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Chapter 11

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.

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Chapter 11

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  1. 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 = 0 Fy = 0 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 = 0 Fy = 0 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 = 1.16 + 1.30 = 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: • 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 tonarrow 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. Ladder Problems! Torques about point 1 1 Groundforces create no torque about point 1! Length L N (ground) Angle q fs (ground)

  27. Ladder Problems! Torques about point 1 Line of Action 1 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 d

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

  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. • Stressis forceper 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 • Stressis 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. • Stressis forceper unit area • Strainis the relative change in size or shape of an object because of externally applied forces. • Strainis the fractionaldeformation 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. • Stressis forceper unit area • Strainis the fractionaldeformation due to the stress. • Elastic modulusis 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 XStrain

  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 XStrain • 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/l0Young’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 stressat the same time.

  49. Bulk stress and strain • Pressure in a fluid is force per unit area: p = F/A. • Bulk stressis pressure change p & bulk strainis 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 stressis F||/A and sheer strainis x/h. • Sheer modulus is sheer stress divided by sheer strain, and is given by S = (F||/A)(h/x).

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