Equilibrium and Elasticity Equilibrium Phys 211 C 11

Equilibrium and Elasticity Equilibrium: Phys 211 C 11 p 1

Center of Gravity: The Center of Gravity of an object The point of an object from which it could be suspended without tending to rotate. The point where all the mass of an object can be considered to be located. CG does not need to be located within the physical object! Horseshoe, for example usually easily identified from symmetry. Phys 211 C 11 p 2

Example: A child walks on a massive (90 kg) 6. 0 m plank which rests on two saw horses separated by D = 1. 5 m equally spaced from the center of the plank. If the plank is not to tip over when the child walks to the end of the plank, what is the most mass the child can have? L D Phys 211 C 11 p 3

How to approach an equilibrium problem: For each object • Draw the forces that act on the object (i. e. draw a free-body diagram) • Choose a convenient set of coordinate axis and resolve all forces into components. Watch carefully for appropriate use of +/- signs. • Pick a convenient axis to calculate your torques about. • Set the sum of the force components along each axis equal to 0. • Set the sum of the torques equal to 0 • Solve the resulting equations for the unknown quantity or quantities. Phys 211 C 11 p 4

Example: A ladder 5. 00 m long is leaning against a frictionless wall with its lower end 3. 00 m away from the wall. The ladder weighs 180 N, and an 800 N man is a third of the way up the ladder. What forces do the wall and the ground exert on the ladder? What is the minimum coefficient of friction necessary for the ladder to not slip? q Phys 211 C 11 p 5

Example : A 60 kg woman stands at the end of a uniform 4 m, 50 kg diving board supported as shown. Determine the forces exerted by the two supports. 4. 00 m . 800 m Phys 211 C 11 p 6

Stress, Strain and Elastic Moduli material property of “stretchiness/springiness” -> how materials respond to stress compression tension shear Elastic modulus = stress/strain (approximately constant) strain: deformation = fractional change stress: force per area property of type of material Phys 211 C 11 p 7

Young’s Modulus: how things stretch (elastically) tensile stress: force per area = F /A A l 0 compression l A l 0 A tension l A strain: fractional change in length = change in length per original length = Dl/l 0 Elastic modulus = stress/strain Young’s modulus (for stretching in one direction) Phys 211 C 11 p 8

Bulk Modulus: compression of solids, liquids and gases bulk stress: force per area = F /A = pressure p ( 1 N/m 2 = 1 Pa) 1 atm = 1. 013 E 5 Pa = 14. 7 lb/in 2 bulk strain: DV/V 0 Bulk Modulus B relates small change in pressure to bulk strain Phys 211 C 11 p 9

Shear Modulus: stress: force per area = F||/A x F A f h F shear strain cannot be supported in fluids (gas, liquid) Phys 211 C 11 p 10

Elastic Moduli Material Liquid Carbon disulfide Ethyl Alcohol Glycerin Mercury Water Y B S 1010 Pa Aluminum 7. 0 7. 5 2. 5 Brass 9. 0 6. 0 3. 5 Copper 11. 0 14. 0 4. 4 Iron 21. 0 16. 0 7. 7 Lead 1. 6 4. 1 0. 6 Steel 20. 0 16. 0 7. 5 Compressibility k = 1/B 1 E-11 Pa 93. 0 110. 0 21. 0 3. 70 45. 8 Phys 211 C 11 p 11

Example: A steel cable 2. 0 m long has a cross section area of 0. 30 cm 2. A 550 kg mass is suspended from the cable. Calculate the stress, the resulting strain, and the elongation of the cable. Example: A hydraulic press contains 0. 25 m 3 of oil is subjected to a pressure increase of Dp = 1. 6 E 7 Pa B = 5. 0 E 9 Pa Find the decrease in volume. Phys 211 C 11 p 12

Elastic Limit: the maximum stress (force) which can be applied to an object without resulting in permanent deformation. Plastic Deformation or Plastic Flow: the permanent deformation which results when a material’s elastic limit has been exceeded. Ultimate strength: greatest tension (or compression or shear) the material can withstand with breaking *snap*, tearing, fracturing etc. a. k. a. Breaking Stress or Tensile Strength. A malleable or ductile material has a large range of plastic deformation. Fatigue: small defects reduce materials strength well below original strength. stress Elastic Hysteresis: strain depends on material “history” Phys 211 C 11 p 13 strain

Example: A copper wire 1. 0 mm in diameter and 2. 0 m long is used to support a mass of 5. 0 kg. By how much does this wire stretch under this load? What is the maximum mass which can be supported without exceeding copper’s elastic limit? Y = 1. 1 x 1011 Pa elastic limit = 1. 5 x 108 Pa Phys 211 C 11 p 14

From Lab Force: suspended weight Area: cross section of wire elongation causes cylinder to rotate d 2 q q DL Phys 211 C 11 p 15