8)2 ] m
R = 240 m
R = 240 m
tan =Ry/Rx = - 201.8 /(- 130) = 1.55
According to my calculator, this means
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Is that correct?
Idatabase 4 3 – organize all your information and data. That depends. Be careful here! The angle (or direction) is,indeed, 57.2oas indicated in the diagram.Normally, tho', we would consider angles counterclockwise aspositive, so we would write this as
Never blindly write down an answer. Always be sureyou understand what it means. This is very important!
3.61 A rectangular parallelepiped has dimensions a, b, andc, as in Figure P3.61.
(a) Obtain a vecor expression for the face diagonalR1. What is the magnitude of this vector?
(b) Obtain a vector expression for the body diagonal vectorR2.
Note that R1, c k, andR2 make a right triangle, and prove that themagnitude of R2 is SQRT( a2 +b2 + c2 ).
------------------------
R1 is the hypotenuse of a right triangle in thexy plane -- or the diagonal of the rectangle in the xy plane. Thesides are a (along x) and b (along y). Therefore,
R2 is the hypotenuse of a right triangle in theplane containing R1 and c k (or the z-axis)-- or the diagonal of the rectangle in that plane. The sides areR1 (along R1) and c (along the z-axis).Therefore,
Solutions to the additional problems from Serway's fourth edition
(4 ed) 3.2 A point is located in a polar coordinate system bythe coordinates r = 2.50 m and = 35.0o .Find the cartesian coordinates of this point, assuming the twocoordinate systems have the same origin.
x = r cos = (2.50 m) cos 35o = (2.50 m) ( 0.819) x = 2.05 m
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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)