chapter 9-static equilibrium - university of...

Chapter 9- Static Equilibrium

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The following changes will take place until the end of the semester

Office-hours:

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- Wednesday, 14:00-15:00h

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can be picked up in my office

(LB-212)

All marks, including assignments, have

been posted on the web.

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Please, verify that all your marks have

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Chapter 9

• The Conditions for Equilibrium

• Solving Statics Problems

Recalling Recalling LastLast LecturesLecturesRecalling Recalling LastLast LecturesLectures

Type Linear Rotational Relation

Displacement x θ x = rθ

Velocity v ω v = rω

Acceleration a αααα a = rαααα

Dynamics F (force) ττττ (torque) ττττ = Fd

Inertia m (mass) I (moment of inertia)

Angular Momentum and Its ConservationAngular Momentum and Its ConservationWe have defined several angular quantities in analogy to linear motion.

Inertia m (mass) I (moment of inertia)

Kinetic Energy

Newton’s 2nd Law

Equations of motion

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

1) Angular momentum

L is called angular momentum.

(8-26)


2) Work in angular momentum

Eq. 8-27 represents the work done by the torque ττττ when rotating an object through an angle ∆θ.

(8-27)

Power P (rate with it the work is done):

(8-28)


3) Impulse in angular momentum

Impulse in rotational motion as

(8-29)


4) Generalized for of Newton’s 2nd Law

General form of the rotational equivalent of Newton’s 2nd law:

Changes in the momentum of inertia ( L = Iωωωω ) are taken into account in the

(8-30)

Changes in the momentum of inertia ( L = Iωωωω ) are taken into account in the formalism of Newton’s 2nd law.


5) Conservation of Angular Momentum

The total angular momentum is conserved if there is NO external net torque

acting on the system.

(8-31)


Since NO external torque is applied on her, her

total angular momentum should be conserved:

Therefore, if her momentum of inertia decreases,

her angular velocity has to increase to maintain

the total angular momentum constant.

TodayTodayTodayToday

Problem 8-62 (textbook): A 4.2-m-diameter merry-go-round is rotating freely with

an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 Kg.m2. Four

people standing on the ground, each of mass 65 kg, suddenly step onto the edge of

the merry-go-round. What is the angular velocity of the merry-go-round now?

What if the people were on it initially and then jumped off in a radial direction (relative

to the merry-go-round)?

Problem 8-62 (textbook):

The angular momentum of the merry-go-round and people combination will be

conserved because there are no external torques on the combination. This situation

is a totally inelastic collision, in which the final angular velocity is the same for both

the merry-go-round and the people.

Subscript 1 represents before the collision, and subscript 2 represents after the

collision. The people have no initial angular momentum.

1 2 1 1 2 2 L L I Iω ω= → = →

.

If the people jump off the merry-go-round radially, then they exert no torque on the

merry-go-round, and thus cannot change the angular momentum of the merry-go-

round. The merry-go-round would continue to rotate at .0.80 rad/s.

1 2 1 1 2 2

( )( )( )

m-g-r m-g-r1

2 1 1 1 2

2 m-g-r people m-g-r person

2

22

4

1760 kg m 0.80 rad s 0.48 rad s

1760 kg m 4 65 kg 2.1 m

I II

I I I I M Rω ω ω ω= = =

+ +

= =+

�

�

The Conditions for EquilibriumThe Conditions for Equilibrium

So far, we have worried about motion (velocity) and change of motion (net

acceleration).

However, there are many situations where we want to make sure objects remain

in their original configuration of motion: at rest or constant velocity.

In this lecture, we will deal with that situation where the original state of motion of

an object is such that it is at rest.an object is such that it is at rest.

As we have seen, for an object to remain in its state of rest � NO net force

should act on it � in other words, it does not have any acceleration.

If external or internal net forces acts on an object, it can deform or fracture it.

�We want to find a state the equilibrium such that the object is not subjected to

fractures or deformations


Static

Static is the area of physics concerned about the calculation of the forces acting

on structures that are in equilibrium.

� Determining these forces allow engineers to build structures that can

significantly minimize effects due to deformation or fracture.

No matter what, there is always at least one force acting on a object: gravity.

If, for example, a book is at rest on a table,

the normal force has to balance the

gravitational force in order for the book to

be found in equilibrium.


It is clear that if an object is to stay in its state of equilibrium

� NO net forces NOR net torque should act on the object

� In other words � NO changes in translational or

rotational motion must occur.

The chandelier should not move in the vertical,

horizontal or circular direction for equilibrium

to be met.


First condition of equilibrium

� No change in translational motion

The first condition for equilibrium is that the forces

along each coordinate axis add to zero.

(9-1)

In the figure, the equilibrium condition implies that:

120o


Second condition of equilibrium

� No change in rotational motion

The second condition of equilibrium is that there be

no torque around any axis; the choice of axis is

arbitrary and we will assume that its direction is

perpendicular to the xy plane of rotation.

(9-2)

In the figure, the sign supported by a beam of

length l mounted by a hinge on a wall is in

equilibrium. Chosen the hinge as the center of

rotation, and counterclockwise as positive:

l

l/2


Recalling an important note:

We consider that the mass of an object with

uniform mass distribution (or uniform object for

short) is such that it can be assumed to be

located at the center of the object.

This point is know as either center of mass or

center of gravity of the object.center of gravity of the object.

In the figure, assuming an uniform beam, its

center of mass will be located at the middle of the

beam (and this is why the force of gravity is

shown to be acting on this point).

l

l/2

For an uniform sphere, its center of mass is located in its center.

Solving Statics ProblemsSolving Statics Problems1. Choose one object at a time, and make a free-body diagram showing all the

forces on it and where they act.

2. Choose a coordinate system and resolve forces into components.

3. Write equilibrium equations for the forces.

4. Choose any axis perpendicular to the plane of the forces and write the torque 4. Choose any axis perpendicular to the plane of the forces and write the torque

equilibrium equation. A clever choice here can simplify the problem

enormously.

5. Make sure you correctly identify the components of the forces that can

contribute to rotational motion. Forces that have zero lever arm do not

contribute to rotational motion.

6. Solve.

Solving Statics ProblemsSolving Statics Problems

If you do not know the direction a force, just assume a generic one.

For example, assume that FB points up and FA is unknown in the figure below.

You initially assume that FA points up and then try to find the conditions of

equilibrium for this problem. You also assume that the axis of rotation passes

through the point of application of FA and that counterclockwise is positive.

FA in your solution comes negative. It just means that it’s in the opposite direction

from the one you chose. This is trivial to fix, so don’t worry about getting all the

signs of the forces right before you start solving.

Solving Statics ProblemsSolving Statics Problems

If there is a cable or cord in the problem, it can support forces only along its

length. Forces perpendicular to that would cause it to bend.

Problem 9-1 (textbook): Three forces are applied to a tree sapling, as shown in

Fig. 9–41, to stabilize it. If and , find in magnitude and

direction.

Problem 9-1 :

If the tree is not accelerating, then the net force in all

directions is 0.

( )A B C

C A B

cos110 0

cos110 310 N 425 N cos110 164.6 N

sin 0

x x

x

F F F F

F F F

F F F

= + + = →

= − − = − − = −

= + = →

∑

∑

And so is 430 N, at an angle of 112o clockwise from .

( )B C

C B

sin 0

sin110 425 N sin110 399.4 N

y y

y

F F F

F F

= + = →

= − = − = −

∑

( ) ( )2 22 2 2

C C C

C1 1 o o o o o

C

164.6 N 399.4 N 432.0 N 4.3 10 N

399.4 Ntan tan 67.6 , 180 67.6 112.4 112

164.6 N

x y

y

x

F F F

F

Fθ φ− −

= + = − + − = ≈ ×

−= = = = − = ≈

−

Problem 9-18 (textbook): Calculate

(a) the tension FT in the wire that supports the 27-kg beam shown in Fig. 9–52, and

(b) the force exerted by the wall on the beam (give magnitude and direction).WF

r

Problem 9-18 :

(a)

The beam is in equilibrium, and so the net force and net torque

on the beam must be zero. From the free-body diagram for the

beam, calculate the net torque (counterclockwise positive) about

the wall support point to find , and calculate the net force in

both the x and y directions to find the components of . WF

rTF

r

L

( )( )

o

T

2

2

T o o

sin 40 2 0

27 kg 9.80 m s205.8 N 2.1 10 N

2sin 40 2sin 40

F L mg L

mgF

τ = − = →

= = = ≈ ×

∑

( )o o o

W T W T

o

W T

cos 40 0 cos 40 205.8 N cos 40 157.7 N

sin 40 0

x x x

y y

F F F F F

F F F mg

= − = → = = =

= + − = →

∑∑

( )( ) ( )o 2 o

W Tsin 40 27 kg 9.80 m s 205.8 N sin 40 132.3 N

yF mg F= − = − =

L

Problem 9-18 :

(b)

( )o o o

W T W T

o

W T

cos 40 0 cos 40 205.8 N cos 40 157.7 N

sin 40 0

x x x

y y

F F F F F

F F F mg

= − = → = = =

= + − = →

∑∑

( )( ) ( )o 2 o

W Tsin 40 27 kg 9.80 m s 205.8 N sin 40 132.3 N

yF mg F= − = − =

( ) ( )2 22 2 2

W W W

W1 1 o

W

157.7 N 132.3 N 205.8 N 2.1 10 N

132.3tan tan 40

157.7

x x

y

x

F F F

F

Fθ − −

= + = + = ≈ ×

= = =

Problem 9-27 (textbook): Consider a ladder with a painter climbing up it (Fig. 9–

61). If the mass of the ladder is 12.0 kg, the mass of the painter is 55.0 kg, and the

ladder begins to slip at its base when her feet are 70% of the way up the length of the

ladder, what is the coefficient of static friction between the ladder and the floor?

Assume the wall is frictionless.

Problem 9-27:

The ladder is in equilibrium, so the net torque and net force must be zero.

By stating that the ladder is on the verge of slipping, the static frictional force at the

ground, , is at its maximum value and so:C xF

C Cx s yF Fµ=

Since the person is standing 70% of the way up Since the person is standing 70% of the way up

the ladder, the height of the ladder is:

The width of the ladder is:

0.7 2.8 m 0.7 4.0 my yL d= = =

0.7 2.1 m 0.7 3.0 mx xL d= = =

Problem 9-27

Torques are taken about the point of contact of the ladder

with the ground, and counterclockwise torques are taken

as positive. The three conditions of equilibrium are as follows:

C W C W0

x x xF F F F F= − = → =∑

G0

y yF F Mg mg= − − = →∑( ) ( )( )

( )

G

2

G

1

W 2

67.0 kg 9.80 m s 656.6 N

0

y y

y

y x x

F M m g

F L mg L Mgdτ

= + = =

= − − =

∑

∑Solving the torque equation gives:

( )( ) ( )( ) ( )11

222

W

12.0 kg 3.0 m 55.0 kg 2.1 m9.80m s 327.1 N

4.0 m

x x

y

mL MdF g

L

++= = =

The coefficient of friction then is found to beG

G

327.1 N0.50

656.6 N

x

s

y

F

Fµ = = =

chapter 9-static equilibrium - university of...

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