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Conditions of Equilibrium

Conditions of Equilibrium. In order for a rigid body to be considered to be in equilibrium, there must be no change in the either the linear or the angular momentum. Therefore:. ∑ F ext = 0 and ∑  ext = 0. If ∑  ext = 0, then there is no change in L either.

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Conditions of Equilibrium

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  1. Conditions of Equilibrium In order for a rigid body to be considered to be in equilibrium, there must be no change in the either the linear or the angular momentum. Therefore: ∑Fext = 0 and ∑ext = 0 If ∑ext = 0, then there is no change in L either. True equilibrium means no linear or angular acceleration! Static equilibrium means no linear or angular acceleration or motion at all!

  2. Is the first condition of equil. satisfied? 2 N Yes: ∑F = 0 Is the second cond. of equil. satisfied? No: ∑ ≠ 0 2 N 2 N 2 N Could a single, fourth force restore equilibrium? If so, what would be the magnitude and direction of that force? Can’t be done!

  3. Center of Gravity i = mig x ri ∑ = (∑miri) x g mi ri cg = Mrcm x g  = rcm x Mg mig Therefore, all torque due to gravity about the cm will be 0!

  4. A uniform beam with a mass of 1.8 kg rests with its ends upon two scales. A 2.7 kg block is placed so that its center is one fourth of the way from the beam’s left end. What will the scales read? L m1 = 1.8 kg m2 = 2.7 kg Fl Fr L /4 m1g m2g

  5. First solution: ∑Fx = 0 ∑Fy = 0 = Fl + Fr - m1g - m2g ∑  = 0 = Fl(0) + Fr(L) - m1g(L /2) - m2g(L /4) Fr = (m1g)/ 2 + (m2g)/ 4 = 15 N Fl = g(m1 + m2) - Fr = 29 N Another possible attack is to use a different axis: ∑  = Fr(0) - Fl(L) + m1g(L /2) + m2g(3L /4) Fl = (m1g)/2 + 3(m2g)/4 = 29 N

  6. A ladder whose mass is 45 kg is 12 m long is leaned against a wall. Its upper end is 9.3 m above the ground. The center of mass of the ladder is one-third of the way up the ladder. A man with a mass of 72 kg climbs halfway up the ladder. Assume the wall, but not he ground, is frictionless. Find the force that the wall and the ground exert on the ladder. L = 12 m m1 = 45 kg m2 = 72 kg h = 9.3 m

  7. N1 ø N2 L h F2 F1 ø f a = L2 - h2 = 7.6 m

  8. ∑Fx = 0 = N1 - f ∑Fy = 0 = N2 - (F1 + F2) N2 = g(m1 + m2) = 1150 N If we resolve torque with respect to the point at which the ladder touches the ground, we conveniently eliminate two forces: ∑z = -N1cosø (12) + F2cosø(6) + F1cosø(4) a/2 a/3 h N1 = m2g(a/2) + m1g(a/3) h = 410 N

  9. In the problem pictured, a square sign with a mass of 52.3 kg is hung from a horizontal rod and cable as shown. A) Find the tension in the cable and B) find the components of the force exerted by the rod on the wall.

  10. Elasticity An material is considered to be elastic if compression or shear forces produce deformation that reverts to 0 when those forces are removed. ∆L F F ∆L L F tensile stress shear stress

  11. Stress is the ratio of force to cross sectional area. Stress produces strain! Strain is the ratio of deformation to original length. The ratio of stress to strain for a given substance will be a constant value up to the point of permanent deformation (elastic limit or yield strength). This value is called the elastic modulus or Young’s modulus: E = stress = F/A strain ∆L /L

  12. The Young’s Modulus of a substance will remain constant up to the point at which it permanently deforms--> The Yield Strength Beyond yielding comes the inevitable rupture. This occurs at a stress which is called the Ultimate Strength (Tensile Strength for elongation stresses). The Ultimate Strength for a substance can vary depending upon compression or tension: • Concrete has great strength when compressed, but is relatively weak when placed in tension.

  13. Human bone has an ultimate strength of 170 x 106 N/m2 in compression. What compressive load would be needed to break a male, adult femur, which has a diameter of 2.8 cm? A = πr2 = π (.014)2 = 6.2 x 10-4 m2 F = Su A = (170 x 106 N/m2)(6.2 x 10-4 m2) = 1.1 x 105 N  11 tons This a very large force, but it does not need to be sustained-- a bad parachute jump landing could do it in a few milliseconds!

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