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chapter2.2r
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à 2.2ïResultant Vector Forces
ä Please find the resultant vector forces in the following
êë problems.
#âïTwo forces, F¬ and F½, of magnitude 20 lbs and 30 lbs, act on
#a point at the origin.ïF¬ is up and to the right at a 45° angle, and
#F½ is horizontal and to the right.ïFind the magnitude and direction of
the resultant vector force.ïBy the Law of Cosines, the magnitude of the
resultant vector is 46.35 lbs.ïAlso, by the Law of Sines, the direction
of the resultant vector is 17.77° from the horizontal.ïPlease see DE-
TAILS for an explanation.
éS
#êêêè In the figure, the effect of forces F¬ and F½
êêêïacting on the point at the origin is exactly the
êêêïsame as the effect of the single resultant force,
@fig2201.bmp,25,118
#êêêïR = F¬ + F½.ïGeometrically, the resultant has
êêêïmagnitude and direction equal to that of the main
êêêïdiagonal of the parallelogram in the figure.
êêêëThus, finding the magnitude and direction of
êêêïthe resultant vector is equivalent to finding the
êêêïlength of side "c" and the measure of angle A in
êêêïtriangle ABC of the next figure.ïSide "c" can be
êêêïfound by using the Law of Cosines.
#êêêêïcìï=ï30ì + 20ì - 2∙30∙20∙cos 135°
êêêêêïcï≈ï46.35 lbs
êêêïAngle A can be found by the Law of Sines.
êêêê 20/sin Aè=è46.35/sin 135°
@fig2202.bmp,25,178
êêêêè sin Aè≈è.30512
êêêêê Aè≈è17.77°
êêêïTherefore, the magnitude and direction of the
#êêêïresultant, R = F¬ + F½, is 46.35 lbs at an angle
êêêïof 17.77° from the horizontal.
1ïFind the magnitude of the resultant vector, R, that is the
#sum of the two forces F¬ and F½.ïF¬ has magnitude 12 lbs in a direction
#of 65°, and F½ has a magnitude of 16 lbs in a direction of 33°.
(The angles are drawn counter-clockwise from the x-axis.)
êê A)ï18.96 lbsêêB)ï26.94 lbs
êê C)ï22.4 lbsêê D)ïå of ç
üêêë The angle at C is seen to be 148°.ïThe Law
êêêïof Cosines is used to find the magnitude of R.
#êêêêcìï=ï16ì + 12ì - 2∙16∙12∙cos 148°
êêêêêïcï≈ï26.94 lbs
@fig2203.bmp,100,600
Ç B
2ïFind the direction of the resultant vector, R, that is the
#sum of the two forces F¬ and F½.ïF¬ has magnitude 12 lbs in a direction
#of 65°, and F½ has a magnitude of 16 lbs in a direction of 33°.
(The angles are drawn counter-clockwise from the x-axis.)
êê A)ï46.65°êêèB)ï37.2°
êê C)ï28.9°êêè D)ïå of ç
üêêë In Problem 1, the magnitude of R was found to
êêêïbe 26.94 lbs.ïThe Law of Sines can be used to
êêêïfind angle A.
êêêêë12/sin Aè=è26.94/sin 148°
êêêêê sin Aè≈è.23604
êêêêêë Aï≈è13.65°
@fig2203.bmp,100,600
êêêïThus, the direction is 13.65° + 33° = 46.65°.
Ç A
3ïFind the magnitude of the resultant vector, R, that is the
#sum of the two forces F¬ and F½.ïF¬ has magnitude 2.73 lbs in a direc-
#tion of 156°, and F½ has a magnitude of 5.2 lbs in a direction of 97°.
(The angles are drawn counter-clockwise from the x-axis.)
êê A)ï8.2 lbsêêïB)ï6.35 lbs
êê C)ï7.01 lbsêê D)ïå of ç
üêêë The angle at B is seen to be 121°.ïThe Law
êêêïof Cosines is used to find the magnitude of R.
#êêêïcìï=ï(5.2)ì + (2.73)ì - 2∙(5.2)∙(2.73)∙cos 121°
êêêêêïcï≈ï7.01 lbs
@fig2204.bmp,100,600
Ç C
4ïFind the direction of the resultant vector, R, that is the
#sum of the two forces F¬ and F½.ïF¬ has magnitude 2.73 lbs in a direc-
#tion of 156°, and F½ has a magnitude of 5.2 lbs in a direction of 97°.
(The angles are drawn counter-clockwise from the x-axis.)
êê A)ï116.52°êêïB)ï161.5°
êê C)ï98.3°êêè D)ïå of ç
üêêë In Problem 3, the magnitude of R was found to
êêêïbe 7.01 lbs.ïThe Law of Sines can be used to
êêêïfind angle A.
êêêêè 5.2/sin Aè=è7.01/sin 121°
êêêêê sin Aè≈è.63584
êêêêêë Aï≈è39.48°
@fig2204.bmp,100,600
êêêïThus, the direction is 156° - 39.48° = 116.52°.
Ç A
# 5ïThe resultant, R, of the sum on two forces F¬ and F½ has a
#magnitude of 12 lbs in a direction of 48°.ïF¬ has a magnitude of 6.5
#lbs at an angle of 18°.ïFind the magnitude of F½.
(The angles are drawn counter-clockwise from the x-axis.)
êêïA)ï7.15 lbsêêè B)ï9.6 lbs
êêïC)ï8.24 lbsêêè D) å of ç
üêêë In the figure, angle A of triangle ABC is seen
êêê to be 30°.ïThe Law of Cosines can be used to find
#êêê the magnitude of F½.
#êêêëaìï=ï(6.5)ì + (12)ì - 2∙(6.5)∙(12)∙cos 30°
êêêêêèaè≈è7.15 lbs
@fig2205.bmp,100,600
Ç A
# 6ïThe resultant, R, of the sum on two forces F¬ and F½ has a
#magnitude of 12 lbs in a direction of 48°.ïF¬ has a magnitude of 6.5
#lbs at an angle of 18°.ïFind the direction of F½.
(The angles are drawn counter-clockwise from the x-axis.)
êêïA)ï96.1°êêêB)ï83.08°
êêïC)ï75.06°êêë D) å of ç
üêêë In Problem 5, side "a" was found to be 7.15.
êêê The Law of Sines can be used to find angle C.
êêêêè12/sin Cè=è7.15/sin 30°
êêêêë sin Cè≈è.8392
êêêêêèCè≈è57.06° or 122.94°
@fig2205.bmp,100,600
êêê Thus, the direction is 180°-(122.94°-18°) = 75.06°
Ç C