Statics of a Particle- 2D & 3D equilibrium analysis

o satisfy this equation we require

3

©Fx = 0- For internal use only ©Fy = 0 (3–5) ©Fz = 0

y x F1

Fig. 3–9

at the algebraic sum of the components of article along each of the coordinate axes e can solve for at most three unknowns, dinate direction angles or magnitudes of free-body diagram.

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/203/MHDQ294/bee02286_disk1of1/0077402286/bee02286_pagefiles

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

Analysis

C H A P T E R

uilibrium problems for a particle can be rocedure.

UNIVERSITI KUALA LUMPUR

Malaysian Institute of Aviation Technology

n any suitable orientation. and unknown force magnitudes and m. aving an unknown magnitude can be

ns of equilibrium, ©Fx = 0, ©Fy = 0, it is easy to resolve each force into its UniKL MIAT – Jan 2015 geometry appears difficult, then first

Uni KL

W

A

FB C

B

FD

FC

D

15

Mulia Minhat

Uni KL

PARTICLE ANALYSIS

Malaysian Institute of Aviation Technology

Forces acting on a particle y

Body under consideration

F1

F3 x

2-D F2

Equilibrium analysis

! ! ! !

F1 + F2 + F3 = FR = 0

UniKL MIAT - MM2015

y

F1

F3 x

(∑ F = 0, ∑ F = 0, ∑ F = 0) x

z

z

y

3-D F2

2

OBJECTIVES

Uni KL Malaysian Institute of Aviation Technology

• 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 ( ∑ F = 0 ) due to the equilibrium condition.

UniKL MIAT - MM2015

3

CASE STUDY 1 (2-D)

Uni KL Malaysian Institute of Aviation Technology

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.

UniKL MIAT - MM2015

4

Uni KL

THEORY

Malaysian Institute of Aviation Technology

- 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.

B

FBD of particle A A

FAB

C y

D 10 Ib UniKL MIAT - MM2015

A

FAC

∑ F = 0,

x

W = 10lb 5

Uni KL

THEORY

Malaysian Institute of Aviation Technology

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

C θ lo = 0.4 m

A

s = - 0.15 m

B T

TAC = TBC UniKL MIAT - MM2015

T

F = k(xf-xi)

s = 0.3 m

F = kΔx

6

Uni KL

THEORY

Malaysian Institute of Aviation Technology

- Equilibrium conditions: Summation of forces in the respective axis of a coordinate system equals zero.

2-D Cartesian system

∑F ∑F

UniKL MIAT - MM2015

x

= 0,

y

= 0.

3-D Cartesian system

∑F ∑F ∑F

x

= 0,

y

= 0,

z

= 0.

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APPROACH TO SOLUTION

Uni KL Malaysian Institute of Aviation Technology

- 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.

UniKL MIAT - MM2015

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Uni KL

SOLUTION

Malaysian Institute of Aviation Technology

-  Draw the FBD of particle A and resolve all forces into vector form to identify the necessary components. y

FAB y

FAB

x

2O

FAB y = FAB cos 20 = 0.9994FAB

A

FAB x

FAB x = FAB sin 20 = 0.0349FAB

FAC y

30O

FAC x

FAC x = FAC cos300 = 0.866 FAC

FAC

FAC y = FAC sin300 = 0.5 FAC

3500 Ib UniKL MIAT - MM2015

9

Uni KL

SOLUTION

Malaysian Institute of Aviation Technology

- Solve the simultaneous equations to get the two unknown forces.

3570 Ib

From eq. (1): 0.866 FAC FAB = 0.0349

A 144 Ib

Substitude in eq. (2) 3500 Ib ⎛ 0.866 FAC ⎞ 0.9994 ⎜ − 0.5FAC − 3500 = 0 ⎟ ⎝ 0.0349 ⎠ Then substitude FAC into eq. (1) again FAC = 144 Ib 0.866 (144 ) FAB = 0.0349 FAB = 3570Ib UniKL MIAT - MM2015

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KL FORCE 3.3Uni COPLANAR

CASE STUDY 2 (2-D)

Malaysian Institute of Aviation Technology

EXAMPLE the 3.4 required length of cord AC so that Determine 8-kg lampthe can be suspended in inthe Determine required length of cord AC Fig.position 3–8a so that the 8-kg lamp can be suspended in the position length shown. The shown below.

The undeformed ofundeformed spring AB length of spring AB is l¿AB = 0.4 m, and the spring has a stiffness of is k0.4 m and the spring stiffness is 300 N/m.

AB = 300 N>m.

y

2m C TAC

kAB ! 300 N/m

30"

30"

A

B

A

(b

(a) UniKL MIAT - MM2015

SOLUTION

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Fi

Uni KL

CASE STUDY 3 (2-D) (e)

Malaysian Institute of Aviation Technology

A load is applied to the pulley C, which can roll on cable ACB. The pulley is held in the position by ing that the free end of second cable CAD, which passes over pulley A and supports a load P. Knowing that P = 750 N, a) tension in cable ACB and magnitude ch can rolldetermine on the cable load Q. hown by aofsecond cable

upports a load P. Knowtension in cable ACB,

ey C, which can roll on position shown by a secpulley A and supports a ACB, (b) UniKLthe MIAT -magnitude MM2015

B

A 25째 D 55째

C

P Q Fig. P2.69 and P2.70

12

Uni KL

CASE STUDY 4 (3-D)

Malaysian Institute of Aviation Technology

3.4crate THREEis -DIMENSIONAL FORCE 1 0 7 one The supported bySYSTEMS two cables and spring-cord. Determine the tensions in all cables and spring cord. Furthermore, how much spring is deformed under its own tension?

pport the 100-kg crate

z C 60! 135!

of the cords can be point A. The free-body eight of the crate is

he free-body diagram is ng Eq. 2–9 for FC and UniKL MIAT - MM2015 ve

D

120!

A B x

2m 2m 1m y

3

k " 1.5 kN/m

13

(a)

Probs. 3–45/46

Uninet KL 3–47. The shear leg derrick is used to haul the 200-kg of Malaysian Institute fish onto the dock. Determine the compressive Aviation forceTechnology alongof each of the legs AB and CB and the tension in the winch cable DB. Assume the force in each leg acts along its axis.

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.

z 5.6 m 4m B

D

4m

C A x

2m

2m y

Prob. 3–47 UniKL MIAT - MM2015

14

3–50 supp

3–51 supp

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

CONCEPTUAL PROBLEM 1

P3–3. The device D as to hold a door clos Uni and theKL horizontal se Malaysian Institute of between DB and the Aviation Technology

3

D A

B

C

15

UniKL MIAT - MM2015

P3–2. The truss is hoisted using cable ABC that passes

P3–4. The two chai

equilibrium.

CONCEPTUAL PROBLEM 2

Uni KL Malaysian Institute of Aviation Technology

F A

B UniKL MIAT - MM2015

B多

C 16

CONCLUSION

Uni KL Malaysian Institute of Aviation Technology

• 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.

UniKL MIAT - MM2015

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