Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 03

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©Fx = 0; F2sin 70° + F1cos 60° - 5 cos 30° - 4

5(7) = 0

The members of a truss are pin connected at joint O.

Determine the magnitudes of and for equilibrium

x O

F2

F1

u

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The members of a truss are pin connected at joint O.

Determine the magnitude of and its angle for

equilibrium Set F2= 6 kN

u

F1

x O

F2

F1

u

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Free-Body Diagram: By observation, the force has to support the entire weight

of the container Thus,

Equations of Equilibrium:

Thus,

Ans.

If the maximum allowable tension in the cable is 5 kN, then

Ans.

l = 1.5cos 29.37° = 1.72 m

u = 29.37°

l = 1.5cos u

u = 29.37°

2452.5 cos u = 5000

FAC = FAB= F = 52.45 cos u6 kN

4905 - 2F sin u = 0 F = 52452.5 cos u6 N+ c ©Fy = 0;

FACcos u - FABcos u = 0 FAC = FAB= F:+ ©Fx = 0;

F1 = 50019.812 = 4905 N.F1

The lift sling is used to hoist a container having a mass of

500 kg Determine the force in each of the cables AB and

AC as a function of If the maximum tension allowed in

each cable is 5 kN, determine the shortest lengths of cables

AB and AC that can be used for the lift The center of

gravity of the container is located at G.

C B

G

F

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Cords AB and AC can each sustain a maximum tension of

800 lb If the drum has a weight of 900 lb, determine the

A

C

uu

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The members of a truss are connected to the gusset plate If

the forces are concurrent at point O, determine the

magnitudes of F and T for equilibrium Take u = 30°

T = 13.32 = 13.3 kN:+ ©Fx = 0; -T cos 30° + 8 + 5 sin 45° = 0

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The gusset plate is subjected to the forces of four members

Determine the force in member B and its proper

orientation for equilibrium The forces are concurrent at

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Equations of Equilibrium: A direct solution for can be obtained by summing

forces along the y axis.

FBC = 2.993 kN

FBC = 2.993 kN = 2.99 kN

3.5 sin 48.37° - FBCsin 60.95° = 0+ c ©Fy = 0;

FBC

The device shown is used to straighten the frames of

wrecked autos Determine the tension of each segment of

the chain, i.e., AB and BC, if the force which the hydraulic

cylinder DB exerts on point B is 3.50 kN, as shown.

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Two electrically charged pith balls, each having a mass of

0.2 g, are suspended from light threads of equal length

Determine the resultant horizontal force of repulsion, F,

acting on each ball if the measured distance between them

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Determine the maximum weight of the flowerpot that can

be supported without exceeding a cable tension of 50 lb in

either cable AB or AC.

SOLUTION

Equations of Equilibrium:

(1)

(2)

Since failure will occur first at cable AC with Then

solving Eqs (1) and (2) yields

FAC = 1.20FAB

FAC sin 30° - FABa35b = 0:+ ©Fx = 0;

30°4

35

B

C

A

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SOLUTION

Equations of Equilibrium: Applying the equations of equilibrium along the x and y

axes to the free-body diagram shown in Fig a,

(1) (2)

Solving Eqs (1) and (2), yields

Ans.

+ c ©Fy = 0; FCB sin 40° + FCA sin 30° - 10(9.81) = 0

:+ ©Fx = 0; FCB cos 40° - FCA cos 30° = 0

Determine the tension developed in wires and

required for equilibrium of the 10-kg cylinder Take CA u = 40°CB

30°

C

u

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Equations of Equilibrium: Applying the equations of equilibrium along the x and

y axes,

(1) (2)

However, it is required that

If cable is subjected to a tension that is twice that of

cable , determine the angle for equilibrium of the 10-kg

cylinder Also, what are the tensions in wires and CA CB?

uCA

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FBC= FBD= F

FBD sin 45° - FBC sin 45° = 0:+ ©Fx = 0;

FBC = FBD

FAB= 400 lb

The concrete pipe elbow has a weight of 400 lb and the

center of gravity is located at point G Determine the force

FAB and the tension in cables BC and BD needed to support it

45⬚ 45⬚

FAB

D C

B

G

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Ans.

s = 4 tan 53.13° = 5.33 ft

tanu = s4

u = sin-1(0.8) = 53.13°

+ c ©Fy = 0; 2(5) sin u - 8 = 0

Blocks D and F weigh 5 lb each and block E weighs 8 lb.

Determine the sag s for equilibrium Neglect the size of the

4 ft 4 ft

s

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If blocks D and F weigh 5 lb each, determine the weight of

block E if the sags = 3 ft.Neglect the size of the pulleys

4 ft 4 ft

s

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The spring has a stiffness of and an unstretched

length of 200 mm Determine the force in cables BC and BD

when the spring is held in the position shown

The Force in The Spring:Thespringstretches

Applying Eq 3–2, we have

Fsp = ks = 800(0.3) = 240 N

s = l - l0 = 0.5 - 0.2 = 0.3 m

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:+ ©Fx = 0; T cos u - T cos u = 0

The 10-lb lamp fixture is suspended from two springs, each

having an unstretched length of 4 ft and stiffness of

Determine the angle for equilibrium.u

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Determine the mass of each of the two cylinders if they

cause a sag of when suspended from the rings at

A and B Note that when the cylinders are removed.s = 0

k 100 N/m k 100 N/mSOLUTION

Ans.

m = 2.37 kg+ c ©Fy = 0; 32.84 sin 45° - m(9.81) = 0

TAC = 100 N>m (2.828 - 2.5) = 32.84 N

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Determine the stretch in springs AC and AB for

equilibrium of the 2-kg block The springs are shown in

the equilibrium position

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The unstretched length of spring AB is 3 m If the block is

held in the equilibrium position shown, determine the mass

T = 67.88 N

:+ ©Fx = 0; Tcos 45° - 60a4

5b = 0

F = kx = 30(5 - 3) = 60 N

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The springs BA and BC each have a stiffness of 500 N>m and an

unstretched length of 3 m Determine the horizontal force F

applied to the cord which is attached to the small ring B so

that the displacement of the ring from the wall is d = 1.5 m

F

B

C d

A

k ⫽ 500 N/m

k ⫽ 500 N/m

6 m

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The springs BA and BC each have a stiffness of 500 N>m and an

unstretched length of 3 m Determine the displacement d of the

cord from the wall when a force F = 175 N is applied to the cord

F

B

C d

A

k 500 N/m

k 500 N/m

6 m

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A vertical force is applied to the ends of the 2-ft

cord AB and spring AC If the spring has an unstretched

length of 2 ft, determine the angle for equilibrium Take

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Determine the unstretched length of spring AC if a force

causes the angle for equilibrium Cord

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The springs on the rope assembly are originally unstretched

when Determine the tension in each rope when

Neglect the size of the pulleys at B and D.

F = 90 lb

u = 0°

F

A B

SOLUTION

(1)

(2)

Substituting Eq (1) into (2) yields:

By trial and error:

l = 2

cos u

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The springs on the rope assembly are originally stretched

1 ft when Determine the vertical force F that must

be applied so that u = 0° u = 30°

F

A B

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= 120(sec u - 1)

Fs = kx; Fs = 10a 12

cos u - 12b+ c ©Fy = 0; Fs sin u - 10 = 0

The 10-lb weight A is supported by the cord AC and roller C,

and by the spring that has a stiffness of k = 10 lb>in If the

unstretched length of the spring is 12 in determine the distance

d to where the weight is located when it is in equilibrium.

d

A C

B

12 in

k

u

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The 10-lb weight A is supported by the cord AC and roller C,

and by spring AB If the spring has an unstretched length of

8 in and the weight is in equilibrium when d = 4 in., determine

the stiffness k of the spring.

d

A C

B

12 in

k

u

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Equations of Equilibrium: Applying the equations of equilibrium along the x and y

axes to the free-body diagram of joint D shown in Fig a, we have

Ans.

Ans.

Using the result F CD= 339.83 N and applying the equations of equilibrium along the

x and y axes to the free-body diagram of joint D shown in Fig b, we have

339.83 - FCAa35b - FCD cos 45° = 0

©Fx = 0;

:+

FCD= 339.83 N = 340 N392.4cos30° - FCD= 0

Determine the tension developed in each cord required for

equilibrium of the 20-kg lamp

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Equations of Equilibrium: Applying the equations of equilibrium along the x and y

axes to the free-body diagram of joint D shown in Fig a, we have

Ans.

Ans.

Using the result F CD = 16.99m and applying the equations of equilibrium along the

x and y axes to the free-body diagram of joint D shown in Fig b, we have

(1) (2)

Solving Eqs (1) and (2), yields

Notice that cord DE is subjected to the greatest tensile force, and so it will achieve

the maximum allowable tensile force first Thus

16.99m - FCAa35b - FCD cos45° = 0

©Fx = 0;

:+

FCD= 16.99m19.62mcos30° - FCD = 0

©Fx = 0;

:+

FDE = 19.62m

FDEsin 30° - m(9.81) = 0+ c ©Fy = 0;

Determine the maximum mass of the lamp that the cord

system can support so that no single cord develops a tension

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SOLUTION

Geometry: The angle u which the surface makes with the horizontal is to be

determined first

Free-Body Diagram: The tension in the cord is the same throughout the cord and is

equal to the weight of block B,

A 4-kg sphere rests on the smooth parabolic surface

Determine the normal force it exerts on the surface and the

mass of block B needed to hold it in the equilibrium

position shown

mB

B

A y

x

0.4 m0.4 m

60

y 2.5x2

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If the bucket weighs 50 lb, determine the tension developed

in each of the wires

A B

E

C

D

4 3 5

3030

SOLUTION

Equations of Equilibrium: First, we will apply the equations of equilibrium along the

x and y axes to the free-body diagram of joint E shown in Fig a.

(1)

(2)

Solving Eqs (1) and (2), yields

Ans.

Using the result and applying the equations of equilibrium to the

free-body diagram of joint B shown in Fig b,

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Determine the maximum weight of the bucket that the wire

system can support so that no single wire develops a tension

exceeding 100 lb

A B

E

C

D

SOLUTION

Equations of Equilibrium: First,we will apply the equations of equilibrium along

the x and y axes to the free-body diagram of joint E shown in Fig a.

(1)

(2)

Solving,

Using the result and applying the equations of equilibrium to the

free-body diagram of joint B shown in Fig b,

From these results, notice that wire BA is subjected to the greatest tensile force.

Thus, it will achieve the maximum allowable tensile force first

Ans.

W = 57.7 lb

FBA= 100 = 1.7320W

FBA= 1.7320W:+ ©Fx = 0; 1.3957W cos 30° + 0.8723Wa35b - FBA = 0

FBC = 1.3957W+ c ©Fy = 0; FBC sin 30° - 0.8723Wa45b = 0

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Determine the tension developed in each wire which is

needed to support the 50-lb flowerpot

SOLUTION

Equations of Equilibrium: First, we will apply the equations of equilibrium along

the x and y axes to the free-body diagram of joint E shown in Fig a.

(1)(2)Solving Eqs (1) and (2), yields

Ans.

Using the result and applying the equations of equilibrium along the x and y axes to

the free-body diagram of joint C shown in Fig b, we have

FED = FEC = 28.87 lb = 28.9 lb

FEDcos30° + FECcos30° - 50 = 0+ c ©Fy = 0;

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Equations of Equilibrium: First, we will apply the equations of equilibrium along

the x and y axes to the free-body diagram of joint E shown in Fig a.

(1) (2)

Solving Eqs (1) and (2), yields

Ans.

Using the result and applying the equations of equilibrium along

the x and y axes to the free-body diagram of joint C shown in Fig b, we have

Due to symmetry,

From this result, notice that cables DB and CA are subjected to the greater tensile

forces Thus, they will achieve the maximum allowable tensile force first

FEC = 0.5774W

FED = FEC = 0.5774W

FEDcos30° + FECcos30° - W = 0+ c ©Fy = 0;

If the tension developed in each of the wires is not allowed

to exceed 40 lb, determine the maximum weight of the

flowerpot that can be safely supported

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Equations of Equilibrium: Since cable ABC passes over the smooth pulley at B, the

tension in the cable is constant throughout its entire length Applying the equation

of equilibrium along the y axis to the free-body diagram in Fig a, we have

2Tsinf - 100(9.81) = 0+ c ©Fy = 0;

Cable ABC has a length of 5 m Determine the position x

and the tension developed in ABC required for equilibrium

of the 100-kg sack Neglect the size of the pulley at B.

A

B

C x

0.75 m

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The single elastic cord ABC is used to support the 40-lb

load Determine the position x and the tension in the cord

that is required for equilibrium The cord passes through

the smooth ring at B and has an unstretched length of 6 ft

and stiffness of k = 50 lb>ft

A

C B

x

5 ft

1 ft

SOLUTION

Equations of Equilibrium: Since elastic cord ABC passes over the smooth ring at B,

the tension in the cord is constant throughout its entire length Applying the equation

of equilibrium along the y axis to the free-body diagram in Fig a, we have

Substituting Eq (4) into Eq (1) yields

Solving the above equation by trial and error

Substituting into Eqs (1) and (3) yields

5 - xcosf =

5cosf

2Tsinf - 40 = 0+ c ©Fy = 0;

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The 200-lb uniform tank is suspended by means of a

6-ft-long cable, which is attached to the sides of the tank and

passes over the small pulley located at O If the cable can be

attached at either points A and B, or C and D, determine

which attachment produces the least amount of tension in

the cable What is this tension?

SOLUTION

Free-Body Diagram: By observation, the force F has to support the entire weight

of the tank Thus, F = 200 lb The tension in cable AOB or COD is the same

throughout the cable

Equations of Equilibrium:

(1)

From the function obtained above, one realizes that in order to produce the least

amount of tension in the cable, hence must be as great as possible Since the

attachment of the cable to point C and D produces a greater

as compared to the attachment of the cable to points A and ,

the attachment of the cable to point C and D will produce the least amount

:+ ©Fx = 0; T cos u - T cos u = 0 ( Satisfied!)

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The sling BAC is used to lift the 100-lb load with constant

velocity Determine the force in the sling and plot its value T

(ordinate) as a function of its orientation , where

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