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Note for PHYS 1321 with Professor Hor at UH chapter 3


Note for PHYS 1321 with Professor Hor at UH chapter 3

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This 10 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 24 views.

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Date Created: 02/06/15
Chapter 3 Concept Question 2 The two vectors shown in Fig 321 lie in an xy plane What are the signs of the x and y components respectively of aEZ wE d 2 oiE gtlt positive y positive gtlt positive y negative gtlt negative y positive AwNi A gtlt negative y negative Solution 3 3 bl4lcl 1 Chapter 3 Concept Question 5 Which of the arrangements of axes in Fig 323 can be labeled righthanded coordinate system As usual each axis label indicates the positive side of the axis I Chapter 3 Concept Question 9 x x i39 3 1 J a b C x z x Z y ll 3 z d P 1 Solution abcdf If F q gtlt and G is perpendicular to E then what is the direction of E in the four situations shown in Fig 324 T l39 1 o l 4 2 HI a situation 1 q positive b situation 1 q negative c situation 2 q positive d situation 2 q negative e situation 3 q positive f situation 3 q negative Solution ax bX cz dz ez fz Chapter 3 Problem 11 a In unitvector notation what is the sum of 21 40 111 30 111 and E 130 111 70 111 What are b the magnitude and c the direction of EE relative to i Solution a The x and the y components of 7 are rx ax bx 40 m 713 m 790 m and ry ay by 30 m 70 m 10 m respectively Thus quot 9 10 b The magnitude of 7 is r i fi lr 12 1l 90 m2 10 m2 13 m c The angle between the resultant and the x axis is given by 0tan 1 ri tan 1 48 or 132 r 90m x Since the x component of the resultant is negative and the y component is positive characteristic of the second quadrant we nd the angle is 1320 measured counterclockwise from x axis The addition of the two vectors is depicted in the gure below not to scale Indeed we expect 7 to be in the second quadrant Chapter 3 Problem 25 Oasis B is a distance d 25 km east of oasis A along the X axis shown in the figure A confused camel intending to walk directly from A to B instead walks a distance W1 24 km west of due south by angle 91 150 It then walks a distance W2 8 km due north If it is to then walk directly to B a how far in km and b in what direction should it walk relative to the positive direction of the X axis l y I 5 Solution a The strategy is to nd where the camel is C by adding the two consecutive displacements described in the problem and then nding the difference between that 9 location and the oasis B Using the magnitudeangle notation 6 244 90W404 11225 1 6 254W7142594 which is efficiently implemented using a vectorcapable calculator in polar mode The distance is therefore 3471 km The direction is 25940 north of due east S0 Chapter 3 Problem 32 In the figure a vector at with a magnitude of 170 m is directed at angle 9 56 counterclockwise from the gtlt axis What are the components a ax and b ay of the vector A second coordinate system is inclined by angle 9 18 with respect to the first What are the components c a X and d a y in this primed coordinate system Solution a With a 170 m and 639 560 we nd ax a cos 639 951m b Similarly ay a sin 639 141 m c The angle relative to the new coordinate system is 639 39 560 7 180 380 Thus a acosH39 134 m d Similarly a a sin 6quot 105 In Chapter 3 Problem 43 The three vectors in the figure have magnitudes a 300 m b 400 m and c 100 m and angle 9 300 What are a the X component and b the y component of a c the X component and d the y component of b and e the X component and f the y component of Z If E pZ1qb what are the values of g p and hq Solution From the gure we note that E J 5 which implies that the angle between E and the x axis is 6 90 In unitvector notation the three vectors can be written as Z axi E bibjbcos 6lbsin 6 2 01 Cy c cos690W0 1 The above expressions allow us to evaluate the components of the vectors a The x component of 21 is ac a cos 0 a 300 m b Similarly the ycomponnet of E is ay a sin 0 0 c The x component of b is bx b cos 30 400 m cos 30 346 m d and the ycomponent is by b sin 30 400 m sin 30 200 m e The x component of E is C 0 cos 120 100 m cos 120 7500 m f and the ycomponent is cy 0 sin 30 100 m sin 120 866 m g The fact that E pg qi implies c cxi cyj pai qb byj W qbi qbyj or c pa qu cy qby Substituting the values found above we have 500mp300mq346 m 866 m q 200 m Solving these equations we ndp 467 h Similarly q 433 note that it s easiest to solve for q rst The numbers p and q have no units Chapter 3 Problem 44 In the cross or vector product I quE we know that q 2 F 401 20oj12012 G 20406012 f3 BXlByjBz In I the X and y components have the same value What is the value of BZ Solution Applying Eq 323 F qt X E where q is a scalar becomes in E13E qvsz szy i qvB vB j qvBy vyB 12 which 7 plugging in values 7 leads to three equalities 4024OBZ 6053 202603 203 122203y 403 Since we are told that Bx By the third equation leads to By 730 Inserting this value into the rst equation we nd E 30i 303 40h Thus our answer is Bz 740 Chapter 3 Problem 51 Rock faults are ruptures along which opposite faces of rock have slid past each other In the figure points A and B coincided before the rock in the foreground slid down to the right The net displacement Z3 is along the plane of the fault The horizontal component of is the strike slip AC The component of AB that is directly down the plane of the fault is the dip slip AD a What is the magnitude of the net displacement E if the strikeslip is 220 m and the dipslip is 170 m b If the plane of the fault is inclined at angle q 52 to the horizontal what is the vertical component of AB Sll39lkt ii illl DipslipA l l leIll plum Solution Although we think of this as a threedimensional movement it is rendered effectively twodimensional by referring measurements to its welldefined plane of the fault a The magnitude of the net displacement is lfBl lADl2 lACl2 J170 m2 220 m2 278m a b The magnitude of the vertical component of AB is lADl sin 520O 134 m Chapter 3 Problem 62 A golfer takes three putts to get the ball into the hole The first putt displaces the ball 366 m north the second 183 m southeast and the third 091 m southwest What area the magnitude and b the angle between the direction of the displacement needed to get the ball into the hole on the first putt and the direction due east Solution We choose x east and y north and measure all angles in the standard way positive ones counterclockwise from x negative ones clockwise Thus vector all has magnitude d1 366 with the unit meter and three significant figures assumed and direction 61 90 Also 512 has magnitude d2 183 and direction 62 45 and vector 633 has magnitude d3 091 and direction 63 135 We add the x and ycomponents respectively x d1 cos 61dzcos 02d3cos 03 065 m y all sin 01d2 sin 02d3 sin 0317 m a The magnitude of the direct displacement the vector sum 62 1 3 is 065 m2 17 m2 18 m b The angle understood in the sense described above is tan 1 17065 69 That is the first putt must aim in the direction 69 north of east Chapter 3 Problem 63 Here are three vectors in meters d1 30i30j20k 1 l0i 40j 201 30i50j10k R What results from a ll31i3r b ll Iglxgq and 514 c d and e forl j and k components respectively Give your answers in standard SI units Solution 3 b C Since all 33 0l7103301t we have d139Lg33 30l3l03l lh390l 10330l 0 306030 111 Using Eq 330 we obtain 3 gt 73 10l60i 201 Thus J J 73 30i30201 10i50201 301s4052 1113 We found 11 in part a Use of Eq 330 then leads to d1Klgwd i 30l30jl01gtlt0t l0j301l 11i9oj301ampm3 So i component is 11ml d component is 90m2 e 1 component is 30m2 Chapter 3 Problem 64 Consider two displacements one of magnitude 30 m and another of magnitude 40 m What angle between the directions of this two displacements give a resultant displacement of magnitude a 70 m b 10 m and c 50 m Solution a The vectors should be parallel to achieve a resultant 7 m long the unprimed case shown below b antiparallel in opposite directions to achieve a resultant l m long primed case shown c and perpendicular to achieve a resultant I32 42 5m long the doubleprimed case shown In each sketch the vectors are shown in a headtotail sketch but the resultant is not shown The resultant would be a straight line drawn from beginning to end the beginning is indicated by A with or without primes as the case may be and the end is indicated by B A o gto gt 3 EN 014 0 B An


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