Mechanical Engineering Chapter 5 Problem The Hydraulic Piston Exerts Hori Zontal Force Support The Weight The

Problem 5.129 The hydraulic piston exerts a hori-

zontal force at Bto support the weight WD1500 lb

of the bucket of the excavator. Determine the magnitude

of the force the hydraulic piston must exert. (The vector

sum of the forces exerted at Bby the hydraulic piston,

the two-forces member AB, and the two-force member

BD must equal zero.)

14 in

CD

A

16 in

4 in

B

Solution: See the solution to Problem 5.23.

Problem 5.130 The member ACG of the front-end

loader is subjected to a load WD2 kN and is supported

by a pin support at Aand the hydraulic cylinder BC.

Treat the hydraulic cylinder as a two-force member.

(a) Draw a free-body diagrams of the hydraulic

cylinder and the member ACG.

(b) Determine the reactions on the member ACG.

0.5 m

1 m

0.75 m

1.5 m 1.5 m

W

G

C

A

B

Solution: This is a very simple Problem. The free body diagrams

Using the given value for Wand solving these equations, we get

BX

W

CX

1.5 m 1.5 m

Problem 5.131 In Problem 5.130, determine the reac-

tions of the member ACG by using the fact that it is a

three-force member.

Solution: The easiest way to do this is take advantage of the fact

that for a three force member, the three forces must be concurrent. The

fact that the force at Cis horizontal and the weight is vertical make

it very easy to ﬁnd the point of concurrency. We then use this point

to determine the direction of the force through A. We can even know

geometry, we can determine the angle between the force Aand the

A

y

x

G

0.75 m

1.5 m 1.5 m

θ

368

Problem 5.132 A rectangular plate is subjected to

two forces Aand B(Fig. a). In Fig. b, the two forces

are resolved into components. By writing equilibrium

equations in terms of the components Ax,Ay,Bx, and By,

show that the two forces Aand Bare equal in magnitude,

opposite in direction, and directed along the line between

their points of application. A

(a)

h

b

B

A

(b)

h

b

B

By

Bx

Axx

y

Ay

Solution: The sum of forces:

from which AXDBX

FYDAYCBYD0,

A

b

B

ByBx

y

Ay

Fig a

Fig b

Problem 5.133 An object in equilibrium is subjected

to three forces whose points of application lie on a

straight line. Prove that the forces are coplanar.

F1

F2

F3

Solution: The strategy is to show that for a system in equilibrium

forces. The sum of the moments about the point P:

where the vectors are the position vectors of the points of the applica-

tion of the forces relative to the point P. (The position vectors lie in

the plane.) Deﬁne

d12,d13 are parallel to the line L.) The component of the moment

parallel to the line Lis

But by deﬁnition, F2lies in the same plane as the line L, hence it is

normal to the cross product d12 ðe6D 0, and the term

F2Ð⊲d12 ðe⊳D0.

But this means that

F3Ð⊲d13 ðe⊳eD0,

which implies that F3also lies in the same plane as F2, since

d13 ðe6D 0.

Thus the two forces lie in the same plane. Since the choice of the point

about which to sum the moments was arbitrary, this process can be

repeated to show that F1lies in the same plane as F2.Thus all forces

lie in the same plane.

PL

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Problem 5.134 The suspended cable weighs 12 lb.

(a) Draw the free-body diagram of the cable. (The

tensions in the cable at Aand Bare not equal.)

(b) Determine the tensions in the cable at Aand B.

(c) What is the tension in the cable at its lowest point?

A

B

32⬚

50⬚

Solution:

(a) The FBD

(b) The equilibrium equations

50°

TA

TB

Problem 5.135 Determine the reactions at the ﬁxed

support.

3m

A

4 kN

20 kN-m

Solving

AxD4kN,A

yD5kN,M

AD26 kN-m

3 kN

Problem 5.136 (a) Draw the free-body diagram of the

50-lb plate, and explain why it is statically indeterminate.

(b) Determine as many of the reactions at Aand Bas

possible.

12 in

A

20 in

y

Solution:

(a) The pin supports at Aand Bare two-force supports, thus there

for the moment. Thus there are four unknowns and only three

372

Problem 5.137 The mass of the truck is 4 Mg. Its

wheels are locked, and the tension in its cable is TD

10 kN.

(a) Draw the free-body diagram of the truck.

(b) Determine the normal forces exerted on the truck’s

wheels by the road.

AL’s

Towing

(003) 676-5942

mg

2 m 2.5 m 2.2 m

30°3 m

T

Solution: The weight is 4000⊲9.81⊳D39.24 kN. The sum of the

moments about B

from which

The sum of the forces:

Problem 5.138 Assume that the force exerted on the

head of the nail by the hammer is vertical, and neglect

the hammer’s weight.

(a) Draw the free-body diagram of the hammer.

(b) If FD10 lb, what are the magnitudes of the forces

exerted on the nail by the hammer and the normal

and friction forces exerted on the ﬂoor by the

hammer?

F

11 in.

65°

Solution: Denote the point of contact with the ﬂoor by B. The

NHD5.077F

NH

FN

HXB

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Problem 5.139 The spring constant is kD9600 N/m

and the unstretched length of the spring is 30 mm. Treat

the bolt at Aas a pin support and assume that the surface

at Cis smooth. Determine the reactions at Aand the

normal force at C.

15 mm C

B

A

k

30 mm

24 mm

30°

Solution: The length of the spring is

Problem 5.140 The engineer designing the release

mechanism shown in Problem 5.139 wants the normal

force exerted at Cto be 120 N. If the unstretched length

of the spring is 30 mm, what is the necessary value of

the spring constant k?

376

Problem 5.141 The truss supports a 90-kg suspended

object. What are the reactions at the supports Aand B?

300 mm

400 mm 700 mm

A

B

Solution: Treat the truss as a single element. The pin support at

The sum of the forces:

Problem 5.142 The trailer is parked on a 15°slope.

Its wheels are free to turn. The hitch Hbehaves like a

pin support. Determine the reactions at Aand H.

x

y

15°

H

2.8 ft

1.4 ft

870 lb

Solution: The coordinate system has the xaxis parallel to the road.

The wheels are a one force reaction normal to the road, the pin His

a two force reaction. The position vectors of the points of the center

of mass and Hare:

The angle of the weight vector realtive to the positive xaxis is

378

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Problem 5.143 To determine the location of the point

where the weight of a car acts (the center of mass),

an engineer places the car on scales and measures the

normal reactions at the wheels for two values of ˛,

obtaining the following results:

˛A

y(kN) B(kN)

10°10.134 4.357

20°10.150 3.677

What are the distances band h?

α

Ay

Ax

W

B

h

b

2.7 m

x

y

The angle between the weight and the positive xaxis is ˇD270 ˛.

The weight vector at each of the two angles is

D013.8272bC5.0327h27.4054 D0

hD0.50 m

Problem 5.144 The bar is attached by pin supports to

collars that slide on the two ﬁxed bars. Its mass is 10 kg,

it is 1 m in length, and its weight acts at its midpoint.

Neglect friction and the masses of the collars. The spring

is unstretched when the bar is vertical (˛D0), and the

spring constant is kD100 N/m. Determine the values

of ˛in the range 0 ˛60°at which the bar is in

equilibrium.

α

k

Solution: The force exerted by the spring is given by FSDk⊲L 

Lcos ˛⊳. The equations of equilibrium, from the free body diagram, are

and MBD⊲Lsin ˛⊳NACL

FS

NB

B

Spring Constant (K) in N/m vs Alpha (deg)

90000

30000

m

Spring Constant (K) in N/m vs Alpha (deg)

116

114

112

110

K

380

Problem 5.145 With each of the devices shown you

can support a load Rby applying a force F. They are

called levers of the ﬁrst, second, and third class.

(a) The ration R/F is called the mechanical advantage.

Determine the mechanical advantage of each lever.

(b) Determine the magnitude of the reaction at Afor

each lever. (Express you answers in terms of F)

R

L

F

L

A

L

F

L

R

A

F

A

R

First-class lever Second-class lever

Solution:

Problem 5.146 The force exerted by the weight of the

horizontal rectangular plate is 800 N. The weight of the

plate acts at its midpoint. If you represent the reactions

exerted on the plate by the three cables by a single

equivalent force, what is the force, and where does its

line of action intersect the plate?

A

B

C

1 m

0.5 m

2 m

Solution: The equivalent force must equal the sum of the

reactions: FEQ DTACTBCTC.FEQ D300 C100 C400 D800 N.

y

TA

Problem 5.147 The 20-kg mass is suspended by cables

attached to three vertical 2-m posts. Point Ais at

(1, 1.2, 0) m. Determine the reactions at the built-in

support at E.B

A

C

D

y

E

1 m

Solution: All distances will be in meters, all forces in Newtons,

and all moments in Newton-meters. To solve the three dimensional

point equilibrium problem at A, we will need unit vectors eAB,eAC,

after making the substitutions related to the force components yields

the tensions in the cables. They are

jTABjD150 N,

Also,jMGjD68.4 N-m

CD

−TAD

382

Problem 5.148 In Problem 5.147, the built-in support

of each vertical post will safely support a couple of

800 N-m magnitude. Based on this criterion, what is the

maximum safe value of the suspended mass?

Problem 5.149 The 80-lb bar is supported by a ball

and socket support at A, the smooth wall it leans against,

and the cable BC. The weight of the bar acts at its

midpoint.

(a) Draw the free-body diagram of the bar.

(b) Determine the tension in cable BC and the reactions

at A.

x

A

C

B

3 ft

3 ft 5 ft

3 ft

z

4 ft

y

Solution: (a) The ball and socket is a three reaction force support;

the cable and the smooth wall are each one force reaction supports.

rAB DrBrAD2iC4j3k.

Solve:

The reactions at Aare found from the sums of forces:

384

Problem 5.150 The horizontal bar of weight Wis

supported by a roller support at Aand the cable BC. Use

the fact that the bar is a three-force member to determine

the angle ˛, the tension in the cable, and the magnitude

of the reaction at A.

L/2

AB

C

α

L/2

W

Solution: The sum of the moments about Bis

2. This equation cannot be satisﬁed