Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 05

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The Significance of Each Force:

W is the effect of gravity (weight) on the dumpster.

and are the pin A reactions on the dumpster.

is the hydraulic cylinder BC reaction on the dumpster.

FBC

Ax

Ay

Draw the free-body diagram of the dumpster D of the

truck, which has a weight of 5000 lb and a center of gravity

at G It is supported by a pin at A and a pin-connected

hydraulic cylinder BC (short link) Explain the significance

B A

C

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Draw the free-body diagram of member ABC which is

supported by a smooth collar at A, rocker at B, and short link

CD Explain the significance of each force acting on the

diagram (See Fig 5–7b.)

SOLUTION

The Significance of Each Force:

is the smooth collar reaction on member ABC.

is the rocker support B reaction on member ABC.

is the short link reaction on member ABC.

2.5 kN is the effect of external applied force on member ABC.

is the effect of external applied couple moment on member ABC.

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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 (See Fig 5–7b.)

SOLUTION

T force of cable on beam.

, force of pin on beam

80(9.81)N force of load on beam

Ay

Ax

4 3 5

1.5 m

B A

C E

D

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F ⫽ 8 lb

A B

1.5 ft

2 ft0.2 ft

*5–4.

Draw the free-body diagram of the hand punch, which is

pinned at A and bears down on the smooth surface at B.

SOLUTION

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Draw the free-body diagram of the uniform bar, which has a

mass of 100 kg and a center of mass at G The supports A, B,

and C are smooth.

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SOLUTION

Draw the free-body diagram of the beam,which is pin

supported at A and rests on the smooth incline at B.

B

0.6 ft

0.6 ft

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Draw the free-body diagram of the beam, which is pin

connected at A and rocker-supported at B.

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SOLUTION

, , force of wood on bar

10 lb force of hand on bar

NC

NB

NA

Draw the free-body diagram of the bar, which has a

negligible thickness and smooth points of contact at A, B,

and C Explain the significance of each force on the

diagram (See Fig 5–7b.)

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Draw the free-body diagram of the jib crane AB, which is pin

connected at A and supported by member (link) BC.

SOLUTION

8 kN

3 m0.4 m

C

B

3 4 5

4 m

A

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Determine the horizontal and vertical components of

reaction at the pin A and the reaction of the rocker B on

the beam

SOLUTION

Equations of Equilibrium: From the free-body diagram of the beam, Fig a, N Bcan

be obtained by writing the moment equation of equilibrium about point A.

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Ax = 103.528 N:+ ©Fx = 0; Ax - 400 sin 15° = 0

By = 586.37 = 586 N+ ©MA = 0; By(12) - (400cos 15°)(12) - 600(4) = 0

Determine the magnitude of the reactions on the beam at A

15⬚

600 N

8 m

3 4 5

4 m

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Determine the components of the support reactions at the

fixed support A on the cantilevered beam.

Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig a,

A x , A y , and M Acan be obtained by writing the moment equation of equilibrium about

Ax = 3.46 kN

4 cos30° - Ax = 0

©Fx = 0;

:+

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The 75-kg gate has a center of mass located at G If A

supports only a horizontal force and B can be assumed as a

pin, determine the components of reaction at these supports G

Equations of Equilibrium: From the free-body diagram of the gate, Fig a, B yand

A x can be obtained by writing the force equation of equilibrium along the y axis and

the moment equation of equilibrium about point B.

Ans.

Ans.

Using the result A x = 919.69 N and writing the force equation of equilibrium along

the x axis, we have

By = 735.75 N = 736 N

By - 75(9.81) = 0+ c ©Fy = 0;

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The overhanging beam is supported by a pin at A and

the two-force strut BC Determine the horizontal and

vertical components of reaction at A and the reaction at B

Equations of Equilibrium: Since line BC is a two-force member, it will exert a force F BC

directed along its axis on the beam as shown on the free-body diagram, Fig a From the

free-body diagram, F BCcan be obtained by writing the moment equation of equilibrium

Ax = 3133.33 N = 3.13 kN

3916.67a45b - Ax = 0+

: ©Fx = 0;

FBC = 3916.67 N = 3.92 kN

FBCa35b(2) - 600(1) - 800(4) - 900 = 0a

+ ©MA = 0;

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Determine the horizontal and vertical components of

reaction at the pin at A and the reaction of the roller at B on

Equations of Equilibrium: From the free-body diagram, and can be obtained

by writing the moment equation of equilibrium about point and the force

equation of equilibrium along the axis, respectively

Ans.

Ans.

Using the result and writing the force equation of equilibrium along

the axis, we have

Ans.

Ay = 110.86 lb = 111 lb

Ay - 50 cos30° - 67.56 = 0+ c ©Fy = 0;

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Equations of Equilibrium: Since the roller at offers no resistance to vertical

movement, the vertical component of reaction at support is equal to zero From

the free-body diagram, , , and can be obtained by writing the force

equations of equilibrium along the and axes and the moment equation of

equilibrium about point , respectively

Ans.

Ans.

Ans.

MA = PL2

PaL2b - MA = 0a

+ ©MB = 0;

By = P

By - P = 0+ c ©Fy = 0;

MA

By

Ax

AA

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a

Ans.

FB = 105 N:+ ©Fx = 0; Bx = 0

By = 105.11 N+ c ©Fy = 0; 378.39 - 60(9.81) + 2By = 0

Ay = 378.39 N+©MB = 0; -Ay(1.4) + 60(9.81)(0.9) = 0

If the wheelbarrow and its contents have a mass of 60 kg and

center of mass at G, determine the magnitude of the resultant

force which the man must exert on each of the two handles in

order to hold the wheelbarrow in equilibrium

0.5 m

B G

A

0.9 m

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Determine the tension in the cable and the horizontal and

vertical components of reaction of the pin A The pulley at

D is frictionless and the cylinder weighs 80 lb.

SOLUTION

Equations of Equilibrium: The tension force developed in the cable is the same

throughout the whole cable The force in the cable can be obtained directly by

summing moments about point A.

Ax = 33.4 lb

Ax - 74.583¢ 1

25≤ = 0:+ ©Fx = 0;

T = 74.583 lb = 74.6 lb

T152 + T¢ 2

25≤1102 - 801132 = 0+ ©MA = 0;

B A

D

C

5 ft 5 ft

2 1

3 ft

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The shelf supports the electric motor which has a mass of

15 kg and mass center at The platform upon which it

rests has a mass of 4 kg and mass center at Assuming that

a single bolt B holds the shelf up and the bracket bears

against the smooth wall at A, determine this normal force

at A and the horizontal and vertical components of reaction

of the bolt on the bracket

50 mm

40 mm

60 mmSOLUTION

:+ ©Fx = 0; Ax = 989.18 = 989 N

Bx = 989.18 = 989 N+ ©MA = 0; Bx(60) - 4(9.81)(200) - 15(9.81)(350) = 0

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The pad footing is used to support the load of 12 000 lb

Determine the intensities and of the distributed

loading acting on the base of the footing for the

w2 = 1.646 kip>ft = 1.65 kip>ft

w2a3512b117.5 - 11.672 - 12114 - 11.672 = 0+ ©MA= 0;

Equations of Equilibrium: The load intensity w can be determined directly by

summing moments about point A.

2

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FB = 36.2 lb+ c ©Fy = 0; FB sin 75° - 5 - 30 = 0

FA = 30 lb+ ©MB = 0; - 5(12) + FA(2) = 0

When holding the 5-lb stone in equilibrium, the humerus H,

assumed to be smooth, exerts normal forces and on

the radius C and ulna A as shown Determine these forces

and the force that the biceps B exerts on the radius for

equilibrium The stone has a center of mass at G Neglect the

weight of the arm

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The smooth disks D and E have a weight of 200 lb and 100 lb,

respectively If a horizontal force of is applied to

the center of disk E, determine the normal reactions at the

points of contact with the ground at A, B, and C.

P = 200 lb

P

1.5 ft

A B

D

E C

3

5 4

1 ft

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The smooth disks D and E have a weight of 200 lb and 100 lb,

respectively Determine the largest horizontal force P that

can be applied to the center of disk E without causing the

1.5 ft

A B

D

E C

3

5 4

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FH= 59.43 = 59.4 lb

-81132 + FH11.752 = 0+ ©MB = 0;

The man is pulling a load of 8 lb with one arm held as

shown Determine the force this exerts on the humerus

bone H, and the tension developed in the biceps muscle B.

Neglect the weight of the man’s arm

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Determine the magnitude of force at the pin and in the

cable needed to support the 500-lb load Neglect the

weight of the boom

SOLUTION

Equations of Equilibrium: The force in cable can be obtained directly by

summing moments about point

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SOLUTION

Equations of Equilibrium: The force in short link AB can be obtained directly by

summing moments about point C.

Cx = 333 lb

400.62¢ 3

213≤ - Cx = 0:+ ©Fx = 0;

500142 - FAB¢ 3

213≤162 = 0 FAB = 400.62 lb+ ©MC = 0;

The winch consists of a drum of radius 4 in., which is pin

connected at its center C 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 keeps the drum from

rotating If the suspended load is 500 lb, determine the

horizontal and vertical components of reaction at the pin C.

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The sports car has a mass of 1.5 Mg and mass center at G If

the front two springs each have a stiffness of

and the rear two springs each have a stiffness of

determine their compression when the car

is parked on the 30° incline Also, what friction force

must be applied to each of the rear wheels to hold the car in

equilibrium? Hint: First determine the normal force at A

and B, then determine the compression in the springs.

FB

kB = 65 kN>m,

kA = 58 kN>m

SOLUTION

Equations of Equilibrium: The normal reaction can be obtained directly by

summing moments about point B.

FB = 3678.75 N = 3.68 kN

2FB - 14 715 sin 30° = 0a+ ©Fx¿ = 0;

NA = 3087.32 N

- 14 715 sin 30°10.42 - 2NA122 = 0

14 715 cos 30°11.22+ ©MB = 0;

NA

0.8 m1.2 m FB

0.4 m

30°

B

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The telephone pole of negligible thickness is subjected to the

force of 80 lb directed as shown It is supported by the cable

BCD and can be assumed pinned at its base A In order to

provide clearance for a sidewalk right of way, where D is

located, the strut CE is attached at C, as shown by the dashed

lines (cable segment CD is removed) If the tension in

is to be twice the tension in BCD, determine the height h for

placement of the strut CE.

d = 4.7434 ft438.178(d) - 80 cos 30°(30) = 0

TCD¿ = 2(219.089) = 438.178lb

TBCD = 219.089 lb+ ©MA= 0;

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NA= 1850.40 lb = 1.85 kip

+ ©MB = 0; 2500(1.4 + 8.4) - 500(15 cos 30° - 8.4) - NA(2.2 + 1.4 + 8.4) = 0

The floor crane and the driver have a total weight of 2500 lb

with a center of gravity at G If the crane is required to lift

the 500-lb drum, determine the normal reaction on both the

wheels at A and both the wheels at B when the boom is in

the position shown

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The floor crane and the driver have a total weight of 2500 lb

with a center of gravity at G Determine the largest weight

of the drum that can be lifted without causing the crane to

overturn when its boom is in the position shown

SOLUTION

Equations of Equilibrium: Since the floor crane tends to overturn about point B, the

wheel at A will leave the ground and From the free - body diagram of the

floor crane, Fig a, W can be obtained by writing the moment equation of

equilibrium about point B.

a

Ans.

W = 5337.25 lb = 5.34 kip+ ©MB = 0; 2500(1.4 + 8.4) - W(15 cos 30° - 8.4) = 0

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The mobile crane has a weight of 120,000 lb and center of

gravity at the boom has a weight of 30,000 lb and center

of gravity at Determine the smallest angle of tilt of

the boom, without causing the crane to overturn if the

suspended load is Neglect the thickness of

the tracks at A and B.

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SOLUTION

The mobile crane has a weight of 120,000 lb and center of

gravity at the boom has a weight of 30,000 lb and center

of gravity at If the suspended load has a weight of

determine the normal reactions at the tracks

A and B For the calculation, neglect the thickness of the

tracks and take u = 30°

RA = 40 931 lb = 40.9 kip+ ©MB = 0; -(30 000)(12cos 30° - 3) - (16 000)(27 cos 30° - 3) - RA(13) + (120 000)(9) = 0

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The woman exercises on the rowing machine If she exerts a

holding force of on handle , determine the

horizontal and vertical components of reaction at pin and

the force developed along the hydraulic cylinder on the

handle

SOLUTION

Equations of Equilibrium: Since the hydraulic cylinder is pinned at both ends, it can

be considered as a two-force member and therefore exerts a force directed

along its axis on the handle, as shown on the free-body diagram in Fig a From the

free-body diagram, can be obtained by writing the moment equation of

equilibrium about point

Cx = 432.29 N = 432 N

Cx + 200 cos30° - 628.42 cos15.52° = 0+

0.25 m

0.75 m0.15 m0.15 m

0.25 m

30⬚

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SOLUTION

The ramp of a ship has a weight of 200 lb and a center of

gravity at G Determine the cable force in CD needed to just

start lifting the ramp, (i.e., so the reaction at B becomes zero).

Also, determine the horizontal and vertical components of

force at the hinge (pin) at A.

4 ft

3 ft

G

A D

C B

6 ft30

Ax = 97.5 lb:+ ©Fx = 0: 194.9 sin 30° - Ax = 0

FCD = 194.9 = 195 lb

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The toggle switch consists of a cocking lever that is pinned to

a fixed frame at A and held in place by the spring which has

an unstretched length of 200 mm Determine the magnitude

of the resultant force at A and the normal force on the peg

at B when the lever is in the position shown.

Ax = 2.3239 NQ+ ©Fx = 0; Ax - 2.3832 cos 12.808° = 0

NB = 2.11327 N = 2.11 N+ ©MA = 0; -(2.3832sin 12.808°)(0.4) + NB(0.1) = 0

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The worker uses the hand truck to move material down the

ramp If the truck and its contents are held in the position

shown and have a weight of 100 lb with center of gravity

at , determine the resultant normal force of both wheels on

the ground and the magnitude of the force required at the

NA= 81.621 lb = 81.6 lb

- 100 sin 30°(3.5) - 100 cos 30°(2.5) = 0

(NA cos 30°)(5.25) + NA sin 30°(0.5)a+ ©MB = 0;

1.5 ft0.5 ft

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The boom supports the two vertical loads Neglect the size

of the collars at D and B and the thickness of the boom,

and compute the horizontal and vertical components of

force at the pin A and the force in cable CB Set

1 m

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The boom is intended to support two vertical loads, and

If the cable CB can sustain a maximum load of 1500 N before

it fails, determine the critical loads if Also, what is

the magnitude of the maximum reaction at pin A?

Ax = 1200 N:+ ©Fx = 0; Ax - 45(1500) = 0

F1 = 1.45 kN

F1 = 2F2 = 1448 N

F2 = 724 N+ 45(1500)(2.5 sin 30°) + 35(1500)(2.5 cos 30°) = 0

1 m

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12 ft

A

B

The jib crane is pin connected at A and supported by a

smooth collar at B If , determine the reactions on

the jib crane at the pin A and smooth collar B The load has a

weight of 5000 lb

x = 8 ft

SOLUTION

Equations of Equilibrium: Referring to the of the jib crane shown in Fig a,

we notice that and can be obtained directly by writing the moment

equation of equilibrium about point and force equation of equilibrium along

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The jib crane is pin connected at A and supported by a

smooth collar at B Determine the roller placement x of

the 5000-lb load so that it gives the maximum and

minimum reactions at the supports Calculate these

reactions in each case Neglect the weight of the crane

By observation, the maximum support reactions occur when

Ay - 5 = 0 Ay = 5.00 kip+ c ©Fy = 0;

NB1122 - 5x = 0 NB = 0.4167x+ ©MA = 0;

x

12 ft

A B

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