problem 2-12 (page 26)

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


HibblerChapter 2-2

EML 3004C


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


HibblerChapter 2-3

EML 3004C


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.


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


HibblerChapter 2-5

EML 3004C


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


HibblerChapter 2-6

EML 3004C


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


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


HibblerChapter 2-8

EML 3004C


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


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


HibblerChapter 2-10

EML 3004C

Problem 2-59 continued 2


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


HibblerChapter 2-11

EML 3004C


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


HibblerChapter 2-12

EML 3004C


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


HibblerChapter 2-13

EML 3004C


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


HibblerChapter 2-14

EML 3004C



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


HibblerChapter 2-15

EML 3004C


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


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


HibblerChapter 2-17

EML 3004C


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

problem 2-12 (page 26)

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