This chapter discusses static equilibrium, which occurs when an object experiences forces but does not accelerate or change velocity. For an object to be in static equilibrium, the net force and net torque on it must be zero. Solving statics problems involves making free body diagrams, choosing coordinate systems, and writing equilibrium equations to solve for unknown forces. An object's equilibrium can be stable, unstable, or neutral depending on whether applied forces return it to its original position or cause it to move further away.

1. Chapter 9 Static Equilibrium; Elasticity and Fracture

4. 9-1 The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium .

11. 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

12. 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

13. 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

14. 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.

15. 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.

16. 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

17. 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)

18. 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

20. 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.

21. 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.

22. 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!

23. 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.