5-static equilibrium

7/27/2019 5-Static Equilibrium

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Lecture #5

Translational Equilibrium

(2D Coplaner Force Systems,

Reference Chapter 3, section 3)

1R. Michael PE 8/14/2012


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Newtons First Law of Motion

A body at rest will stay at rest and a

body in motion will stay in motion

unless acted upon by an unbalancedforce.

Therefore, sum of all forces must be

zero:

F= 0

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Newtons Second Law of Motion

The acceleration a body undergoes

when experiencing an unbalanced

force is proportional to the magnitudeof the unbalanced force, in the

direction of the unbalanced force, and

inversely proportional to the mass of

the object

Basis of Dynamics

F= ma

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Motion Definitions

Rectilinear Motion - Motion in a

straight line.

Translational Motion - Motion wherethe body does not rotate

Again, this is studied in dynamics!

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2D Particle Static Equilibrium (i.e. Section 2.3,

Coplanar Force Systems)

When a body is in this state there areno unbalanced forces acting on it.

The Resultant Force has a magnitude

of zero.

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In general, for a particle in equilibrium,

F= 0 or Fxi+ Fyj = 0 = 0 i+ 0j (a vector equation)

Or, written in a scalar form,

Fx= 0 and Fy= 0

These are two scalar equations of equilibrium (E-of-E). They can be

used to solve for up to two unknowns.


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3D Particle Static Equilibrium

Because the resultant force is

balanced, the following situation

occurs Fx = 0

Fy = 0

Fz = 0

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Summary: 2D Static

Equilibrium:

Using Components:

Graphically:

0

0; 0; 0x y z

F

F F F

1 2 30F F F F

2F1

F

3F7


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Static Equilibrium Problems

In 2D, have 2 equations, so can solve for 2

unknowns

Find magnitudes of two forces with known

directions

Find magnitude and direction of one force,

knowing magnitude and direction of other

force(s) In 3D have 3 equations, so can solve for 3

unknowns

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Do simple 2D


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2D COPLANAR FORCE SYSTEMS (Section 3.3)

To determine the tensions in the

cables for a given weight of the

cylinder, you need to learn how to

draw a free body diagram and applyequations of equilibrium.

This is an example of a 2-D or

coplanar force system.

If the whole assembly is in equilibrium,

then particle A is also in equilibrium.

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THE WHAT, WHY AND HOW OF A FREE BODY

DIAGRAM (FBD)

Free Body Diagrams are one of the most important things for you to know

how to draw and use.

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

Why ? - It is key to being able to write the equations of equilibrium

which are used to solve for the unknowns (usually forces or angles).

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How ?

Active forces: They want to move the particle. Reactive

forces: They tend to resist the motion.

Note : Cylinder mass = 40 Kg

1. Imagine the particle to be isolated or cut free from its surroundings.

A

3. Identify each force and show all known magnitudes and directions. Showall unknown magnitudes and / or directions as variables .

FC = 392.4 N (What is

this?)

FB

FD

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2. Show all the forces that act on the particle.

FBD at A

A

y

x

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EQUATIONS OF 2-D EQUILIBRIUM

Or, written in a scalar form,Fx= 0 and Fy= 0

These are two scalar equations of equilibrium (E-of-E). They can be

used to solve for up to two unknowns.

Since particle A is in equilibrium, the net force

at A is zero.So FB + FC + FD = 0

or F= 0

FBD at A

A

In general, for a particle in equilibrium,

F= 0 or Fxi+ Fyj = 0 = 0 i+ 0j (a vector equation)

FBD at A

A

FB

FDA

FC = 392.4 N

y

x30

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EXAMPLE

Write the scalar E-of-E:

+ Fx = FB cos 30 FD = 0

+ Fy = FB sin 30 392.4 N = 0

Solving the second equation gives: FB= 785 N

From the first equation, we get: FD= 680 N

Note : Cylinder mass = 40

Kg

FBD at A

A

FB

FDA

FC = 392.4 N

y

x30

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SPRINGS, CABLES, AND PULLEYS

With a frictionless pulley,T1 = T2= T

Spring Force = spring constant *deformation, or

F = k * s

T1

14

T1 T2

See HO


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EXAMPLE

1. Draw a FBD for Point C.

2. Apply E-of-E at Point C to solve for the unknowns (FCB & FCD).

3. Knowing FCB , repeat this process at point B.

Given: Cylinder E weighs 30 lb

and the geometry is as

shown.

Find: Forces in the cables and

weight of cylinder F.

Plan:

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EXAMPLE(continued)

The scalar E-of-E are:+ Fx = FBC cos 15 FCD cos 30 = 0

+ Fy = FCDsin 30 FBC sin 15 30 = 0

A FBD at C should look like the one at the left.

Note the assumed directions for the two cable

tensions.

Solving these two simultaneous equations for the two

unknowns FBC and FCD yields:

FBC = 100.4 lb

FCD = 112.0 lb

FCD

FBC30 lb

y

x3015

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EXAMPLE(continued)

Fx = FBA cos 45 100.4 cos 15 = 0

Fy = FBA sin 45 + 100.4 sin 15 WF = 0

The scalar E-of-E are:

Now move on to ring B.A FBD for B should look like the

one to the left.

Solving the first equation and then the second yields

FBA = 137 lb and WF = 123 lb

FBC =100.4 lb

FBA

WF

y

x15

45

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GROUP PROBLEM SOLVING

Plan:

1. Draw a FBD for point A.

2. Apply the E-of-E to solve for the forces in

ropes AB and AC.

Given: The box weighs 550 lb and

geometry is as shown.

Find: The forces in the ropes AB and AC.

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GROUP PROBLEM SOLVING (continued)

Applying the scalar E-of-E at A, we get;

+ F x = FB cos 30 FC (4/5) = 0

+ F y = FB sin 30 + FC (3/5) - 550 lb = 0

Solving the above equations, we get;

FB = 478 lb and FC = 518 lb

FBD at point AFCFB

A

FD = 550 lb

y

x30

3

4

5

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ATTENTION QUIZ

A

30

40

100 lb

1. Select the correct FBD of particle A.

A) A

100

lb

B)

30

40

A

F1

F2

C) 30A

F

100 lb

A

30 40F1 F2

100 lb

D)

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ATTENTION QUIZ

F2

20 lb

F1

C

50

2. Using this FBD of Point C, the sum of forces in the

x-direction ( FX) is ___ . Use a sign convention

of + .A) F2 sin 50 20 = 0

B) F2 cos 50 20 = 0

C) F2 sin 50 F1 = 0

D) F2 cos 50 + 20 = 0

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READING QUIZ

1) When a particle is in equilibrium, the sum of forces acting

on it equals ___ . (Choose the most appropriate answer)

A) A constant B) A positive number C) Zero

D) A negative number E) An integer

2) For a frictionless pulley and cable, tensions in the cable(T1 and T2) are related as _____ .

A) T1 > T2

B) T1

= T2

C) T1 < T2

D) T1 = T2 sin

T1

T2


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Determine the forces in cablesACandAB needed to hold the 20-kg

ball D in equilibrium. Take F= 300 N and d= 1 m

Practice Problem 1

Reference 3-18/19.


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Determine the tensions developed in wires CD, CB and BA and the

angle required for equilibrium of the 30 lb cylinderEand 60 lbcylinderF.

Practice Problem 2

Reference 3-26.


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Determine the stretch in springsACandAB for equilibrium of the 2-

kg block. The springs are shown in the equilibrium position.

Practice Problem 3

Reference 3-14.

5-static equilibrium

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