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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(LB-212)

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)

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

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)

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

3) Impulse in angular momentum

Impulse in rotational motion as

(8-29)

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

4) Generalized for of Newton’s 2 nd 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.

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

5) Conservation of Angular Momentum

The total angular momentum is conserved if there is NO external net torque acting on the system.

(8-31)

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

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-r12 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 EquilibriumSo 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 remainin their original configuration of motion : at rest or constant velocity .

In this lecture, we will deal with the 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 net acceleration .

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

� We want to find a state of equilibrium such that the object is not subjected to fractures or deformations

The Conditions for EquilibriumThe Conditions for EquilibriumStatic

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.

The Conditions for EquilibriumThe Conditions for EquilibriumIt 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 its translational or rotational motion must occur.

The chandelier should not move in the vertical, horizontal or circular directions for equilibrium to be met.

The Conditions for EquilibriumThe Conditions for EquilibriumFirst 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

The Conditions for EquilibriumThe Conditions for EquilibriumSecond 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:

ll/2

The Conditions for EquilibriumThe Conditions for Equilibrium

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 centre of the object.

This point is know as either centre of mass or centre of gravity of the object.centre of gravity of the object.

In the figure, assuming an uniform beam, its centre 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).

ll/2

For a uniform sphere, its centre 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 equations. A clever choice here can enormously simplify the problem.

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 of 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. Also assume that the point CG (center of gravity) is 25 m from this axis y

x

FA in your solution comes negative. It just means that it’s in the opposite direction from the one you had originally chosen. This is trivial to fix, so don’t worry about getting all the directions of the forces right before you start solving.

x

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).WFr

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

rTFr

L

( )( )o

T

2

2

T o o

sin 40 2 0

27 kg 9.80 m s205.8 N 2.1 10 N

2 sin 40 2 sin 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 T sin 40 27 kg 9.80 m s 205.8 N sin 40 132.3 NyF 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 T sin 40 27 kg 9.80 m s 205.8 N sin 40 132.3 NyF 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= − = → =∑

G 0 y yF F Mg mg= − − = →∑( ) ( )( )

( )

G

2

G

1W 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

222W

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

4.0 mx 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 = = =