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.

Usually: Interested in maintaining a Stable Equilibrium  “BALANCE”. General: Object with Center of Gravity (CG) below its support point is in Stable Equilibrium. CG above base of support? Stable as long as remains above base. Unstable if displaced so CG is no longer above base. Critical point = point where CG is just above edge of base.

Stable (BALANCED): Vertical line from CG falls within support base. Unstable: Vertical line falls outside support base. Critical point in changing from stable to unstable = point where CG is above edge of support base.

Stability: A relative concept Stability: A relative concept. 4 legged animals are more stable than humans.

Section 9-5: Elasticity, Stress, Strain One effect of forces on objects: DEFORMATION = Change of size or shape. Suppose force F pulls on object. Find (L0 >> L) F  L . Write: F = kL “Hooke’s Law” (small forces only!) k = constant which depends on material

Hooke’s “Law” F = kL holds only for small L! For larger L, material will permanently deform & possibly break.

Elastic Modulus F = kL. L depends on applied force & also on material composition. The constant k can be written to account for this. Experiment: Object, cross sectional area A pulled by force F (L << L0) Write: F  EA(L/L0)  k L E  “Elastic Modulus” (Young’s Modulus) (Depends on material) Another form (fractional length change or strain): (L/L0) = (1/E)(F/A) F/A = Force/area (Stress). Strain  Stress

(L/L0) = (1/E)(F/A)

Strain & Stress (L/L0) = (1/E)(F/A)  Strain  Stress External force  Internal stress (tension) This is tensile stress (tension)

3 Types of Stress

Shear Modulus Object under shear stress is not in equilibrium. A net torque exists. ∑τ  0 Shear modulus G.

Bulk Modulus Object subjected to inward forces from all sides  Volume decreases. (Ch. 10: Object submersed in a fluid). Initial volume V0. Change in volume V. Write: (V/V0)  -(1/B) (F/A) = -(1/B) P B  Bulk modulus. P = F/A = pressure - sign indicates volume shrinks under pressure

Section 9-6: Fracture If stress is too great, object breaks or cracks = “Fractures”

Section 9-6: Fracture

Example 9-11 Beam sagging under its own weight:

Conceptual Example 9-12 Tragic substitution!