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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:
∑ F x = 0
∑ F y = 0
∑ F z = 0
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9-1 The Conditions for Equilibrium The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary .
Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act.
Choose a coordinate system and resolve forces into components .
Write equilibrium equations for the forces.
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.
Solve .
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Example 9-4 ∑ F y = 0 = F N – m A g – m B g – Mg ∑ τ = 0 about point P m A g*2.5 – m B gx + Mg*0 + F N *0 = 0 x = 3 m
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Problem 16 m 1 = 50kg, m 2 = 35 kg, m 3 = 25 kg, L = 3.6m Find x so the see-saw balances. Use ∑ τ = 0 (Take rotation axis through point A ) ∑ τ = m 2 g(L/2) + m 3 g x - m 1 g(L/2) = 0 Put in numbers, solve for x : x = 1.1 m
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Example 9-5 ∑ τ = 0 about F A line of application (about point P) -1500*10*g -15000*15*g+F B *20=0 F B = 118000 N ∑ F y = 0 F A – 1500*g – 15000*g + F B = 0 F A = 44100 N ∑ τ = 0 about F A line of application (about point P) to get F B and then ∑ τ = 0 about F B line of application to get F A
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9-2 Solving Statics Problems If a force in your solution comes out negative (as F A 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.
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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.
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Problem 9-11 m = 170 kg, θ = 33º . Find tensions in cords ∑ F x = 0 = F T1 - F T2 cos θ (1) ∑ F y = 0 = F T2 sin θ - mg (2) (2) F T2 = (mg/sin θ ) = 3058.9 N Put into (1) . Solve for F T1 = F T2 cos θ = 2565.4 N
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Prob. 9- 20 Mg =245 N, mg =155 N θ = 35º, L =1.7 m, D =1.35m F T , F hV , F hH = ? For ∑ τ = 0 take rotation axis through point A: ∑ τ = 0 = -(F T sin θ )D +Mg(L)+mg(L/2) F T = 708 N ∑ F x = 0 = F hH - F T cos θ F hH = 580 N ∑ F y = 0 = F hV + F T sin θ -mg -Mg F hV = - 6 N (down)
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Prob. 9-21 M = 21.5 kg, m = 12 kg θ = 37º, L = 7.5 m, H = 3.8 m F T , F AV , F AH = ? For ∑ τ = 0 take rotation axis through point A: ∑ τ = 0 = F T H - Mg(Lcos θ ) - mg(L/2) cos θ F T = 425 N. ∑F x = 0 =F AH - F T F AH = 425 N ∑ F y = 0 = F AV -mg -Mg F AV = 328 N
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 .
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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.
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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.
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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!
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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.
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