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
urntits 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 shoen. 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 TilP.
The two possible sets of coordinates of point B are rm; 2 lllIllﬁ cosTi". 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 HESLT 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 cosl‘l = 0.535, cost}.L = [1.802. and
cosf}; : H.261, and the person at B determines that the
direction cosines of 1‘5}; are cosﬂ. : —D.ﬁ?6, cosfix :
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 1111! = (In — ‘.—t)i + (In — 11” + (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?Tk.
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 ftlb
213 —436 —1C‘9 rm x F =(2180i+153Dj—1T50kj rem N H‘.U.4Iﬂ 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 Emil4T! = 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.
 Spring '08
 mr.

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