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Unformatted text preview: Problem 2.2 ﬂ —p R, and report
183 mm
655° Add the two vectors shown to form a resultant vector,
your result using polar vector representation. Solution Part (a) Note that a is given by a = 180° — 55° = 125°. With this known, the law of cosines may be used to ﬁnd R, i.e., R 183 mm
R = «(101)2 + (183 )2 — 2(101 mm)(183mm) cos 1250 = 254.7mm. (1) ' 550
Next, the law of Sines may be used to obtain the value of ,6, i.e., 1 mm
R 183mm . _1 183mm , o _ 0
Sin a — Sin'B ﬁ ﬁ — Sln Sln > — . Using these results, we may report the vector R by polar vector representation as —o R = 255 mm @ 36.0°A. (3) Problem 2.3 ﬂ Solution Part (a) Refer to the ﬁgure to the right to determine R and R R = (35 kN)2 + (l8kN)2 = 39.36 kN, ﬂ = tan_1 (35 kN l8kN The angle for R from the horizontal is given by 180° — 6278" = using the polar vector representation, we obtain _. R = 39.4kN @ 117°A. ,5: > = 62.78°. (1) 35 kN 1 172°. Therefore, R A
18 kN (2) IR} mg Problem 2.7 l —. Add the three vectors shown to form a resultant vector, R, and report 4011» son) 8mm '15 mm
your result using polar vector representation. 45% l $20.
8 rum ' (b)
Solution Part (a) Refer to the ﬁgure to the right. Although we are interested in
ﬁnding R, we will begin by ﬁnding P. The magnitude of P is given by P = «(60113)? + (801b)2 = 1001b. (1) The angle a is found by 601b _1 60 lb 0
tana — m => a — tan — 36.87 .
Next, use the law of cosines to ﬁnd R:
P2 + (40 lb)2 — 2P(401b) cos(45° + a) = 102.3 lb. (3)
Use the law of Sines to ﬁnd 7:
R _401b _ , _1 401bsin,6 _ o
sin(45° + a) _ sin'y => 7 _ 8m ( R > — 22.77 ' (4) In polar vector representation, the angle associated with R is given by the sum of a and 7, such that
R = 1021b @ 59.6°A. (5) ropes attached to its bow. Determine the magnitude and orientation 6
of the force F so that the resultant force is in the direction of line a and
the magnitude of F is as small as possible. Solution The known force vectors are sketched ﬁrst, relative Possible Choices for F .
to the horizontal line a; there are many possible
choices of 13 such that the resultant force will lie
parallel to line a. The smallest value of F occurs
when F“ is perpendicular to line a, i.e., when i ‘ I " 0.
2 kN sin 30°
m
_‘ 3 kN sin 60°
The magnitude of F may also be found by consid—
ering the force polygon, such that
E = (3 kN) sin 60° 7 (2 kN) sin 30° = 1.60 kN. (2) ...
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This note was uploaded on 04/27/2009 for the course ENGR 111 taught by Professor Walker during the Spring '07 term at Texas A&M.
 Spring '07
 walker

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