Download Chapter 9, Part I

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
and
∑F = 0
∑τ = 0
Equilibrium
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
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
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 = 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!
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 & Wire
M = 28 kg
Example 9-7: Ladder & Wall
Example
y
FT2
FT1

x
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 N
Put into (1). Solve for FT1 = FT2 cosθ =
2211 N
Problem 16
L
A
m1g
x
m3g
FN
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
m2g
Prob. 20: Mg =245 N, mg =155 N
θ = 35º, L =1.7 m, D =1.35m
L
D
y
FhingeV
FhingeH
A
FT
mg
x

B
Mg
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)
Prob. 21: M = 21.5 kg, m = 12 kg
θ = 37º, L = 7.5 m, H = 3.8 m
FT
B

FAV
H

A
FAH
mg
y
Mg
x
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
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)