# Chapter 9: Equilibrium, Elasticity

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23-Feb-2016Category

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### Transcript of Chapter 9: Equilibrium, Elasticity

Chapter 9: Equilibrium, ElasticityThis 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 = 0This does NOT imply no forces, torques act. Only that we have a special case of Newtons 2nd Law F = 0 and = 0

Equilibrium

Example 9-1: Braces!FT = 2.0 N, FW = ?

FWx = FT sin(70) - FT sin(70) = 0FWy = FT cos(70) + FT cos(70) = 2FT cos(70) = 1.36 N

Example: Traction!mg = (20)(9.8) = 196 N 200 N

Fy = mg sin(37) - mg sin(37) = 0Fx = mg cos(37) + mg cos(37) = 2mg cos(37) =320 N

Sect. 9-1: Conditions for Equilibrium

STATICS (Equilibrium):

Body at rest (a = 0) Net force = 0 or F = 0 (Newtons 2nd Law) OR, in component form: Fx = 0, Fy = 0, Fz = 0 FIRST CONDITION FOR EQUILIBRIUM

STATICS (Equilibrium):

Body at rest ( = 0) Net torque = 0 or = 0 (Newtons 2nd Law, rotations) (Torques taken about any arbitrary point!)

SECOND CONDITION FOR EQUILIBRIUM

Example

Example 9-2: Chandelier

Example

- Conceptual Example 9-3: A Lever = 0 About pivot point mgr -FPR = 0OR: FP = (r/R)mg Since r
Section 9-2: Problem Solving Fx = 0, Fy = 0, = 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 = 0.5. Carefully solve the equations (ALGEBRA!!)

Example 9-4

Example 9-5 = 0 (About point of application of F1)Fy = 0

Example: Cantilever

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

Example 9-6: Beam & WireM = 28 kg

Example 9-7: Ladder & Wall

Examplem = 170 kg, = 37. Find tensions in cordsFx = 0 = FT1 - FT2 cos (1)Fy = 0 = FT2 sin - mg (2)

(2) FT2 = (mg/sin) = 2768 NPut into (1). Solve for FT1 = FT2 cos = 2211 N

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

Prob. 20: Mg =245 N, mg =155 N = 35, L =1.7 m, D =1.35mFT, FhV, FhH = ? For = 0 take rotation axis through point A: = 0 = -(FTsin)D +Mg(L)+mg(L/2) FT = 708 NFx = 0 = FhH - FTcos FhH = 580 NFy = 0 = FhV + FTsin -mg -Mg FhV = - 6 N(down)

Prob. 21: M = 21.5 kg, m = 12 kg = 37, L = 7.5 m, H = 3.8 mFT, FAV, FAH = ? For = 0 take rotation axis through point A: = 0 = -FTH + Mg(Lcos) + mg(L/2) cos FT = 425 N. Fx = 0 =FAH - FT FAH = 425 NFy = 0 = FAV -mg -Mg FAV = 328 N

Section 9-3: Application to Muscles & Joints Fx = 0, Fy = 0, = 0

Example 9-8: Elbow

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

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