Chapter 2Chapter 2

Equilibrium of Particle in 2-D

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2

Objectives of Chapter 2

1. Draw free-body diagram (FBD) for a particle.

2. Apply equations of equilibrium (EoE) to solve 2-D problems.

Students will be able to:

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Definition1. A particle will be in equilibrium if it continues

to be at rest if originally at rest or moves with a constant velocity if originally in motion.

2. Static equilibrium indicates a body at rest.3. Newton’s 1st law states that a body at rest

does not have any unbalanced forces, i.e.

F = 0

For given weight of the traffic lights, what are the forces in the cables? What size of cable must you use?

For a spool of given weight, what are the forces in cables AB and AC ?

Application: Equilibrium Of A Particle

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This is an example of a 2-D (coplanar) force system. If the whole assembly is in equilibrium, then particle A is also in equilibrium. To determine the tensions in the cables for a given weight of the engine, we need to know how to draw a free body diagram and apply equations of equilibrium.

Equilibrium of a Particle (2D) Equilibrium of a Particle (2D)

250 kg

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THE WHAT, WHY AND HOW OF A FREE BODY DIAGRAM (FBD)

Free Body Diagram is one of the most important things you must know how to draw and use.

What is FBD? - It is a drawing that shows all the external forces acting on the particle or body.

Why draw? – FBD helps you to write the equations of equilibrium used to solve for the unknowns (usually forces or angles)

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2. Imagine the particle isolated or cut “free” from surroundings or system of which it is part.

3. Show all the forces that act on the particle – these forces can be “active forces” (tend to move the particle), or “reactive forces” (tend to resist motion; caused by supports or constraints)

4. Identify each force, and show all known magnitude and direction. Show all unknown forces as variables. Direction of a force of unknown magnitude can be assumed first.

How to draw FBD for a Particle?

1. Identify the point you wish to analyze. This is often dictated by the forces you wish to determine.

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NOT a free-body diagram!!

FBD of Point A

W = (250 kg)(9.81 ms-2)

EQUATIONS OF 2-D EQUILIBRIUM

Since particle A is in equilibrium, the net force at A will be zero.

So FAB + FAB + FAC = 0

i.e. ∑F = 0

In general, for a particle in equilibrium, ∑F = 0

Written in a scalar form:∑Fx = 0 and ∑Fy = 0

Both are scalar equations of equilibrium (EoE) in 2-D.They can be used to solve up to 2 unknowns. 9

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Exercise: Draw Free-Body Diagram

The sphere has a mass of 6 kg and is supported as shown. Draw FBD of the sphere, the cord CE and the knot at C.

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(FCE, force of cord CE acting on sphere) FCE

W = mg = 58.9 N (W, weight or gravity

acting on sphere)

FBD of sphere: only two forces acting on sphere: its weight W and force of the cord CE.

FBD of cord CE: When isolated, only two forces act on cord CE: force of sphere and force of knot

FCE

(FCE, force of sphere acting on cord CE)

(Force of knot acting on cord CE) FEC

C

E

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FBD of knot C: Three forces act on the knotas shown below.

FCBA

FCD

(Force of spring acting on knot)

FCE (Force of cord CE acting on knot)

(Force of cord CBA acting on knot)

Note: Weight of sphere, W does not directly act on the knot. Instead, the cord CE subjects the knot to this force.

EXAMPLE: FBD of a particle on an inclined surface

MP

θ

β

α

P

θ

β

α

mg

N

FBD

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Types of Connections Encountered in Particle

Equilibrium

1. Cables and Pulleys

2. Springs

All cables are assumed to be weightless and non-stretchable (i.e. not extensible).

All forces acting from cables direct away from point of analysis, i.e. cables are ALWAYS in tension (you can’t push on a rope).

When a cable passes a pulley, the tension is the same as along as it is the same cable.

Pulleys are assumed to be smooth (frictionless) unless stated otherwise.

Dimensions of a pulley are usually neglected in calculations, except stated otherwise.

Cables and Pulleys

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Cables and Pulleys

Cable is always in tension. Tension is the same as long as it is the same cable.

m

T

mg

2T

m

T

mg

2T

mg

3T

m

T m

B

A

β

P

θ

mg

2P P

βθ

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Springs

• Magnitude of force exerted on a spring of stiffness k, deformed a distance s = l – lo, measured from its unloaded position is F = ks.

• If s is positive (elongation), F must pull on the spring; if s is negative (compressed), then F must push on it.

• Example: if spring with an unstretched length lo of 0.4m and k = 500 N/m is to be stretched to 0.6m, s = l – lo = 0.2m and F = ks = 100N is needed.

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Springs

Procedure for Using FBD to Solve forEquilibrium of Particle

Free-Body Diagram Establish x and y axes in any suitable orientation. Show all known and unknown force magnitudes and directions on FBD diagram.

Equation of Equilibrium Apply EoE: Fx = 0 and Fy = 0 Components are positive if they are directed along the positive axis and negative if directed along negative axis. If more than 2 unknowns exist and problem involves spring, apply F = ks to relate spring force F to the deformation s of the spring. If the solution for a force is negative, this means the sense of this force is the opposite (reverse) to what you initially assumed.

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Example 2.11

Determine the tension in cables AB and BC for the system to be in equilibrium

200 kg

20°40°

A

B

C

Ans: TAB = 2129N, TBC = 1735N 21

Solution 1: Trigonometry and force triangle

1962N

TBC

TAB

40°60°

70°

50°

Start, end

Sine Law:

1962

sin60o

TAB

sin70o

TBC

sin50o

TAB

1962sin70o

sin60o2129 N

TBC

1962sin50o

sin60o1735 N

Example 2.11

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Solution 2: Rectangular component

B

C

1962 N

TBCcos20°

TBCsin20°TABsin40°

TABcos40°

Example 2.11

Fx 0

TBC cos 20o ( TAB cos 40o) 0

TBC 0.8152 TAB ...... (1)

+ Fy 1962NTAB sin 40o TBC sin 20o = 1962N ...... (2)

Substitute equ. (1) into equ. (2):

TAB = 2129N TBC = 1735N

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Solution 3: Rectangular component method (slanting axes)

x’

y’

B

1962 N

TBC

1962 N sin20°

TABsin30°

TABcos30°

x’

1962 N cos20°

y’

All forces are resolved into their component x’ and y’ component. Usually, the axis is positioned such that one axis lies on an unknown quantity ( TBC )

Advantage: directly solve one unknownDisadvantage: must be good in determining angles

+ve : 0

1962cos20 cos30 0

2129 N

+ve: 0

1962sin20 sin3

1735

0 0

N

y

AB

x

BC

A

BC

B

B

A

F

T

F

T T

T

T

o o

o o

]

Z

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Given: Bag A weighs 20 lb. and geometry is as shown.

Find: Forces in the cables and weight of bag B.

Plan:

1. Draw a FBD for Point E.

2. Apply EoE at Point E to solve for the unknowns (TEG & TEC).

3. Repeat this procedure at C to find WB

Example 2.12

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: Fx TEG sin 30o TEC cos 45o 0

: Fy TEG cos 30o TEC sin 45o 20lb 0

A FBD of point E should look like the one to the right. Note the assumed directions for both cables: tension.

Solving these two simultaneous equations for the two unknowns yields:

TEC = 38.6 lb and TEG = 54.6 lb

Eqn. of equilibrium (in scalar form):

Example 2.12

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: Fx 38.64lb cos 45o 45 TCD 0

: Fy 35 TCD 38.64lb sin 45o WB 0

Solving the first equation, then the second yields:

TCD = 34.2 lb and WB = 47.8 lb .

Next, move on to ring C. FBD of C should look like the one on the left.

Eqn. of equilibrium (in scalar form):

Example 2.12

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600 N

C

BA 300500

Example 2.13

Given: System in equilibrium.

Find: Tension in cables AC and BC

Plan:1. Draw a FBD for point equilibrium.

2. Apply EoE to solve for the angle.

Soln: Same as in Example 2.11.

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600 N

C

TBC

TAC

300500

Point C as FBD.

Fx 0;

TAC

kos50o = TBC

kos30o

TBC

2T

ACkos50o

3 .........(1)

Fy 0;

TAC

sin50o + TBC

sin30o 600

TAC

sin50o + T

ACkos50o

3 600

TAC

600

(sin50o kos50o

3)

527.63 N

TBC

391.62 N

Example 2.13

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OR

Polygon (triangle) method

180o 60o 40o 80o

600

sin80o =

TAC

sin60o =

TBC

sin40o

TAC

= 527.63 N

TBC

= 391.62 N

600N

TBC

TAC

600

400

θ

Example 2.13

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GROUP PROBLEM SOLVING

Given: The car is towed at a constant speed by the 600-lb force, given the angle θ is 25°.

Find: Forces in ropes AB and AC.

Plan:

1. Draw FBD for point A.

2. Apply EoE to solve for the forces in ropes AB and AC.

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GROUP PROBLEM SOLVING (continue)

30°25°

600 lb

FABFAC

AFBD at point A

Applying the scalar EoE at point A, we get;+ → ∑Fx = FAC cos 30° – FAB cos 25° = 0

+ ↑ ∑Fy = -FAC sin 30° – FAB sin 25° + 600 = 0

Solve the above equations, we get;FAB = 634 lb

FAC = 664 lb

Alternative soln.: Force triangle and trigonometry; see Example 2.1132

4m s

10m

Example 2.14

Given: A 50-kg boy is in equilibrium condition as shown. Length of cable is 20m.

Find: Forces in the cable and distance s.

Plan:

1. Draw FBD for point equilibrium.

2. Apply EoE at to solve for the angle.

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50 g

TT

β θ

Resolve force into x-y componentsFind angle θ and tension T

C

θθ

A

B

L 20-L

10 m

4 m

Distance s:

Example 2.14

10 m

4 (L sin ) (20 L) sin

2 L sin 20 sin 4

Substitute 60o and solve for L:

2 L sin 60o 20 sin 60o 4

L 20 sin 60o 4

2 sin 60o7.69 m

s L cos 60o 3.85 m

Fx 0

T cos = T cos

=

Fy 0

2T sin = 50 (g)

If AB = L and BC = (20 L), then:

L cos + (20 L) cos = 10 m

L cos + 20 cos L cos = 10 m

cos = 10

20; = 60o

Therefore, T =50 (9.81)

2 sin 60o283.2 N

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A

BCk = 200 N/m

θ = 60°

1000 mm

750 mm

Example 2.15a

Given: Unstretched length of spring is 500 mm.

Find: a. Stretch of spring, sb. Tension in springc. Tension in cord ABd. Mass of ball

System is in equilibrium as shown

Answer: s = 125mm; T = 25N; TAB = 50N; m = 4.41kg 35

a) Final stretch of spring:

x1 = 1000 mm – 750 mm cos 60° = 625 mms = 625 mm – lo = 125 mm

b) Tension in spring, FBC = ks = 200N/m(0.125m) = 25N

c) Tension in cord AB, TAB = 25N/cos 60o = 50 N

d) Mass of ball, W = mg = 50 N sin 60o

m = 4.41 kg

Example 2.15a

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Example 2.15b

Given: Determine the required length of

cord AC so that the 8-kg lamp can be hung as shown. The undeformed length of spring AB is 0.4m; spring stiffness is kAB = 300 N/m

Find: a. Tension in spring ABb. Tension in cord ACc. Stretch of springd. Length of cord ACSolution:

1. Draw a FBD for point A.

2. Apply EoE to solve for the forces in AB and AC.

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Fx 0; TAB TAC

cos 30o = 0

+ Fy 0; TACsin 30o 78.5 N = 0

Solving: TAC 157.0 N and TAB = 135.9 N

FBD: Point A

Stretch of spring AB:

TAB kABsAB; 135.9 N (300 N/m)(sAB)

sAB 0.453m

Stretched length: LAB = Lo, AB + sAB

LAB = 0.4 m + 0.453 m = 0.853 m

Required length of cord AC:2 m = LAC cos 30° + 0.853 m

LAC = 1.32 m (ANSWER)

If the force in the spring AB is known, then the stretch of spring can be found using F = ks.

10 kg

60°

500 mm

C

B

D

A

Example 2.16

Given: System is in equilibrium.Find: a. Tension in cable AB

b. Length of cable AB

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10 g

60° θ

10 g TAB

B

Length of BC:

Example 2.16

Fx 0; TAB cos -10(g) cos 60o 0

TAB 5(g)

cos ..... (1)

+ Fy 0; 10(g) sin 60o TAB sin 10(g)0 ....(2)

Substitute eqn.(1) into eqn.(2) and rearrange:

tan = 10(g) - 5(g) 3

5(g)

tan = 0.26795

= 15o

Tension in rope AB:

TAB = 5(g)

cos =

5(9.81)

cos 15o = 50.78N

Length of AB:

500 mm

sin 60o577.35 mm

577.35 mm

sin 15o

LAB

sin 60o

LAB = 1931.85 mm40

OR Resolved forces along line of action TAB (x’) and y’:

10 g

60° θ

10 g

θ

x’

B

Length of cable AB o

AB =

500sin

1931.8

15

mAB

5 m

y

o

o

'

o

x

o

'

10gsin(60 ) 10 sin(90 )

(60 ) (90 )

2

+ve F 0;

15

+ve F 0;

= 0

3

^

Z

o

o

g

A

oAB

B

T 20

T 50.78

gkos(75

N

)

Example 2.16

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The beam has a weight of 3.5kN. Determine the shortest cable ABC that can be used to lift the beam if the maximum force that the cable can sustain is 7.5kN.

Example 2.17

= 13.5° LABC = 3.09m42

Example 2.18

Fx 0; T cos - T cos = 0

=

Fy 0; 2T sin - 49.05N = 0

= tan-1 0.15m

0.20m

36.87o

T= 40.9N

If the 5-kg block is suspended from the pulley B and the sag of the cord is d = 0.15m, find the tension in the cord ABC. Neglect size and weight of pulley.

Note: Similar to Example 2.14W = 5 g

TT

β θ

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Ujian 1 18/10/2010

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Fx 0; - TAC cos + TAB = 0

TAC = TAB

cos ...... (1)

Fy 0; TAC sin 2452.5 N = 0 ...... (2)

FBD: Ring A

Eq.(1) shows TAC is always ≥ TAB since cos ≤ 1. Thus, rope AC will reach maximum tensile force of 12.5 kN before rope AB.

Substitute TAC = 12.5 kN into equ.(2):

(12500 N) sin - 2452.5 N = 0

sin = 2452.5 N

12500 N = 0.1962

= sin-1 0.1962 = 11.31o

Solve for tension in rope AB, use equ.(1):

TAC = TAB

cos

TAB = 12.5 kN cos 11.31o 12.557 kN

Solution

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Fx 0; - TAC cos + TAB = 0

- TAC cos + 12500 N = 0

TAC cos = 12500 N ...... (1)

Fy 0; TAC sin (250)(9.81) = 0

TAC sin = 2452.5 N ...... (2)

Solve (2)(1): TAC sin TAC cos

tan 2452.5

125000.1962

11.10o

Substitute 11.10o into either eqn.(1) or eqn.(2):

TAC cos = 12500 N

TAC = 12500N

cos 11.10o = 12738.3N (>12500N)

FBD: Ring A

Case 1: Assume TAB = 12.5 kN

Alternative Solution

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Fy 0; TAC sin (250)(9.81) = 0

12500 N sin = 2452.5 N

sin = 2452.5

12500 = 0.1962

= sin-1 0.1962 = 11.315o

Fx 0; - TAC cos + TAB = 0

-12500 cos 11.315o +TAB = 0 TAB = 12257 N (< 12500 N)

Therefore, smallest angle: 11.31o

FBD: Ring A Case 2: Assume TAc = 12.5 kN

Alternative Solution (continue)

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