1. Equilibrium, Elasticity, and Hooke’s Law
Conditions for equilibrium:
Exam Example 26: Ladder against wall (example 11.3, problem 11.10)
d/2 h x
Static equilibrium: y
State with
is equilibrium but is not static.
Data: m, M, d, h, y, μs
Find: (a) F2, (b) F1, fs, d θ
Strategy of problem solution:
(c) yman when ladder starts to slip
(0)
Choice of the axis of rotation:
arbitrary - the simpler the better.
(ii) Free-body diagram:
identify all external forces and
their points of action.
(iii) Calculate lever arm and
torque for each force.
(iv) Solve for unknowns.
Solution: equilibrium equations yield
(a) F2= Mg + mg ; (b) F1 = fs
Choice of B-axis (no torque from F2 and fs)
F1h = mgx + Mgd/2 → F1= g(mx+Md/2)/h = fs
(c) Ladder starts to slip when μsF2 = fs, x = yd/h
→ μsg (M+m) = g (mymand/h+Md/2)/h →

2. Center of Gravity Definition. The center of gravity of a rigid body is
the point at which its weight can be considered to act
when calculating the torque due to gravity:
Derivation:
(if has the same value at all points on a body)
Center of gravity
can be located by
suspending object
from different points.
Note: Center of gravity may lie
“outside” the object.

3. The larger the area of support and the lower the center of gravity,
the more difficult it is to overturn a body.

4. Stress, Strain, and Elastic Moduli
Stress = (Elastic modulus) · Strain
Units: [stress F/A]=[pressure p] = 1 N/m2 = 1 pascal = 1 Pa = 1.45·10-4 psi (lb/in2)
Young’s modulus Y~1011 N/m2
Shear modulus S~Y/3
Bulk modulus B~Y
(only for solids, S=0 for fluids)

5. Stress versus Strain and Hooke’s Law
Modulus
Strain
Tensile F┴/A
or compressive =Y
Δl/l0
Shear F║/A =S
Δx/h
Bulk Δp =B
-ΔV/V0
Stress = (Elastic modulus) · Strain
F = - kx
Robert Hooke ( )
(a contemporary of Newton)
Elastic
hysteresis

6. Interesting property: shorter springs are stiffer springs
Proof:
Force F is constant along the spring and
x = x/2 + x/2 → F = - kx = - 2k (x/2)
L+x
L/2+x/2
L/2+x/2