Foundation course - PHYSICS
Lecture 7-1: Equilibrium and elasticity
Gravitational force
● Equilibrium
● Elasticity and deformations
● Gravitational force
Naďa Špačková spackova@physics.muni.cz
τ = r F sinϕ
Two equivalent ways of computing the torque:
τ =(r)(F sinϕ)= r Ft
τ =(r sinϕ)(F)= r⊥ F
τ = r×F
Torque
Couple of forces
Object pivoted about an axis through its
center of mass:
Two forces of equal magnitude form a couple if their lines of action are
different parallel lines. Because each force produces the same torque Fd,
the net torque has a magnitude of 2Fd.
Equivalent forces:
● two forces F1
and F2
are equivalent if and only if F1
= F2
and if
and only if the two produce the same torque about any axis
These forces are
not equivalent
τ = F xx
Example:
F1
F2
P
F1
F2
F1
F2
P P
The figure shows (in the overhead view) two forces of the same magnitude of
30 N acting on a square of 0.6 m side that can rotate about point P. What is the
net torque about the pivot point?
Momentum
Linear momentum p:
p = mv
Newton’s second law expressed in terms of momentum:
Fnet =
d p
dt
Fnet = ma = m
d v
d t
τ net = I α = I Δω
Δt
=
Δ L
Δt
Angular momentum L:
L = I ω
τ net =
d L
d t
Law of conservation of momentum:
Momentum of any closed, isolated system does not change.
Equilibrium
Two requirements for the object’s equilibrium:
● the linear momentum P of the center of mass is constant
● the angular momentum L about the center of mass, or about any
other point, is constant
P = constant L = constant
Examples of objects in equilibrium:
● a book resting on a table
● a hockey puck sliding with a constant velocity across a frictionless
surface
● the rotating blades of a ceiling fan
● the wheel of a bicycle traveling along a straight path at constant
speed
Static equilibrium
Static equilibrium
● objects do not move in a given reference frame
P = 0 L = 0
A balancing rock in static equilibrium
An object displaced by a force
from its equilibrium position
● returns to the state of static
equilibrium – stable static equilibrium
● moves to another state – unstable
static equilibrium
Static equilibrium
Requirements of equilibrium:
Fnet =
Δ P
Δt
Fnet = 0
τ net =
Δ L
Δt
τ net = 0
The vector sum of all the external forces that act on the body must be zero.
The vector sum of all external torques that act on the body, measured
about any possible point, must also be zero.
Forces in one xy plane: Fnet , x = 0 Fnet , y = 0 τnet , z = 0
Static equilibrium
Center of gravity
The gravitational force Fg
on a body effectively acts at a single point,
called the center of gravity (cog) of the body.
If g is the same for all elements
of a body, then the body’s
center of gravity is coincident
with the body’s center of mass.
τ i = xi Fgi τ = xcog Fg
(m1 g + m2 g +...+ mi g)xcog = x1 m1 g + x2 m2 g +...+ xi mi g
Homogeneous gravitational field:
xcog =
x1 m1 + x2 m2 +...+ xi mi
m1 + m2 +...+ mi
= xcom
xcog = xcom
Example of static equilibrium:
A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N,
respectively, as shown in the figure. If the support (called the fulcrum) is under the center
of gravity of the board and if the father is 1.00 m from the center,
(a) determine the magnitude of the upward force n exerted on the board by the support.
(b) Determine where the child should sit to balance the system.
The seesaw
The figure gives an overhead view of a uniform rod in static equilibrium.
(a) Can you find the magnitudes of unknown forces F1
and F2
by balancing the forces?
(b) If you wish to find the magnitude of force F2
by using a balance of torques equation,
where should you place a rotation axis to eliminate F1
from the equation?
(c) The magnitude of F2
turns out to be 65 N. What then is the magnitude of F1
?
Example of static equilibrium:
Tiptoe standing
Figure shows the anatomical structures in the lower leg and foot that
are involved in standing on tiptoe, with the heel raised slightly off the
floor so that the foot effectively contacts the floor only at point P.
Assume distance a = 5.0 cm, distance b = 15 cm, and the person's
weight W = 900 N. Of the forces acting on the foot, what are the
(a) magnitude and direction of the force at point A from the calf muscle,
(b) magnitude and direction of the force at point B from the lower leg
bones?
calf muscle
lower leg bones
(fbula and tibia)
Achilles
tendon
Example of static equilibrium:
Elasticity
Solid matter
● three-dimensional lattice – repetitive arrangement in which each atom
is a well-defined equilibrium distance from its nearest neighbors
● atoms are held together by interatomic forces that are modeled as tiny
springs
● rigid bodies are to some extent elastic (their dimensions can be
slightly changed by pulling, pushing, twisting, or compressing them)
Deformations of solids
Stress
● quantity that is proportional to the force causing a deformation
● the external force acting on an object per unit cross-sectional area
Strain
● measure of the degree of deformation
Strain is proportional to stress
(for sufficiently small stresses)
Elastic modulus
● the constant of proportionality which
depends on the material being deformed
and on the nature of the deformation
stress = elastic modulus × strain
Three types of deformations:
● elasticity in length
● elasticity of shape
● elasticity of volume
Deformations of solids
Tension and compression
σ = E
Δ L
L
σ = Eε
σ tensile stress
E Young’s modulus
ε tensile strain (fractional change of length)
σ =
F
S
Force per unit area = pressure
units: 1 pascal = 1 Pa = 1 kg.m-1
.s-2
Δ L
L
relative elongation
● dimensionless
● fractional (sometimes
percentage) change in a length
Hooke’s law:
● linear dependence of a tensile
stress on a relative elongation
Deformations of solids
σ shear stress
G shear modulus
Shearing
σ =G
Δ x
L
Hydraulic stress
p = B
ΔV
V
p pressure (hydraulic stress)
B bulk modulus
Both solids and liquids have a bulk modulus.
No shear modulus and no Young’s modulus are
given for liquids because they do not sustain a
shearing stress or a tensile stress (it flows instead).
The reciprocal of the bulk modulus
is called the compressibility
Example:
What is the magnitude of the force acting on a steel guitar string of a length
L = 0.65 m and a cross-section area S = 0.325 mm2
, if it elongates by 5 mm?
Young’s modulus of steel is 220 GPa.
Gravitational force
● the force of attraction between two objects which is proportional to
the objects’ masses
● this force acts between any two objects in the universe
Law of universal gravitation
Objects attract other objects with a force that is proportional to the product
of their masses and inversely proportional to the square of the distance
between them.
Fg = G
m1 m2
r
2
Newton’s law of universal gravitation
G is the gravitational constant: G = 6.67 × 10-11
N.m2
/kg2
Example:
How would change the gravitational acceleration on the Earth’s surface, if
the Earth’s radius is a half of its value and (a) the Earth’s mass is the same,
and (b) the Earth’s density is the same?
Gravitational potential energy
● it characterizes a system of two particles
● we set Ep
= 0 for r = ∞
Ep =−
G m M
r
Gravitational potential energy
Escape speed
● a minimal speed of the particle to escape Earth’s gravitational
field from its surface
Ek + Ep =
1
2
mv
2
−
Gm M
r
= 0 v =
√2G M
r
Newton’s thought experiment:
● a cannon on a high mountain, firing a
cannonball horizontally with a given
horizontal speed
● in case of very high horizontal speed the
cannonball would travel all the way around
Earth
● to ignore air resistance, the distance from
the surface must be more than 150 km
Satellite in an orbit: Fcentripetal = Fgravitational
mv2
r
=
G M m
r
2
v =
√G M
r
Satellite’s speed: Satellite’s orbital period:
m4π2
r
T
2
=
G M m
r
2
T = 2π
√ r
3
G M
Orbits of planets and satellites
The gravitational field
Gravitational acceleration:
F =
G M m
r
2
= mag
On Earth’s surface ag
= g, r = rE
g =
G M
rE
2 M =
grE
2
G
ag = g(rE
r )
2
free-fall acceleration is
distance-dependent
The strength of Earth’s gravitational field varies
inversely with the square of the distance from
Earth’s center.
Earth’s gravitational field depends on Earth’s mass
but not on the mass of the object experiencing it.
ag =
G M
r
2
The gravitational field
Influence of the centripetal force to the freefall
acceleration:
FN − mag =−mω2
R
g = ag − ω
2
R
m g = mag − mω2
R
Difference in free-fall acceleration and
gravitational acceleration is in centripetal
acceleration (g is latitude-dependent).
FN − mag =−mac