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 2 nd Law ∑F = 0 and ∑τ = 0
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Chapter 9: Equilibrium, Elasticity This chapter: Special case of motion. That is NO MOTION! Actually, no acceleration! Everything we say would hold if.
Example 9-1: Braces! F T = 2.0 N, F W = ? F Wx = F T sin(70º) - F T sin(70º) = 0 F Wy = F T cos(70º) + F T cos(70º) = 2F T cos(70º) = 1.36 N
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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
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!!!
(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