lecture ch 9

Chapter 9

Static Equilibrium; Elasticity and Fracture

Units of Chapter 9•The Conditions for Equilibrium

•Solving Statics Problems

•Applications to Muscles and Joints

•Stability and Balance

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 a field concerned with a study of the forces acting on objects in 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

9-1 The Conditions for Equilibrium

An object with forces acting on it, but that is not moving, is said to be in equilibrium.

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

9-1 The Conditions for Equilibrium

The first condition for equilibrium is that the forces along each coordinate axis add to zero.

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

9-1 The Conditions for EquilibriumThe second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.

Body at rest (α = 0) Net torque = 0 or ∑τ = 0

(Newton’s 2nd Law, rotations)

(Torques taken about any arbitrary point!)

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!

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9-2 Solving Statics Problems

1. Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act.

2. Choose a coordinate system and resolve forces into components.

3. Write equilibrium equations for the forces.

4. Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously.

5. Solve.

Example 9-4

∑Fy = 0 = FN – mAg – mBg – Mg

∑τ = 0 about point PmAg*2.5 – mBgx + Mg*0 + FN*0 = 0x = 3 m

Problem 16

x

L

m2g

FN

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

Example 9-5

∑τ = 0 about FA line of application(about point P)

-1500*10*g -15000*15*g+FB*20=0FB = 118000 N

∑Fy = 0

FA – 1500*g – 15000*g + FB = 0FA = 44100 N

∑τ = 0 about FA line of application (about point P) to get FB and then ∑τ = 0 about FB line of application to get FA

9-2 Solving Statics ProblemsIf a force in your solution comes out negative (as FA will here), it just means that it’s in the opposite direction from the one you chose. This is trivial to fix, so don’t worry about getting all the signs of the forces right before you start solving.

9-2 Solving Statics Problems

If there is a cable or cord in the problem, it can support forces only along its length. Forces perpendicular to that would cause it to bend.

Problem 9-11

FT2

FT1x

y

mg

m = 170 kg, θ = 33º. Find tensions in cords

∑Fx = 0 = FT1 - FT2 cosθ (1)∑Fy = 0 = FT2 sinθ - mg (2)

(2) FT2 = (mg/sinθ) = 3058.9 NPut into (1). Solve for FT1 = FT2 cosθ = 2565.4 N

Prob. 9- 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. 9-21 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 9-4: Stability & Balance

• STATICS (Equilibrium) ∑F = 0 and ∑τ = 0• Now: A body initially at equilibrium. Apply a small force & then take that force

away. The body moves slightly away from equilibrium.

3 Possible Results:1. Object returns to the original position.

The original position was a STABLE EQUILIBRIUM.

2. Object moves even further from the original position.

The original position was an UNSTABLE EQUILIBRIUM.

3. Object remains in the new position.

The original position was a NEUTRAL EQUILIBRIUM.

9-4 Stability and Balance

If the forces on an object are such that they tend to return it to its equilibrium position, it is said to be in stable equilibrium.

9-4 Stability and Balance

If, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium.

9-4 Stability and Balance

An object in stable equilibrium may become unstable if it is tipped so that its center of gravity is outside the pivot point. Of course, it will be stable again once it lands!

9-4 Stability and Balance

People carrying heavy loads automatically adjust their posture so their center of mass is over their feet. This can lead to injury if the contortion is too great.

Summary of Chapter 9

• An object at rest is in equilibrium; the study of such objects is called statics.

• In order for an object to be in equilibrium, there must be no net force on it along any coordinate, and there must be no net torque around any axis.

• An object in static equilibrium can be either in stable, unstable, or neutral equilibrium.