Chapter 11

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman

Chapter 11

Equilibrium and Elasticity

Goals for Chapter 11

• To study the conditions for equilibrium of a body

• To understand center of gravity and how it relates to a body’s stability

• To solve problems for rigid bodies in equilibrium

Goals for Chapter 11

• To analyze situations involving tension, compression, pressure, and shear

• Real materials are not truly rigid. They are elastic and do deform to some extent.

• To investigate what happens when a body is stretched so much that it deforms or breaks

Conditions for equilibrium

• First condition: The sum of all external forces is equal to zero:

Fx = 0 Fy = 0 Fz = 0

• Second condition: The sum of all torques about any (and ALL!) given point(s) is equal to zero.

Conditions for equilibrium

Conditions for equilibrium• First condition: The sum of all external forces is equal to zero:

Fx = 0 Fy = 0 Fz = 0

• Second condition: The sum of all torques about any (and ALL!) given point(s) is equal to zero.

Center of gravity

• Treat body’s weight as though it all acts at a single point—

the center of gravity…

Center of gravity

• If ignore variation of gravity with altitude…

Center of Gravity (COG) is same as center of mass.

Center of gravity

• We can treat a body’s weight as though it all acts at a single point—the center of gravity…

i ii

ii

mm m mm m m m

1 1 2 2 3 3cm

1 2 3

r

r r rr

Center of gravity

Center of gravity

Center of gravity & Stability

Walking the plank

• Uniform plank (mass = 90 kg, length = 6.0 m) rests on sawhorses 1.5 m apart, equidistant from the center.

• IF you stand on the plank at the right edge, what is the maximum mass m so that the plank doesn’t move?

Walking the plank

• Ask WHAT WILL HAPPEN?

• It will rotate CLOCKWISE!

• If it rotates, where will it rotate around? ID Axis!

• It will rotate around S!

Walking the plank

• What is the maximum mass “m” for stability?

Origin at c

Center of Gravity at s

Walking the plank

Rcm = [M(@0) + m (@ ½ L)] = m( ½ L) = ½ D

M + m (M+m)

Center of Gravity is the point where the gravitational torques are balanced.

i ii

ii

mm m mm m m m

1 1 2 2 3 3cm

1 2 3

r

r r rr

Walking the plank

Rcm = [M(@0) + m (@ ½ L)] = m( ½ L) = ½ D

M + m (M+m)

So… m = M D = 90 kg (1.5/6-1.5) = 30 kg

(L – D)

i ii

ii

mm m mm m m m

1 1 2 2 3 3cm

1 2 3

r

r r rr

Solving rigid-body equilibrium problems

• Make a sketch; create a coordinate system and draw all normal, weight, and other forces.

• Specify a positive direction for rotation, to establish the sign for all torques in the problem. Be consistent.

• Choose a “convenient” point as your reference axis for all torques. Remember that forces acting at that point will NOT produce a torque!

• Create equations for Fx = 0 Fy = 0 and


• Car with 53% of weight on front wheels and 47% on rear. Distance between axles is 2.46 m. How far in front of the rear axle is the center of gravity?


• Suppose you measure from the REAR wheels….

• Forces at the axis of rotation create NO torque!

= 0 =>

[0.47w x 0 meters] – wLcg +[0.53w x 2.46m] = 0

So..

Lcg = 1.30 m


• Suppose you measure from the FRONT wheels….

• Forces at the axis of rotation create NO torque!

= 0 =>

- [0.47w x 2.46 meters] + w L’cg +[0.53w x 0 meters] = 0

So..

L’cg = 1.16 m

L’ + L = 1.16 + 1.30 = 2.46 m

L’cg

Ladder Problems!

• Lots of variables:

• Length of ladder L

• Mass of ladder (and its center of gravity)

• Angle of ladder against wall

• Normal force of ground

• Static Friction from ground

• Normal force of wall

• Static Friction of wall

• Weight of person on ladder

• Location of person on ladder

Ladder Problems!

• Lots of variables:

• Length of ladder L

• Mass of ladder (and its center of gravity)

• Angle of ladder against wall

• Normal force of ground

• Static Friction from ground

• Normal force of wall

• Static Friction of wall

• Weight of person on ladder

• Location of person on ladder

Wall

(normal

force and

friction!)

Ground

(normal

force and

friction!)

cog of

ladder

Ladder Problems!

• Equilibrium involves Fx = 0 Fy = 0 z = 0

• 3 Equations!

• Solve for up to 3 unique unknowns!

• Use key words/assumptions tonarrow choices

• “uniform ladder”

• “just about to slip”

• “frictionless”

Wall

(normal

force and

friction!)

Ground

(normal

force and

friction!)

Ladder Problems!

Wall

(normal

force and

friction!)

Ground

(normal

force and

friction!)

Length

L

Angle

N (ground)

fs (ground)

N (wall)

fs (wall)

Weight (ladder)

1

Ladder Problems!

Ground forces create no torque

about point 1!

Torques about

point 1

Length

L

Angle

N (ground)

fs (ground)

Ladder Problems!

Weight drawn from center of gravity

Torque = Wladder x d

Clockwise!

= - Wladder x (1/2 L sin

Torques about

point 1

Length

½ L

Angle

Weight (ladder)

1

Lever Arm

Line of Action

d

+

Ladder Problems!

Torques about

point 1

Length

L

Angle

N (wall)

fs (wall)

1

Line of Action

Line of Action

Lever Arm = Lsin

Lever Arm = Lcos

Angle

Ladder Problems!

Fx = 0 = +fs (ground) – N (wall)

Fy = 0 = N (ground) + fs (wall) – Wladder

• 1 = 0 = +[N (wall) Lsin CCW

+ [fs (wall) Lcos CCW

- [Wladder Lsin CW

Will the ladder slip?

• 800 N man climbing ladder 5.0 m long that weighs 180 N. Find normal and frictional forces that must be present at the base of the ladder.

• What is the minimum coefficient of static friction at the base to prevent slipping?

• What is the magnitude and direction of the contact force on the base of the ladder?





Fx = 0 =>

fs – n1 = 0


Fx = 0 =>

fs – n1 = 0

Fy = 0 =>

n2 – wperson – wladder = 0


Fx = 0 =>

fs – n1 = 0

Fy = 0 =>


(about ANY point) = 0 =>

+n1(4m) – wp(1.5m) –wl(1.0m) = 0


Fx = 0 =>

fs – n1 = 0

Fy = 0 =>



+n1(4m) – wp(1.5m) –wl(1.0m) = 0

So

n1 = 268 N


Fx = 0 =>

fs – n1 = 0

Fy = 0 =>



+n1(4m) – wp(1.5m) –wl(1.0m) = 0

So

n1 = 268 N & s(min) = fs/n2 = .27

Will the ladder slip? – Example 11.3



• What is the magnitude and direction of the contact force on the base of the ladder?

Example 11.4: Equilibrium and pumping iron

Given weight w and angle between tension and horizontal, Find T and the two components of E

Strain, stress, and elastic moduli

• Stretching, squeezing, and twisting a real body causes it to deform.

• Stress is force per unit area

• Measured in N/m2 or “Pascals” or lbs/sq. in. or “PSI”•1 PSI ~ 7000 Pa•Typical tire pressure ~ 32 PSI ~ 200 kPA• Same units as PRESSURE in fluids/gases

• Stress is an internal force on an object that produces (or results from) a Strain




• Strain is the relative change in size or shape of an object because of externally applied forces.

• Strain is the fractional deformation due to the stress.

• Strain is dimensionless – just a fraction.

• Linear strain = l/l0 (tensile strain)




• Strain is the fractional deformation due to the stress.

• Elastic modulus is stress divided by strain.

• The direct (linear) proportionality of stress and strain is called Hooke’s law:

STRESS / STRAIN = Elastic Modulus


•The direct (linear) proportionality of stress and strain is called Hooke’s law.

• Similar to spring/rubber bands stretching:

• Apply external force F, see spring/band deform a distance x from the original length x.

•Apply that force F over the entire cross sectional area of the rubber band (or thickness of the spring wire)

•F/A creates a STRESS




•Apply that force F over the entire area of the rubber band (or thickness of the spring wire) F/A creates a STRESS

•Deformation of band in length x/x is the resulting Strain

• F = kx

•F/A = Stress = kx/A = Yx/x = Y X Strain




• Tensile Stress/Strain = Y = Young’s Modulus

•F/A = Stress = kx/A = Yx/x = Y X Strain

•Y = kx/A = Force/Area (Pascals)

Tensile and compressive stress and strain

• Tensile stress = F /A & tensile strain = l/l0

(stretching under tension)

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

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

Tensile and compressive stress and strain

• Tensile stress = F /A & tensile strain = l/l0

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

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

• Compressive stress & compressive strain defined similarly.

Some values of elastic moduli

Tensile stress and strain

• Body can experience both tensile & compressive stress at the same time.

Bulk stress and strain

• Pressure in a fluid is force per unit area: p = F/A.

• Bulk stress is pressure change p & bulk strain is

fractional volume change

V/V0.

• Bulk modulus is bulk stress divided by bulk strain and is given by B = –p/(V/V0).

Sheer stress and strain

• Sheer stress is F||/A and sheer strain is x/h.

• Sheer modulus is sheer stress divided by sheer strain, and is given by S = (F||/A)(h/x).

Elasticity and plasticity

• Hooke’s law applies up to point a.

• Table 11.3 shows some approximate breaking stresses.

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Chapter 11