problem 2-12 (page 26)
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Problem 2-12 (page 26). The cable exerts a force of 600 N on the frame. Resolve this force into components acting (a) along the x and y axes and (b) along the u and v axes. What is the magnitude of each component?. Solution:. Problem 2-16 (page 26). - PowerPoint PPT PresentationTRANSCRIPT
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-1
EML 3004C
Problem 2-12 (page 26)
The cable exerts a force of 600 N on the frame. Resolve this force into components acting (a) along the x and y axes and (b) along the u and v axes.
What is the magnitude of each component?
Solution:a )
Fx 600cos 75deg( ) N Fx 155.291N
Fy 600sin 75deg( ) N Fy 579.555N
b )
Fu 600cos 45deg( ) N Fu 424.264N
Fv 600sin 45deg( ) N Fv 424.264N
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-2
EML 3004C
Problem 2-16 (page 26)
If the resultant FR of the two forces acting on the jet aircraft is to be directed along the positive x axis and have a magnitude of 10 kN, determine the angle of the cable attached to the truck at B such that the force FB in this cable in a minimum. What is the magnitude of force in each cable when this occurs?
Solution:
90deg 20deg 70 deg
FB 10 sin 20deg( ) 103 N FB 3.42 10
3 N
Fc 10 cos 20deg( ) 103 N Fc 9.397 10
3 N
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-3
EML 3004C
Problem 2-29 (page 35)
Three forces act on the bracket. Determine the magnitude and orientation of F2 so that the resultant force is directed along the positive u axis and has a magnitude of 50 lb.
Solution:
Summation of Forces in the X direction.
1( )50cos 25deg( ) 80 Fs cos 25deg 5
1352
F2 cos 25deg 54.68Summation of Forces in the Y direction.
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-4
EML 3004C
Problem 2-29 continuedSummation of Forces in the X direction.
50 cos 25deg( ) 80 Fs cos 25deg 5
1352
F2 cos 25deg 54.68Summation of Forces in the Y direction.
2( )50 sin 25deg( ) Fs cos 25deg 5
1352
F2 sin 25deg 69.13
Eq 1 and 2 yields:
tan 25deg 1.2642 25deg 128.34
103deg
Substituting into Eqn 1 or 2 yields:
F2 88.1 lb
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-5
EML 3004C
Problem 2-42 (page 47)
The pipe is subjected to the force F which has components acting along the x, y, z axes as shown. If the magnitude of F is 12 kN, and α = 120 deg and γ = 45 deg, determine the magnitudes of its three components.
Solution:2 42
cos2 cos
2 cos2 1
cos2120deg cos
2 cos245deg 1 cos 0.5 plus or minus
From the Figure, cos = + 0.5 0 deg
Fx F cos Fx 12 cos 120 deg Fx 6 kN
Fy F cos Fy 12 cos 60 deg Fx 6 kN
Fz F cos Fz 12 cos 45 deg Fx 8.49 kN
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-6
EML 3004C
Problem 2-48 (page 55)
Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles.
Solution:
Position vector: R = (4 - 0) i +[ - 4 - ( - 2)] j + ( 6 - 3) k = { 4i - 2j + 3k} m
Magnitude : r 42
2( )2 3
2 r 5.39 m
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-7
EML 3004C
Problem 2-48 continued
Coordinate direction angles:
urR
r ur
4i 2j 3k5.385
ur 0.74281i 0.3714j 0.5571k
cos 0.7428 42 deg
cos 0.3714 112 deg
cos 0.5517 56.1 deg
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-8
EML 3004C
Problem 2-59 (page 57)
Express each of the two forces in Cartesian vector form and then determine the magnitude and coordinate direction angles of the resultant force.
Solution:2 59rab 0 4( ) i 8 8( ) j 0 12( )[ ] k
rab 4i 0j 12k( )ft
rAB 12.649ftrAB 4( )2
02 12( )
2
F1v F1
rab
rAB F1v 12
4i 0j 12k12.649
F1v 3.79i 11.38k( ) lb
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-9
EML 3004C
Problem 2.59 continued 1
rac 5 4( )i 8 8( ) j 4 12( )[ ]k
rac 9i 16j 16k( )ft
rAC 24.352ftrAC 9( )2
16( )2 16( )
2
F2v F2
rac
rAC F1v 18
9i 16j 16k24.352
F2v 6.65i 11.8j 11.8k( ) lb
FR F1 F2 FR 3.79i 11.38k( ) 6.65i 11.82j 11.82k( )
FR 10.44 i 11.82j 23.21k( ) lb
Fr 10.442 11.82
2 23.212 FR 28.067 lb FR 28.1 lb
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-10
EML 3004C
Problem 2-59 continued 2
Coordinate direction angles:
ur
FR
Fr ur
10.44i 11.82j 23.21k28.067
ur 0.37i 0.42j 0.82k
cos 0.37 112 deg
cos 0.42 115 deg
cos 0.82 34.2 deg
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-11
EML 3004C
Problem 2-65 (page 59)
The cylindrical vessel is supported by three cables which are concurrent at point D. Express each force which the cables exert on the vessel as a Cartesian vector, and determine the magnitude and coordinate direction angles of the resultant force.
Solution:2 65ra 0 0.75( ) i 0 0( ) j 3 0( ) k
ra 0.75i 0j 3k( )m
rA 3.0923mrA 0.75( )2
02 3( )
2
FAv FA
ra
rA FAv 6
0.75i 0j 3k3.0923
FAv 1.461i 5.82k( ) kN
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-12
EML 3004C
Problem 2-65 continued 1
rc 0 0.75 sin 45 deg( )[ ]i 0 0.75 cos 45 deg( )[ ] j 3 0( )k
rc 0.5303i 0.5303j 3k( )m
rC 3.0923rC 0.5303( )2
0.5303( )2 3( )
2
FCv FC
rc
rC FCv 5
0.5303i 0.5303j 3k3.0923
FCv 0.857i 0.857j 4.85k( ) kN
rb 0 0.75 sin 30 deg( )[ ]i 0 0.75 cos 30 deg( )[ ] j 3 0( )k
rb 0.375i 0.6495j 3k( )m
rB 3.0923rB 0.375( )2
0.6495( )2 3( )
2
FBv FB
rc
rC FBv 8
0.375i 0.6495j 3k3.0923
FBv 0.970i 1.68j 7.76k( ) kN
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-13
EML 3004C
Problem 2-65 continued 2
Resultant Force Vector:
FR FA FB FC
FR 1.45i 5.8208k( ) 0.97i 1.68j 7.76k( ) 0.85i 0.85j 4.85k( )
FR 0.3724i 0.8228j 18.4326k( ) kN
Fr 0.37242
0.8228( )2 18.4326
2 FR 18.4547 lb FR 18.5 lb
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-14
EML 3004C
Problem 2-65 continued 3
Coordinate direction angles:
ur
FR
Fr ur
0.3724i 0.8228j 18.4326k18.4547
ur 0.02018i 0.04458j 0.9988k
cos 0.02018 88.8 deg
cos 0.04458 92.6 deg
cos 0.9988 2.81 deg
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-15
EML 3004C
Problem 2-78 (page 65)
Determine the magnitudes of the projected components of the force F={-80i + 30j + 20k} lb in the direction of the cables AB and AC.
Solution:
uAC4 0( )i 3 0( ) j 0 8( )k
4 0( )2
3 0( )2 0 8( )
2 uAC
4i 3j 8k
89
uAB0 5( )i 0 4( )[ ] j 8 0( )k
0 5( )2
0 4( )[ ]2 8 0( )
2 uAC
5i 4j 8k
105
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-16
EML 3004C
Problem 2-78 continued
FAC F uAC FAC 80 i 30j 20k( )4i 3j 8k
89
FAC80( ) 4( ) 30 3( ) 20 8( )
89lb
FAC 26.5 lb
FBA F uBA FBA 80 i 30j 20k( )5i 4j 8k
105
FBA80( ) 5( ) 30 4( ) 20 8( )
105lb
FAC 26.5 lb
Namas ChandraIntroduction to Mechanical engineering
HibblerChapter 2-17
EML 3004C
Problem 2-95 (page 68)
Express each of the three forces acting on the column in Cartesian vector form and determine the magnitude of the resultant force.
Solution:
F1 140 sin 30deg( ) i 140 cos 30deg( ) j( F1 70i 121j( ) lb
F2 180j( ) lb
F3 125 cos 45deg( ) i 125 sin 45deg( ) j( ) F3 88.4i 88.4j( ) lb
FR F1 F2 F3 FR 70i 121.1j( ) 180j( ) 88.4i 88.4j( )
FR 18.38i 389.63j( ) lb
Magnitude :
FR 18.38( )2
398.63( )2
FR 390 lb