1. Equilibrium and Elasticity
Chapter 11
Equilibrium and Elasticity

2. 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

3. 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

4. Conditions for equilibrium
First condition: The sum of all external forces is equal to zero:
Fx = Fy = Fz = 0
Second condition: The sum of all torques about any (and ALL!) given point(s) is equal to zero.

5. Conditions for equilibrium

6. Conditions for equilibrium
First condition: The sum of all external forces is equal to zero:
Fx = Fy = Fz = 0
Second condition: The sum of all torques about any (and ALL!) given point(s) is equal to zero.

7. Center of gravity
Treat body’s weight as though it all acts at a single point— the center of gravity…

8. Center of gravity
If ignore variation of gravity with altitude… Center of Gravity (COG) is same as center of mass.

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

10. Center of gravity

11. Center of gravity

12. Center of gravity & Stability

13. 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?

14. 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!

15. Walking the plank What is the maximum mass “m” for stability?
Origin at c
Center of Gravity at s

16. 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.

17. 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)

18. 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 SFx = 0 SFy = 0 and St = 0

19. Solving rigid-body equilibrium problems
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?

20. Solving rigid-body equilibrium problems
Suppose you measure from the REAR wheels….
Forces at the axis of rotation create NO torque!
St = 0 => [0.47w x 0 meters] – wLcg +[0.53w x 2.46m] = 0
So..
Lcg = 1.30 m

21. Solving rigid-body equilibrium problems
Suppose you measure from the FRONT wheels….
Forces at the axis of rotation create NO torque!
St = 0 => - [0.47w x 2.46 meters] + w L’cg +[0.53w x 0 meters] = 0
So..
L’cg = 1.16 m
L’ + L = = 2.46 m
L’cg

22. 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

23. Ladder Problems! Lots of variables: Wall (normal force and friction!)
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!)
cog of ladder
Ground (normal force and friction!)

24. Ladder Problems! Equilibrium involves Fx = 0 Fy = 0 tz = 0
3 Equations!
Solve for up to 3 unique unknowns!
Use key words/assumptions to narrow choices
“uniform ladder”
“just about to slip”
“frictionless”
Wall (normal force and friction!)
Ground (normal force and friction!)

25. Ladder Problems! fs (wall) Wall (normal force and friction!) N (wall)
Length L
N (ground)
Weight (ladder)
Ground (normal force and friction!)
Angle q
fs (ground)

26. Ground forces create no torque about point 1!
Ladder Problems!
Torques about point 1
Ground forces create no torque about point 1!
Length L
N (ground)
Angle q 1
fs (ground)

27. Weight drawn from center of gravity Torque = Wladder x d Clockwise!
Ladder Problems!
Torques about point 1
Line of Action
Weight drawn from center of gravity
Torque = Wladder x d Clockwise!
= - Wladder x (1/2 L sinq)
Length ½ L
Weight (ladder) +
Angle q
Lever Arm 1 d

28. Torques about point 1 1 Ladder Problems! Line of Action fs (wall)
N (wall)
Angle q
Lever Arm = Lsinq
Line of Action
Length L
Lever Arm = Lcosq
Angle q 1

29. Ladder Problems! Fx = 0 = +fs (ground) – N (wall)
Fy = 0 = N (ground) + fs (wall) – Wladder
t1 = 0 = +[N (wall) Lsinq] CCW [fs (wall) Lcosq] CCW
- [Wladder Lsinq] CW

30. 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?

31. 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?

32. Will the ladder slip?
SFx = 0 =>
fs – n1 = 0

33. Will the ladder slip? SFx = 0 => fs – n1 = 0 SFy = 0 =>
n2 – wperson – wladder = 0

34. Will the ladder slip? SFx = 0 => fs – n1 = 0 SFy = 0 =>
n2 – wperson – wladder = 0
St (about ANY point) = 0 =>
+n1(4m) – wp(1.5m) –wl(1.0m) = 0

35. Will the ladder slip? SFx = 0 => fs – n1 = 0 SFy = 0 =>
n2 – wperson – wladder = 0
St (about ANY point) = 0 =>
+n1(4m) – wp(1.5m) –wl(1.0m) = 0 So
n1 = 268 N

36. Will the ladder slip? SFx = 0 => fs – n1 = 0 SFy = 0 =>
n2 – wperson – wladder = 0
St (about ANY point) = 0 =>
+n1(4m) – wp(1.5m) –wl(1.0m) = 0 So
n1 = 268 N & ms(min) = fs/n2 = .27

37. Will the ladder slip? – Example 11.3
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?

38. Example 11.4: Equilibrium and pumping iron
Given weight w and angle between tension and horizontal, Find T and the two components of E

39. 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

40. Strain, stress, and elastic moduli
Stretching, squeezing, and twisting a real body causes it to deform.
Stress is force per unit area
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 = Dl/l0 (tensile strain)

41. Strain, stress, and elastic moduli
Stretching, squeezing, and twisting a real body causes it to deform.
Stress is force per unit area
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

42. Strain, stress, and elastic moduli
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 Dx 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

43. Strain, stress, and elastic moduli
The direct (linear) proportionality of stress and strain is called Hooke’s law.
Similar to spring/rubber bands stretching:
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 Dx/x is the resulting Strain
F = kDx
F/A = Stress = kDx/A = YDx/x = Y X Strain

44. Strain, stress, and elastic moduli
The direct (linear) proportionality of stress and strain is called Hooke’s law.
Similar to spring/rubber bands stretching:
Tensile Stress/Strain = Y = Young’s Modulus
F/A = Stress = kDx/A = YDx/x = Y X Strain
Y = kx/A = Force/Area (Pascals)

45. 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)

46. 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.

47. Some values of elastic moduli

48. Tensile stress and strain
Body can experience both tensile & compressive stress at the same time.

49. 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).

50. 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).

51. Elasticity and plasticity
Hooke’s law applies up to point a.
Table 11.3 shows some approximate breaking stresses.