Chap. 11: Static

Equilibrium

2

It doesn’t move!

0

0

net

net

F

Static Equilibrium

Equations

Calculate the torque along with the rotational axis!

3

Static Equilibrium

4

Real Life Phys 218

Static Equilibrium

5

Real Life Phys 218

Static Equilibrium

Static Equilibrium Analysis

0 n e t

i

Identify each force on a rigid body or a system

Calculate the torque (magnitude and direction) by each force

Not rotating

9

Static Equilibrium

We assume:

Center of mass (CM or cm)

= Center of gravity

(if uniform gravity)

? F r

10

Torque due to Gravity

Static Equilibrium

What is the unknown mass? BALANCEING

in mass x distance

12

Static Equilibrium

torque = zero

17

What is the unknown mass?

Static Equilibrium

M

m

FT = ?

A metal advertising sign (mass M) is suspended from

the end of a horizontal rod of length L and mass m.

ISEE – Where is physics?

19

Static Equilibrium

FRx – FT cosq = 0

FRy + FT sinq – mg – Mg = 0

FT sinq L – mg (L/2) – Mg L= 0

0P

M, m, L, h , q : given FRy?, FRx?, FT?

ISEE – Static Equilibrium Eqs.

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Static Equilibrium

Now You Can Do This!

M = 30.61 kg m = 5.00 kg

You design this and use a string that sustain up to 50 N.

Will your customer be happy?

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Static Equilibrium

Try This!

M

m

l

FT = ?

You design this and use a string that sustain up to 500 N.

Will your customer be happy?

M = 30.61 kg m = 5.00 kg

l = 4.00 m

33

Static Equilibrium

You design this and use a string that sustain up to 50 N.

Will your customer be happy?

34

Static Equilibrium

M, m, l, q , ms: given

x?

Sm

oo

th su

rface

Rough surface

y

Note: sinq = 4/5, cosq = 3/5

l

Particle

(mass M)

x

x

q

… Model and F.B.D.

Mg

ISEE – F.B.D. for Ladder

38

Static Equilibrium

M, m, l, q , ms: given

x? (FW?)

Sm

oo

th su

rface

Rough surface x

y

FGx – FW = 0

FGy – mg – Mg = 0

FGx = ms Fgy

FW sinq l

– mg (l/2) cosq – Mg x = 0 q

Note: sinq = 4/5, cosq = 3/5

l

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ISEE – Equilibrium Eqs.

Static Equilibrium

41

)about (rotation 0

0

0

P

F

F

ii,P

ii,y

ii,x

ISEE – Equilibrium Eqs.

P P

P

Static Equilibrium

Try this!

42

Static Equilibrium

Try this!

43

Try this!

44

You want to raise a bicycle wheel (mass m and radius R) up over a curb of height h. What is the smallest magnitude of F that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel and (b) at the top of the wheel?

r

r

F

(cw) )((cw) Fh-RFr F r F

mg

mg

hR R

22)( hRR

(ccw) )((ccw) 22

mghRRmgr gm r g

Chap. 11: Static

Equilibrium

46

Strain, stress, and elastic moduli

Strain, Stress, and Elastic Moduli • Stretching, squeezing, and twisting a real body causes it to

deform, as shown in Figure 11.12 below. We shall study the

relationship between forces and the deformations they cause.

• Stress is the force per unit area and strain is the fractional

deformation due to the stress. Elastic modulus is stress divided

by strain.

• The proportionality of stress and strain is called Hooke’s law.

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Tensile and Compressive Stress and Strain

• Tensile stress = F /A and Tensile strain = l/l0.

• Compressive stress and Compressive strain are defined in a

similar way. (See Figures 11.13 and 11.14 below.)

• Young’s modulus is tensile stress divided by tensile strain, and is

given by Y = (F/A)/(l / l0).

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Tensile Stress and Strain

• In many cases, a body can experience both tensile and

compressive stress at the same time, as shown in Figure 11.15

below.

• Follow Example 11.5.

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Stress = F/A

A

Bulk Stress and Strain • Pressure in a fluid is force per

unit area: p = F/A.

• Bulk stress is pressure change

p and bulk strain is fractional

volume change V/V0. (See

Figure 11.16.)

• Bulk modulus is bulk stress

divided by bulk strain and is

given by B = –p/(V/V0).

• Follow Example 11.6.

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Sheer Stress and Strain

• Sheer Stress is F||/A and

Sheer Strain is x/h, as shown

in Figure 11.17.

• Sheer modulus is sheer stress

divided by sheer strain, and

is given by S = (F||/A)/(x/h).

• Follow Example 11.7.

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Some Values of Elastic Moduli

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Problem 1 A wire of length l0 and cross-sectional area A supports a hanging

weight W. (a) Show that if the wire obeys Eq. (11.7), it behaves like a

spring of force constant AY/l0, where Y is Young’s modulus for the

material of which the wire is made. (b) What would the force constant

be for a 75.0-cm length of 16-gauge (diameter = 1.291 mm) copper

wire? See Table 11.1. (c) What would W have to be to stretch the wire

in part (b) by 1.25 mm?

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Problem 2 In lab tests on a 9.25-cm cube 0f a certain

material, a force of 1375 N directed at 8.50o to

the cube causes the cube to deform through an

angle of 1.24o. What is the shear modulus of

the material?

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qq tanh

x

q

S = (F||/A)/(x/h).

Elasticity and Plasticity

• Hooke’s law applies up to point a in Figure 11.18 below.

• Table 11.3 shows some approximate breaking stresses.

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Quizzes

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A. more stress and more strain.

B. the same stress and more strain.

C. the same stress and less strain.

D. less stress and less strain.

E. the same stress and the same

strain.

Q11.5

Two rods are made of the

same kind of steel and have

the same diameter.

F F length 2L

F F length L

A force of magnitude F is applied to the end of each rod.

Compared to the rod of length L, the rod of length 2L has

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Y = (F/A)(l0/l)

A. more stress and more strain.

B. the same stress and more strain.

C. the same stress and less strain.

D. less stress and less strain.

E. the same stress and the same

strain.

Q11.6

Two rods are made of the

same kind of steel. The

longer rod has a greater

diameter. F F length 2L

F F length L

A force of magnitude F is applied to the end of each rod.

Compared to the rod of length L, the rod of length 2L has

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Y = (F/A)(l0/l)

Which of the following situations satisfies both the first

condition for equilibrium (net force = 0) and the second

condition for equilibrium (net torque = 0)?

A. an automobile crankshaft turning at an increasing

angular speed in the engine of a parked car

B. a seagull gliding at a constant angle below the

horizontal and at a constant speed

C. a thrown baseball that does not rotate as it sails

through the air

D. more than one of the above

E. none of the above

Q11.1

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A. T = w sin q

B. T = w cos q

C. T = w/(sin q)

D. T = w/(cos q)

E. none of the above

Q11.3

A metal advertising sign (weight w)

is suspended from the end of a

massless rod of length L. The rod is

supported at one end by a hinge at

point P and at the other end by a

cable at an angle q from the

horizontal.

What is the tension in the cable?

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A. the weight of the sign

B. the tension in the cable

C. the vertical force component exerted on the rod by hinge

P

D. two or more of these are tied for least

A11.4

A metal advertising sign (weight w)

is suspended from the end of a

massless rod of length L. The rod is

supported at one end by a hinge at

point P and at the other end by a

cable at an angle q from the

horizontal.

Which of these forces is least?

73