Lecture11 (Topic10)
Equilibrium and Elasticity
Conditions for Equilibrium
First Condition
Components of net force must each be zero
ΣF=0
Second Condition
Zero net force but torque presents
Στ must be zero to ensure equilibrium
Stability and Balance
If object is displaced slightly, 3 possibilities:
(1) stable equilibrium - returns to original position
(2) unstable equilibrium - moves further from original position
(3) neutral equilibrium - remains in new position
Generally,
normal force can be exerted only within contact area,
so if Fg beyond this area,
net torque will topple the object.
Larger base→Lower CG→More stable object
Elasticity (Stress & Strain)
Hooke's Law
Young's Modulus
E = σ/ε
stress, σ=F/A (N/m²)
strain, ε=e/𝑙 (no unit)
∴ E=F𝑙/eA
Tension, Compression, Shear Stress
compression is opposite to tension
(force inwardly on object)
shear stress
(there are equal opposite forces applied across its opposite faces)
G=F𝑙/eA
A=surface area parallel to applied force
Bulk Modulus
Inward force from all sides
Volume changes (decreases)
Fracture
Stress too great, objects break or fracture
Example:
concrete bad tensile strength
reinforced concrete
(concrete + iron) much stronger
Truss
- device to support large spans
- framework of rods/struts joined tgt at ends by pins/rivets
- arranged as triangle (relatively stable)
- a strut has only 2 forces on it, at the ends
- forces are equal and opposite (equilibrium)
- (real case) strut with mass, there are 3 forces
Arches
pointed arch, steeper arch
less force horizontal component
more nearly vertical is the force exerted at arch base
reduce load on wall
more openness and light
NG HUI YUN
A20SC0187