Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-1

EML 3004C

Problem 4-4 (page 138)

Draw the free body diagram of the beam which supports the 80-kg load and is supported by the pin at A and a cable which wraps around the pulley at D. Explain the significance of each force on the diagram.

Solution:

80(9.81)N is the effect of the cable(the weight of the object) on the beam.T is the effect of the cable on the beam.Ax and Ay are the effect of the pin Support on the beam.

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-2

EML 3004C

Problem 4-8 (page 139)

Draw the free-body diagram of the winch, which consists of a drum radius 4 in. It is pin-connected at it center C, and at its outer rim is a ratchet gear having a mean radius of 6 in. The pawl AB serves as a two-force member (short link) and holds the drum for rotating.

Solution:

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-3

EML 3004C

Problem 4-23 (page 155)

The uniform rod AB has a weight of 15 lb. Determine the force in the cable when the rod is in the position shown.

Solution:

M

a

0 NB 5sin 40deg( ) 15 2.5cos 40deg( )([ 0 NB 8.938lb

Fx 0 T cos 10deg( ) 8.938 0 T 9.08lb

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-4

EML 3004C

Problem 4-32 (page 156)

Determine the resultant normal force acting on each set of the wheels of the airplane. There is a set of wheels in the front, A, and a set of wheels under each wing, B. Both wings have total weight of 50 kip and center of gravity at Gw, the fuselage has a weight of 180 kip and center of gravity at Gf, and both engines(one on each side) have a weight 22 kip and center of gravity at Ge.

Solution:

M

a

0 2 NB 40( ) 18000031( ) 2200038( ) 5000040( ) 0

NB 105200lb

Fy 0NA 2 105200( ) 180000 22000 50000 0

NA 41600lb

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-5

EML 3004C

Problem 4-57 (page 172)

The triangular plate is supported by a ball-and-socket joint at B and rollers at A and C. Determine the x, y, z components of each reaction at these supports due to the loading shown.

Solution:

Mx 0Cx 0.6( ) 800 0.3( ) 400 0.3( ) 0 Cx 600 N

My 0 Bx 0.8( ) 600 0.4( ) 400 0.4( ) 0 Bx 500 N

Fx 0Az 600 500 800 400 600 0 Az 700 N

Fx 0 Bx 0 Fy 0 By 0

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-6

EML 3004C

Problem 4-70 (page 174)

Cable BC and DE can support a max. tension fo 700 lb before it breaks. Determine the greatest weight W that can be suspended from the end of the boom. Also, determine the x, y, z components of reaction at the ball-and-socket joint A.

Solution:Assuming failure at cable BC

TBC 7002i 3j 6k

22

3( )2 6

2

TBC 200i 300j 600k( ) lb

TDE TDE3i 6j 2k

3( )2

6( )2 2

2

TDE3

7TDE i

6

7TDE j

2

7TDE k

FA Ax i Ay j Az k W W k( )lb

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-7

EML 3004C

Problem 4-70 (continued)

Force Summation

F 0

200i 300j 600k( )3

7TDE i

6

7TDE j

2

7TDE k

Ax i Ay j Az k 0

2003

7TDE Az

i 300

6

7TDE Ay

j 6002

7TDE Az W

0

Equilizing i, j, k components

Fx 0 2003

7TDE Ax 0

Fy 0 3006

7TDE Ay 0

Fz 0 6002

7TDE Az W 0

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-8

EML 3004C

Problem 4-70 (continued)

Ay 500lbAx 100 lbAz 392 lbW 275lb

so assumption OK233.33 700 lbTDE 233.33lb

Solving Equations

18

7TDE 600 0Mz 0

180012

7TDE 8W 0Mx 0

Equating i and k components

180012

7TDE 8W

i

18

7TDE 600

k 0

3j( )3

7TDE i

6

7TDE j

2

7TDE k

8j( ) W k( ) 0Ma 0

Moment Summation

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-9

EML 3004C

Problem 4-78 (page 189)

The car has a weight of 4000 lb and a center of gravity at G. If it pulls off the side of a road, determine the greatest angle of tilt, , it can have without slipping or tipping over. The coeffiecient of static friction between its wheels and the ground is 0.4.

Solution:Equilibrium

Fx. 0 4000sin Fa Fb 0

Fy. 0 NA NB 4000cos 0 Assuming Slipping occurs. Therefore:

Ma 0 NB 8( ) 40002 sin 40004 cos 0 FA 0.4NA FB 0.4NB

Substituting

NA 1484.6lb NB 2228.3lb

21.8deg

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-10

EML 3004C

Problem 4-85 (page 190)

The two stone blocks have weights of Wa=600lb and Wb=500lb. Determine the smallest horizontal force P that must be applied to block A in order to move it. The coeffeicient of the static friction between the blocks is 0.3 and between the floor and each block is 0.4.Solution:

Case I: Block A and Block B move together:

Fx 0 N 600 500( ) 0 N 1100 lb

Fy 0 0 0.5 1100( ) P P 550lb

Case II: Only block A moves

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-11

EML 3004C

Problem 4-85 continuedFy 0

Case II: Only block A moves

Fx. 0 N 600cos 20 deg( ) P cos 70 deg( ) 0

P sin 70 deg( ) 600sin 20 deg( ) 0.3N 0Fy. 0

Solving

P 447.2lb N 716.8lb

Choose the smalles P among the two cases:

P 447 lb

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-12

EML 3004C

Problem 4-110 (page 194)

Determine the angle at which the applied force P should act on the log so that the magnitude of P is as small as possible for pulling the log up the incline. What is the corresponding value of P? The log weighs W and the slope is known. Express the answer in terms of the angle of kinetic friction, = atan ().

Solution:

Fx. 0 N P sin W cos 0 N W cos P sin

Fy. 0 P cos W sin tan W cos P sin 0

PW sin tan cos

cos tan sin

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-13

EML 3004C

Problem 4-110 (continued)

PW cos sin sin cos

cos cos sin sin

PW sin

cos

Pd

d

W sin sin

cos2

W sin sin 0 W sin 0

sin 0 0 0

PW sin

cos P W sin