statics of a particle- 2d & 3d equilibrium analysis

2C H A P T E R

15

Statics of Particles

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Whenever cables are used for hoisting loads, they must be selected so that they donot fail when they are placed at their points of attachment. In this chapter, we willshow how to calculate cable loadings for such cases.

UniKLUNIVERSITIKUALA LUMPUR

UniKL MIAT – Jan 2015 Mulia Minhat

Malaysian Institute of Aviation Technology

Statics of a Particle: 2-D & 3-D Equilibrium Analysis

3.4 THREE-DIMENSIONAL FORCE SYSTEMS 103

3

3.4 Three-Dimensional Force Systems

In Section 3.1 we stated that the necessary and sufficient condition forparticle equilibrium is

(3–4)

In the case of a three-dimensional force system, as in Fig. 3–9, we canresolve the forces into their respective i, j, k components, so that

. To satisfy this equation we require

(3–5)

These three equations state that the algebraic sum of the components ofall the forces acting on the particle along each of the coordinate axesmust be zero. Using them we can solve for at most three unknowns,generally represented as coordinate direction angles or magnitudes offorces shown on the particle’s free-body diagram.

©Fz = 0©Fy = 0©Fx = 0

©Fxi + ©Fy j + ©Fzk = 0

©F = 0

Procedure for Analysis

Three-dimensional force equilibrium problems for a particle can besolved using the following procedure.

Free-Body Diagram.

• Establish the x, y, z axes in any suitable orientation.

• Label all the known and unknown force magnitudes anddirections on the diagram.

• The sense of a force having an unknown magnitude can beassumed.

Equations of Equilibrium.

• Use the scalar equations of equilibrium,in cases where it is easy to resolve each force into its

x, y, z components.

• If the three-dimensional geometry appears difficult, then firstexpress each force on the free-body diagram as a Cartesian vector,substitute these vectors into and then set the i, j, kcomponents equal to zero.

• If the solution for a force yields a negative result, this indicatesthat its sense is the reverse of that shown on the free-bodydiagram.

©F = 0,

©Fz = 0,©Fy = 0,©Fx = 0,

F3F2

F1

x

y

z

Fig. 3–9

The ring at A is subjected to the force fromthe hook as well as forces from each of thethree chains. If the electromagnet and its loadhave a weight W, then the force at the hookwill be W, and the three scalar equations ofequilibrium can be applied to the free-bodydiagram of the ring in order to determine thechain forces, , and FD.FCFB,

A

D

CB

FCFDFB

W

- For internal use only -

UniKL MIAT - MM2015 2

PARTICLE ANALYSIS

Body under consideration

Forces acting on a particle

!F1 +!F2 +

!F3 =

!FR = 0

Equilibrium analysis

3-D x

y

z

F3 F1

F2

x

y

F3 F1

F2

2-D

Fx∑ = 0, Fy∑ = 0, Fz∑ = 0( )

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OBJECTIVES

•  To draw the free body diagram of a particle under equilibrium condition in 2-D and 3-D system.

•  To determine the unknown forces acting on a particle based on the fact that summation of forces in the respective coordinate system is zero ( ) due to the equilibrium condition.

F∑ = 0



CASE STUDY 1 (2-D) In a ship unloading operation, a 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. Determine the tension in the rope AC and cable AB.



THEORY - Free body diagram (FBD): a diagram, which focus on

specific body that can be idealized as a particle or rigid body and is in equilibrium condition. The summation of forces in FBD is always equal to zero.

10 Ib

A

B

C

D

FBD of particle A

x

y A FAC

FAB

W = 10lb

F∑ = 0,



THEORY

θ

T

T

lo = 0.4 m

s = 0.3 m

s = - 0.15 m

F = k(xf-xi)

- Typical connections used in the FBD such as cable and pulley, and also spring.

TAC = TBC

A

C

B

F = kΔx



THEORY - Equilibrium conditions: Summation of forces in the

respective axis of a coordinate system equals zero.

2-D Cartesian system 3-D Cartesian system

Fx∑ = 0,

Fy∑ = 0.

Fx∑ = 0,

Fy∑ = 0,

Fz∑ = 0.



APPROACH TO SOLUTION

- Construct FBD surrounding particle A and label the necessary information.

- Resolve all forces into vector form and identify the x- and y-components.

- Express all equilibrium equations appropriately, where the summations of forces are equal to zero in the respective axis system.

- Solve the equilibrium (simultaneous) equations to determine the magnitudes of unknown forces.



SOLUTION - Draw the FBD of particle A and resolve all forces

into vector form to identify the necessary components.

A

FAB

FAC

3500 Ib

FAC x

FAC y

FAB y

FAB x 30O

2O x

y

FAB x = FAB sin20 = 0.0349FAB

FAB y = FAB cos20 = 0.9994FAB

FAC x = FAC cos300 = 0.866 FAC

FAC y = FAC sin300 = 0.5FAC



SOLUTION - Solve the simultaneous equations to get the two

unknown forces.

From eq. (1):

FAB =0.866 FAC

0.0349

Substitude in eq. (2)

0.99940.866 FAC

0.0349⎛⎝⎜

⎞⎠⎟− 0.5FAC − 3500 = 0

FAC = 144 Ib

Then substitude FAC into eq. (1) again

FAB =0.866 144( )

0.0349FAB = 3570Ib

A

3570 Ib

144 Ib

3500 Ib



CASE STUDY 2 (2-D) Determine the required length of cord AC so that 8-kg lamp can be suspended in the position shown below. The undeformed length of spring AB is 0.4 m and the spring stiffness is 300 N/m.

EXAMPLE 3.4

3.3 COPLANAR FORCE SYSTEMS 93

3

Determine the required length of cord AC in Fig. 3–8a so that the 8-kg lamp can be suspended in the position shown. The undeformedlength of spring AB is and the spring has a stiffness ofkAB = 300 N>m.

l¿AB = 0.4 m,

(a)

A B

! 300 N/m30"

2 m

C

kAB

Fig. 3–8

SOLUTIONIf the force in spring AB is known, the stretch of the spring can befound using From the problem geometry, it is then possible tocalculate the required length of AC.

Free-Body Diagram. The lamp has a weight and so the free-body diagram of the ring at A is shown in Fig. 3–8b.

Equations of Equilibrium. Using the x, y axes,

Solving, we obtain

The stretch of spring AB is therefore

so the stretched length is

The horizontal distance from C to B, Fig. 3–8a, requires

Ans.lAC = 1.32 m

2 m = lAC cos 30° + 0.853 m

lAB = 0.4 m + 0.453 m = 0.853 m

lAB = l¿AB + sAB

sAB = 0.453 m

135.9 N = 300 N>m1sAB2TAB = kABsAB;

TAB = 135.9 N

TAC = 157.0 N

TAC sin 30° - 78.5 N = 0+ c©Fy = 0;TAB - TAC cos 30° = 0:+ ©Fx = 0;

W = 819.812 = 78.5 N

F = ks.

y

x

W ! 78.5 N

A

(b)

30"

TAC

TAB



CASE STUDY 3 (2-D) A load is applied to the pulley C, which can roll on cable ACB. The pulley is held in the position by second cable CAD, which passes over pulley A and supports a load P. Knowing that P = 750 N, determine a) tension in cable ACB and magnitude of load Q.

45Problems

A

D

B

C

P

25°

55°

Q

Fig. P2.69 and P2.70

2.66 A 200-kg crate is to be supported by the rope-and-pulley arrange-ment shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chap. 4.)

2.4 m

P

A

α

200 kg

0.75 m

B

Fig. P2.66

T

TT T T

(a) (b) (c) (d) (e)

Fig. P2.67

2.67 A 600-lb crate is supported by several rope-and-pulley arrange-ments as shown. Determine for each arrangement the tension in the rope. (See the hint for Prob. 2.66.)

2.68 Solve parts b and d of Prob. 2.67, assuming that the free end of the rope is attached to the crate.

2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Know-ing that P 5 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.

2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a sec-ond cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P.

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CASE STUDY 4 (3-D) The crate is supported by two cables and one spring-cord. Determine the tensions in all cables and spring cord. Furthermore, how much spring is deformed under its own tension?


3

EXAMPLE 3.8

y1 m2 m

z

60! 135!2 m

D

120!

x

(a)

B

A

k " 1.5 kN/m

C

Fig. 3–13

y

x

z

W " 981 N

A

FC

(b)

FD

FB

Determine the tension in each cord used to support the 100-kg crateshown in Fig. 3–13a.

SOLUTIONFree-Body Diagram. The force in each of the cords can bedetermined by investigating the equilibrium of point A.The free-bodydiagram is shown in Fig. 3–13b. The weight of the crate is

Equations of Equilibrium. Each force on the free-body diagram isfirst expressed in Cartesian vector form. Using Eq. 2–9 for andnoting point for we have

Equilibrium requires

W = 5-981k6 N= -0.333FD i + 0.667FDj + 0.667FDk

FD = FD c -1i + 2j + 2k

2 1-122 + 1222 + 1222 d= -0.5FC i - 0.707FC j + 0.5FCk

FC = FC cos 120°i + FC cos 135°j + FC cos 60°k

FB = FB i

FD,D1-1 m, 2 m, 2 m2 FC

W = 10019.812 = 981 N.

Equating the respective i, j, k components to zero,(1)

(2)

(3)

Solving Eq. (2) for in terms of and substituting this into Eq. (3)yields is then determined from Eq. (2). Finally, substituting theresults into Eq. (1) gives Hence,

Ans.

Ans.

Ans.FB = 694 N

FD = 862 N

FC = 813 N

FB.FDFC.

FCFD

0.5FC + 0.667FD - 981 = 0©Fz = 0;

-0.707FC + 0.667FD = 0©Fy = 0;

FB - 0.5FC - 0.333FD = 0©Fx = 0;

- 0.333FD i + 0.667FD j + 0.667FDk - 981k = 0

FB i - 0.5FC i - 0.707FC j + 0.5FCk

FB + FC + FD + W = 0©F = 0;



CASE STUDY 5 (3-D) The shear leg derrick is used to haul the 200-kg net of fish onto the deck. Determine the compressive force along each of the leg AB and CB, and the tension in the winch cable BD.


3

All problem solutions must include an FBD.

•3–45. Determine the tension in the cables in order tosupport the 100-kg crate in the equilibrium position shown.

3–46. Determine the maximum mass of the crate so that thetension developed in any cable does not exceeded 3 kN.

*3–48. Determine the tension developed in cables , ,and required for equilibrium of the 300-lb crate.

•3–49. Determine the maximum weight of the crate so thatthe tension developed in any cable does not exceed 450 lb.

ADACAB

3–47. The shear leg derrick is used to haul the 200-kg net offish onto the dock. Determine the compressive force alongeach of the legs AB and CB and the tension in the winchcable DB.Assume the force in each leg acts along its axis.

2.5 m2 m

2 m

2 m

1 mA

z

D

yx

B

C

Probs. 3–45/46

4 m

4 m

2 m2 m

5.6 m

D

B

C

Ax

y

z

Prob. 3–47

A

D

C

x

1 ft

3 ft

2 ft 1 ft2 ft

2 fty

z

2 ft

B

Probs. 3–48/49

3 ft d yx

C

D

B

A

3500 lb

4 ft

3 ft

10 ft

4 ft

2 ft

z

Probs. 3–50/51

PROBLEMS

3–50. Determine the force in each cable needed tosupport the 3500-lb platform. Set .

3–51. Determine the force in each cable needed tosupport the 3500-lb platform. Set .d = 4 ft

d = 2 ftUniKLMalaysian Institute of Aviation Technology


CONCEPTUAL PROBLEM 1

102 CH A P T E R 3 EQ U I L I B R I U M O F A PA RT I C L E

3

A

B C

B

A C

A

CB

D

F

B

A

B¿C

CONCEPTUAL PROBLEMS

P3–1. The concrete wall panel is hoisted into position usingthe two cables AB and AC of equal length. Establishappropriate dimensions and use an equilibrium analysis toshow that the longer the cables the less the force in each cable.

P3–4. The two chains AB and AC have equal lengths andare subjected to the vertical force F. If AB is replaced by ashorter chain , show that this chain would have tosupport a larger tensile force than in order to maintainequilibrium.

ABAB¿

P3–2. The truss is hoisted using cable ABC that passesthrough a very small pulley at B. If the truss is placed in atipped position, show that it will always return to thehorizontal position to maintain equilibrium.

P3–3. The device DB is used to pull on the chain ABC soas to hold a door closed on the bin. If the angle between ABand the horizontal segment BC is 30º, determine the anglebetween DB and the horizontal for equilibrium.UniKL

Malaysian Institute of Aviation Technology

102 CH A P T E R 3 EQ U I L I B R I U M O F A PA RT I C L E

3

A

B C

B

A C

A

CB

D

F

B

A

B¿C

CONCEPTUAL PROBLEMS

P3–1. The concrete wall panel is hoisted into position usingthe two cables AB and AC of equal length. Establishappropriate dimensions and use an equilibrium analysis toshow that the longer the cables the less the force in each cable.

P3–4. The two chains AB and AC have equal lengths andare subjected to the vertical force F. If AB is replaced by ashorter chain , show that this chain would have tosupport a larger tensile force than in order to maintainequilibrium.

ABAB¿

P3–2. The truss is hoisted using cable ABC that passesthrough a very small pulley at B. If the truss is placed in atipped position, show that it will always return to thehorizontal position to maintain equilibrium.

P3–3. The device DB is used to pull on the chain ABC soas to hold a door closed on the bin. If the angle between ABand the horizontal segment BC is 30º, determine the anglebetween DB and the horizontal for equilibrium.


CONCEPTUAL PROBLEM 2 UniKLMalaysian Institute of Aviation Technology


CONCLUSION •  The concept of FBD is very important in

equilibrium analysis.

•  When particle is in equilibrium, the summation of forces in the respective axis of specific coordinate system must equal to zero.

•  The equilibrium equations will allow us to solve the magnitudes of unknown forces.


statics of a particle- 2d & 3d equilibrium analysis

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