Mechanical Engineering Chapter 5 Problem The Rocket Launcher Supported Byy The Hydraulic Jack And The Bearings

Problem 5.109 The rocket launcher is supported by

the hydraulic jack DE and the bearings Aand B. The

bearings lie on the xaxis and support shafts parallel to

the xaxis. The hydraulic cylinder DE exerts a force on

the launcher that points along the line from Dto E. The

coordinates of Dare (7, 0, 7) ft, and the coordinates of

Eare (9, 6, 4) ft. The weight WD30 kip acts at (4.5,

5, 2) ft. What is the magnitude of the reaction on the

launcher at E?3 ft

W

AB D

E

3 ft

y

x

Solution: The position vectors of the points D,Eand Ware

The vector parallel to DE is

The unit vector parallel to DE is

Since the bearings cannot exert a moment about the xaxis, the sum of

the moments due to the weight and the jack force must be zero about

the xaxis. The sum of the moments about the xaxis is:



From which

348

Problem 5.110 Consider the rocket launcher described

in Problem 5.109. The bearings at Aand Bdo not exert

couples, and the bearing Bdoes not exert a force in the

xdirection. Determine the reactions at Aand B.

Solution: See the solution of Problem 5.109. The force FDE can

be written

The equilibrium equations are

M⊲origin⊳D



ijk



ij k



The components of the moment eq. are

5.9997FDE C60 D0,

Problem 5.111 The crane’s cable CD is attached to a

stationary object at D. The crane is supported by the

bearings Eand Fand the horizontal cable AB. The

tension in cable AB is 8 kN. Determine the tension in

the cable CD.

Strategy: Since the reactions exerted on the crane by

the bearings do not exert moments about the zaxis, the

sum of the moments about the zaxis due to the forces

exerted on the crane by the cables AB and CD equals

zero. (See the discussion at the end of Example 5.9.)

Solution: The position vector from Cto Dis

The moment about the origin due to the forces exerted by the two

cables is



ijk



ij k



2 m 2 m

3 m

x

3 m

D

C

y

D

3 m

z

350

c

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Problem 5.112 In Example 5.9, suppose that the cable

CE is shortened and its attachment point Eis moved to

the point (0, 80, 0) mm. The plate remains in the same

position. Draw a sketch of the plate and its supports

showing the new position of cable CE. Draw the free-

body diagram of the plate and apply equilibrium to deter-

mine the reactions at the hinges and the tension in the

cable.

y

B

C

D

A

⫺400j (N)

200 mm

80 mm

100

mm

z

x

E

Solution: The sketch and free-body diagram are shown.

The vector from Cto Eis

The force exerted by cable CE can be expresses as

The equilibrium equations for the plate are

Problem 5.113 The plate is supported by hinges at A

and Band the cable CE, and it is loaded by the force at

D. The edge of the plate to which the hinges are attached

lies in the yzplane, and the axes of the hinges are

parallel to the line through points Aand B. The hinges

do not exert couples on the plate. What is the tension in

cable CE?

2i – 6j (kN)

Dx

C

B

A

E

z

y

3 m

1 m

2 m

Solution:

However, we just want tension in CE. This quantity is the only

unknown in the moment equation about the line AB. To get this, we

need the unit vector along CE.

y

FD = 2i – 6j

352

Problem 5.114 In Problem 5.113, the hinge at Bdoes

not exert a force on the plate in the direction of the hinge

axis. What are the magnitudes of the forces exerted on

the plate by the hinges at Aand B?

Solution: From the solution to Problem 5.113, TCE D8.15 kN

Also, from that solution,

y

F = + 2i – 6j (kN)

Problem 5.115 The bar ABC is supported by ball and

socket supports at Aand Cand the cable BD. The

suspended mass is 1800 kg. Determine the tension in

the cable.

y

x

z

A

B

C

(⫺2, 2, ⫺1) m

2 m 4 m

4 m

D

Solution: We take moments about the line AC to eliminate the

reactions at Aand C.

We have

TBD

Az

354

Problem 5.116* In Problem 5.115, assume that the

ball and socket support at Ais designed so that it

exerts no force parallel to the straight line from Ato

C. Determine the reactions at Aand C.

Solution: We have

TBD

Ay

Ax

Az

Cz

Cx

17.66 kN

Problem 5.117 The bearings at A,B, and Cdo not

exert couples on the bar and do not exert forces in the

direction of the axis of the bar. Determine the reactions

at the bearings due to the two forces on the bar.

x

y

C

B

200 i (N)

300 mm

180 mm

Solution: The strategy is to take the moments about Aand solve

the resulting simultaneous equations. The position vectors of the bear-

ings relative to Aare:

The sum of the moments about Ais:

MADrA1ðF1CrAB ðBCrA2ðF2CrAC ðCD0



x

z

CY

BZ

BX

CX

200 N

356

Problem 5.118 The support that attaches the sailboat’s

mast to the deck behaves like a ball and socket support.

The line that attaches the spinnaker (the sail) to the top

of the mast exerts a 200-lb force on the mast. The force

is in the horizontal plane at 15°from the centerline of the

boat. (See the top view.) The spinnaker pole exerts a 50-

lb force on the mast at P. The force is in the horizontal

plane at 45°from the centerline. (See the top view.)

The mast is supported by two cables, the back stay AB

and the port shroud ACD. (The forestay AE and the

starboard shroud AFG are slack, and their tensions can

be neglected.) Determine the tensions in the cables AB

and CD and the reactions at the bottom of the mast.

A

x

z

A

P

F

G

B

Top View (Spinnaker not shown)

15°

45°C

D

200 lb

50 lb

6 ft

Aft View

D

EB

P

y

x

Spinnaker

50 ft

Side View 21 ft 15 ft

A

C

G

P

CF

A

E

Solution: Although the dimensions are not given in the sketch,

rAC DrCrAD25j7.5k.

The forces acting on the mast are: (1) The force due to the spinnaker

(4) The force acting on the cross spar CE:

(5) The force due to the spinnaker pole:



FTX FTY FTZ

C



000.2873jTACj



C



ijk

066



7.1829 D389.81 lb,

5.118 (Continued )

The tension in cable CD is the vertical component of the tension

in AC,

jTCDjDjTACj⊲jÐeAC⊳DjTACj⊲0.9578⊳D373.37 lb.

QD31.14iC823.26j87.12k(lb)

z

FA

TAB

TCE

TCD

FA

FPFP

QYQB

358

Problem 5.119* The bar AC is supported by the cable

BD and a bearing at Athat can rotate about the axis AE.

The person exerts a force FD50j(N) at C. Determine

the tension in the cable.

Strategy: Use the fact that the sum of the moments

about the axis AE due to the forces acting on the free-

body diagram of the bar must equal zero.

(0.3, 0.4, 0.3) m

(0.3, 0.5, 0) m

(0.82, 0.60, 0.40) m

(0.46, 0.46, 0.33) m

B

C

E

A

x

y

Problem 5.120* In Problem 5.119, determine the

reactions at the bearing A.

component of MAparallel to the axis AE to equal zero.

Solution: See the previous problem for setup. We add the reactions

(force, moment)

We have the following 6 equilibrium equations:





Fx:AxC0.440TBD D0

MADMACrAB ðTBD CrAC ðFD0





MAx :MAx 5 N-m C⊲0.0440 m⊳TBD D0

MAy :MAy ⊲0.0366 m⊳TBD D0

Problem 5.121 In Active Example 5.10, suppose that

the support at Ais moved so that the angle between the

bar AB and the vertical decreases from 45°to 30°. The

position of the rectangular plate does not change. Draw

the free-body diagram of the plate showing the point

Pwhere the lines of action of the three forces acting

on the plate intersect. Determine the magnitudes of the

reactions on the plate at Band C.

4 ft

B

45

AC

Solution: The equilibrium equations are

360

Problem 5.122 The magnitude of the reaction exerted

on the L-shaped bar at Bis 60 lb. (See Example 5.11.)

(a) What is the magnitude of the reaction exerted on the

bar by the support at A?

(b) What are the xand ycomponents of the reaction

exerted on the bar by the support at A?

14 in

17 in

x

y

B

A

Solution: The angle between the line AB and the xaxis is

Problem 5.123 The suspended load weighs 1000 lb.

The structure is a three-force member if its weight is

neglected. Use this fact to determine the magnitudes of

the reactions at Aand B.

5 ft

B

A

10 ft

Solution: The pin support at Ais a two-force reaction, and the

roller support at Bis a one force reaction. The moment about Ais

MAD5B10⊲1000⊳D0, from which the magnitude at Bis BD

2000 lb. The sums of the forces:

FXDAXCBDAXC2000 D0,from which AXD2000 lb.

5 ft

A

10 ft

Problem 5.124 The weight WD50 lb acts at the

center of the disk. Use the fact that the disk is a three-

force member to determine the tension in the cable and

the magnitude of the reaction at the pin support.

60°

Solution: Denote the magnitude of the reaction at the pinned joint

362

Problem 5.125 The weight WD40 N acts at the

center of the disk. The surfaces are rough. What force

Fis necessary to lift the disk off the ﬂoor?

50 mm

F

150 mm

W

Solution: The reaction at the obstacle acts through the center of

the disk (see sketch) Denote the contact point by B. When the moment

F

150 mm

Problem 5.126 Use the fact that the horizontal bar is

a three-force member to determine the angle ˛and the

magnitudes of the reactions at Aand B. Assume that

0˛90°.

2 m

B

a30⬚

60⬚

364

Problem 5.127 The suspended load weighs 600 lb.

Use the fact that ABC is a three-force member to

determine the magnitudes of the reactions at Aand B.

45⬚

30⬚

3 ft

4.5 ft

A

B

C

Solution: All of the forces must intersect at a point.

Problem 5.128 (a) Is the L-shaped bar a three-force

member?

(b) Determine the magnitudes of the reactions at Aand B.

(c) Are the three forces acting on the L-shaped bar

concurrent?

250

mm

500

mm

700 mm

300 mm

150 mm

2 kN

3 kN-m

A

B

Solution: (a) No. The reaction at Bis one-force, and the reaction

at Ais two-force. The couple keeps the L-shaped bar from being a three

force member.(b) The angle of the member at Bwith the horizontal is

366

c

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