1. Equilibrium and Elasticity

2. 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.

3. Example: A child walks on a massive (90 kg) 6
Example: A child walks on a massive (90 kg) 6.0 m plank which rests on two saw horses separated by D = 1.5m 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? D L

4. 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.

5. 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 180N, 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

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

7. 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

8. Young’s Modulus: how things stretch (elastically)
tensile stress: force per area = F/A
compression
tension l A l0
strain: fractional change in length
= change in length per original length = Dl/l0
Elastic modulus = stress/strain
Young’s modulus (for stretching in one direction)

9. Bulk Modulus: compression of solids, liquids and gases
bulk stress: force per area = F/A = pressure p ( 1 N/m2 = 1 Pa)
1 atm = E5 Pa = 14.7 lb/in2
bulk strain: DV/V0
Bulk Modulus B relates small change in pressure to bulk strain

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

11. Elastic Moduli
Material Y
1010 Pa B S
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
Liquid
Compressibility k = 1/B
1E-11 Pa
Carbon disulfide
93.0
Ethyl Alcohol
110.0
Glycerin
21.0
Mercury
3.70
Water
45.8

12. Example: A steel cable 2. 0 m long has a cross section area of 0
Example: A steel cable 2.0 m long has a cross section area of 0.30 cm2. 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 m3 of oil is subjected to a pressure increase of Dp = 1.6E7 Pa B = 5.0E9 Pa Find the decrease in volume.

13. Elastic Hysteresis: strain depends on material “history”
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.
Elastic Hysteresis: strain depends on material “history”
stress
strain

14. Example: A copper wire1. 0 mm in diameter and 2
Example: A copper wire1.0 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 = 1.1x1011 Pa elastic limit = 1.5x108 Pa

15. Force: suspended weight Area: cross section of wire
From Lab
Force: suspended weight
Area: cross section of wire
elongation causes cylinder to rotate d 2q q DL