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Chapter 9: Equilibrium, Elasticity

Chapter 9: Equilibrium, Elasticity. This chapter: Special case of motion . That is NO MOTION ! Actually, no acceleration ! Everything we say would hold if the velocity is constant! STATICS ( Equilibrium ): Net (total) force = 0 AND net (total) torque = 0

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Chapter 9: Equilibrium, Elasticity

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  1. Chapter 9: Equilibrium, Elasticity • This chapter: Special case of motion. That is NO MOTION! • Actually, no acceleration! Everything we say would hold if the velocity is constant! • STATICS (Equilibrium): Net (total) force = 0 AND net (total) torque = 0 This does NOT imply no forces, torques act. Only that we have a special case of Newton’s 2nd Law ∑F = 0 and ∑τ = 0

  2. Equilibrium

  3. Example 9-1: Braces! FT = 2.0 N, FW = ? FWx = FT sin(70º) - FT sin(70º) = 0 FWy =FT cos(70º) + FT cos(70º) = 2FT cos(70º) = 1.36 N

  4. Example: Traction! mg = (20)(9.8) = 196 N  200 N Fy = mg sin(37º) - mg sin(37º) = 0 Fx =mg cos(37º) + mg cos(37º) = 2mg cos(37º) =320 N

  5. Sect. 9-1: Conditions for Equilibrium • STATICS (Equilibrium): • Body at rest (a = 0)  Net force = 0 or ∑F = 0 (Newton’s 2nd Law) OR, in component form: ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 FIRST CONDITION FOR EQUILIBRIUM

  6. STATICS (Equilibrium): • Body at rest (α = 0)  Net torque = 0 or ∑τ= 0 (Newton’s 2nd Law, rotations) (Torques taken about any arbitrary point!) SECOND CONDITION FOR EQUILIBRIUM

  7. Example

  8. Example 9-2: Chandelier

  9. Example

  10. Conceptual Example 9-3: A Lever ∑τ = 0 About pivot point  mgr -FPR = 0 OR: FP = (r/R)mg Since r << R FP << mg • Can lift a heavy weight with a small force! Mechanical advantage of a lever!

  11. Section 9-2: Problem Solving ∑Fx = 0, ∑Fy = 0, ∑τ = 0 (I) 1. Choose one body at a time to consider. Apply (I). 2.DRAW free body diagrams, showing ALL forces, properly labeled, at points where they act. For extended bodies, gravity acts through CM. 3. Choose convenient (x,y) coordinate system. Resolve forces into x,y components! 4. Use conditions (I). Choose axis about which torques are taken for convenience (can simplify math!). Any forces with line of action through axis gives τ = 0. 5.Carefullysolve the equations (ALGEBRA!!)

  12. Example 9-4

  13. Example 9-5 ∑τ = 0 (About point of application of F1) ∑Fy = 0

  14. Example: Cantilever

  15. NOTE!!! • IF YOU UNDERSTAND EVERY DETAIL OF THE FOLLOWING TWO EXAMPLES, THEN YOU TRULY UNDERSTAND VECTORS, FORCES, AND TORQUES!!!

  16. Example 9-6: Beam & Wire M = 28 kg

  17. Example 9-7: Ladder & Wall

  18. Example m = 170 kg, θ = 37º. Find tensions in cords ∑Fx = 0 = FT1 - FT2 cosθ (1) ∑Fy = 0 = FT2 sinθ - mg (2) (2)  FT2 = (mg/sinθ) = 2768N Put into (1). Solve for FT1 = FT2 cosθ = 2211N

  19. Problem 16 m1 = 50kg, m2 = 35 kg, m3 = 25 kg, L = 3.6m Find x so the see-saw balances. Use ∑τ = 0 (Take rotation axis through point A) ∑τ = m2g(L/2) + m3g x - m1g(L/2) = 0 Put in numbers, solve for x: x = 1.1 m

  20. Prob. 20:Mg =245 N, mg =155 N θ= 35º, L =1.7 m, D =1.35m FT, FhV, FhH = ? For ∑τ = 0 take rotation axis through point A: ∑τ= 0 = -(FTsinθ)D +Mg(L)+mg(L/2) FT = 708 N ∑Fx = 0 = FhH - FTcosθ FhH = 580 N ∑Fy = 0 = FhV + FTsinθ -mg -Mg  FhV = - 6 N (down)

  21. Prob. 21:M = 21.5 kg, m = 12 kg θ = 37º, L = 7.5 m, H = 3.8 m FT, FAV, FAH = ? For ∑τ = 0 take rotation axis through point A: ∑τ = 0= -FTH + Mg(Lcosθ) + mg(L/2) cosθ FT = 425 N. ∑Fx = 0 =FAH - FT FAH = 425 N ∑Fy = 0 = FAV -mg -Mg  FAV = 328 N

  22. Section 9-3:Application to Muscles & Joints ∑Fx = 0, ∑Fy = 0, ∑τ = 0

  23. Example 9-8: Elbow

  24. Example 9-9: Forces on Your Back ∑Fx = 0, ∑Fy = 0, ∑τ = 0 (axis at spine base)

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