HW 1 Solutions

# HW 1 Solutions - Problem 2.34 A sun'es-‘or measures the...

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Unformatted text preview: Problem 2.34 A sun-'es-‘or measures the location of point A and determines that rm : 4|]EIi + Sﬂﬂj [111]. He urn-tits to determine the location of a point B so that |1‘._15| : 4a: In and |rm + 11ml : JEDEI In. What are the eartesian coordinates of point 3'.“ Solution: Two possibilities are: The point 3 lies west of point .11: or point B lies east of point A: as shoe-n. The strategy is to detem'Llne the unknossn angles u. ,3. and (Jr. The magnitude of 021 is _ fr 2 r 1 _ .- |ru_.1| _ x. HUD) + lEEID) _ 39—.4. The angle ,8 is detesnsinetl Dj.‘ SUD Fail 2 — = 2. 2 6.3.4:. ‘8 4GB ' ﬂ The angle a is determined from the cosine law: case: = = ﬂgﬁgg 2(3q4.é)(12DD) u = 14.33 The angle it is it = ﬂit: = 49.121 Til-P. The two possible sets of coordinates of point B are rm; 2 lllIllﬁ cos-Ti". T + sin 71?) = 154.6?i+1172.6sj I'_1TL| I'm; = lllIllﬂ cos 49.1 +5i1'L—1P.1}I = 785. Hi + Q07. iij :ij The re'o possibilities lead to HES-LT n1, 11?2.F nsz or Sufi-'85.! n2. QDTJ n25 Proposed roadway Problem 2.81 In Prohlem 2.30, suppose that the coor— dinates of the i'ocket’s position ate unknown. At a given instant, the person at A determines that the direction cosines of 11m are cos-l‘l- = 0.535, cos-t}.L = [1.802. and cos-f}; : H.261, and the person at B determines that the direction cosines of 1‘5}; are cos-ﬂ. : —D.ﬁ?6, cos-fix : 0.?93, and cost“)z = —U.1W.What are the coordinates of the 1'oclret‘s position at that instant. Solution: The vector fromﬁi to B is given by 11-11! = (In — -‘.—t)i + (In — 1-1” + (in — Lt)" Elf rm 2 (5 — o)i + (0— on + (2 — EDIE: 5i + 2k The magnitude of 11.”; is given 133' |r,.ug| = 33(5); + (I 2 = 5.39 km. The unit vector along .43, U_.1_fg,iS given by a” 2 “Equal = nszsi + oj + 0.3.111: km. The unit vector along the Line AR. 1.11 = :‘oseﬁzi +cos£iyj + cosﬁak = U.535i+ USUEJ + U. 26?]; Similarly. the \ector along ER, on =—ﬂ.5?ei+ﬂ.T-‘PS—D.1?T-k. From the diagram in the problem statement. we see that it.“ = 11.”; +13“. ‘L'sing the unit '-.'e:ror5. the wares 11.13 and ['53 can be written on TAR = 0.5351ﬁgi + USOETARj + DQCIFLLRKI and r” = —O.E.Tt5rni+ Elma-“j — El. ITTrnk. Substituting into the ‘L‘Bi‘tﬂf addition 11.”; :11“; +13“ and equating components. we get. in the .t direction. 0.53511.“ = —0.576rn. and in the y direction. 0.302111 2 ll T9313“. Solving. we get dint r.“ = :43? km. Calculating the components. we get r” = rum = D.535(4.239)i+ carnelian + coast—1.43m. Hence. the coordinates of the racket, R. are {2.4). 3.60. 1.2m km. Problem 2.12.5 The two segments of the L—shzrped bar are parallel to the x and z axes. The rope AB exerts a forte of magnitude |F| : EDD 11:: on the he: at A. Determine the cross product rm x F, where 1‘“ is the position vector form point C to point A. Solution: We need to determine the force F in terms of its components. The vector fromA to B is used to deﬁne F. ranzﬂi—élj—kjft F = (5m 11:) m = (500 ﬂow Intel t£(1}‘+|{—4}I‘+(—1}|' F 2(2131' — 436j — 109k) 11: Also we have rm. 2 (4i+ 5k) ft Therefore i j k rm x 1' = 4 I] S =(318Di+ lSEﬂj — 11'5ij ft-lb 213 —4-36 —1C|‘9 rm x F =(2180i+153Dj—1T50kj rem N H‘.U.4I|ﬂ Problem 2.131 The force F : lﬂi — 4j [N]. DEIEI- 3: min»: the cross 1'DE111ISI r x F. P ‘3 (6.1mm Solution: The position wanton: is A [6. 3. III r” = (ﬁ—ﬁ)i+I{D—3)j +I{—1—D)k= Oi— 3j+4}: The aims product: i j k mxf: u —3 4 =i(15)—j(—4Dj+k(3l]j ll] —4 D Emil-4T! = 151+ 401' + 30k IN—m} ...
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## This note was uploaded on 04/29/2008 for the course MAE 212 taught by Professor Mr. during the Spring '08 term at ASU.

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