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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

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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 • To analyze situations involving tension, compression, pressure, and shear • To investigate what happens when a body is stretched so much that it deforms or breaks

  3. Introduction • Many bodies, such as bridges, aqueducts, and ladders, are designed so they do not accelerate. • Real materials are not truly rigid. They are elastic and do deform to some extent. • We shall introduce concepts such as stress and strain to understand the deformation of real bodies.

  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 • 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.

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

  8. Center of gravity • We can treat a body’s weight as though it all acts at a single point—the center of gravity… • If we can ignore the variation of gravity with altitude, the center of gravity is the same as the 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 • What is the maximum mass “m” for stability? Origin at c Center of Gravity at s

  15. 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.

  16. 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)

  17. 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

  18. 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?

  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? St = 0 => 0.47w(0) – wLcg +0.53w(2.46m) = 0 So.. Lcg = 1.30 m

  20. 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?

  21. 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?

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

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

  24. 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

  25. 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

  26. 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

  27. 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?

  28. Equilibrium and pumping iron • Follow Example 11.4 using Figure 11.10 below.

  29. 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.

  30. 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)(l0/l).

  31. Some values of elastic moduli

  32. 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.

  33. 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.

  34. 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)(h/x). • Follow Example 11.7.

  35. 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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