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 = 0This does NOT imply no forces, torques act. Only that we have a special case of Newton’s 2nd Law ∑F = 0

and ∑τ = 0

Equilibrium

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

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

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

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

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

• 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

Example

Example 9-2: Chandelier

Example

Conceptual Example 9-3: A Lever ∑τ = 0

About pivot point mgr -FPR = 0OR:

FP = (r/R)mg Since r << R

FP << mg • Can lift a heavy

weight with a small force!Mechanical advantage of a lever!

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. Carefully solve the equations (ALGEBRA!!)

Example 9-4

Example 9-5 ∑τ = 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

Example

FT2FT1

x

y

mg

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) = 2768 NPut into (1). Solve for FT1 = FT2 cosθ =

2211 N

Problem 16

x

L

m2gFN

Am3gm1g

m1 = 50kg, m2 = 35 kg, m3 = 25 kg, L = 3.6mFind x so the see-saw balances. Use ∑τ = 0 (Take rotation axis through point A) ∑τ = 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 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)

A B

mg

FTFhingeV

L

FhingeH

x

y

Mg

D

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

H

B

FAH

MgFAV

FT

x

y

A

mg

Section 9-3: Application to Muscles & Joints

∑Fx = 0, ∑Fy = 0, ∑τ = 0

Example 9-8: Elbow

Example 9-9: Forces on Your Back

∑Fx = 0, ∑Fy = 0, ∑τ = 0 (axis at spine base)