Dynamics
• Force - a quantitative description of the
interaction between two physical bodies;
that which produces or prevents motion;
that which can impose a change of velocity
on a material
Types of forces –
Weight (Fg) – force due to gravity (down)
Normal Force (FN) – force pressing the surfaces together
(perpendicular to the surface)
Tension (FT) – force in a rope (pull)
Friction (f) – force due to the nature of the surfaces in contact
(opposite the direction of motion or the direction it would
tend to move)
Force Calculations
• Fg - always equal to mg
• FN – usually equal to Fg
• FT – determined by problem solving
• f ≤ μFN
where: μ = coefficient of friction
If the object is moving
its an = sign!!!!!!
Dynamics Problem Solving
Draw a free body diagram labeling all the
forces acting on the object
Outside the free body diagram label the
direction of the acceleration.
Write an expression for the sum of the forces in
and opposite to the direction of the acceleration.
Set that expression equal to mּa and solve for the
unknown
Consider a mass, m, hanging
from a rope accelerating upwards
with acceleration, a.
Draw the free body diagram:
F= FT – Fg
FT
a
F = ma
ma = FT - Fg
Fg
ma = FT – mg
ma + mg = FT
What is the tension in the rope in this scenario?
FT = m(a + g)
Consider a mass m, pulled along a frictionless
surface with acceleration, a.
Draw the free body diagram:
FN
F = FT
FT
F = ma
Fg
ma = FT
a
What is the tension the rope in this scenario?
FT = ma
Consider mass, m, pulled with a force of FT
across a surface. The coefficient of friction
between the surfaces is μ.
f = μFN
Draw the free body diagram:
FN = Fg
FN
F = FT - f
FT
Fg = mg
f = μmg
F = ma
f
Fg
ma = FT - f
a
ma = FT - μmg
What is the acceleration of the mass in this scenario?
FT
a   g
m
More involved dynamics
problems
• Multiple body
• Incline plane
• Push/Pull
Best reviewed with an
example of each to follow
Determine a in the following in terms
of m1, m2, and g.
Draw two free body diagrams, one for each object:
FN
m1
m2
m1
Fg
1
FT
FT
f
a
m2
Fg
Write a ΣF
expression for
each mass
a
mass 1
F = FT - f
F = m1a
m1a = FT - f
2
mass 2
F = Fg - FT
2
F = m2a
m2a = Fg - FT
2
m1a = FT – μm1g
m2a = Fg2 – FT
f =μFN
FN = Fg
m2a = m2g - FT
1
Fg = m1g
1
FT = m2g – m2a
f = μm1g
FT = m1a + μm1g
Solve both for FT and
set them equal
Solve for a
(m1 + m2)a = (m2 – μm1)g
m1a + μm1g = m2g – m2a
+ m2a – μm1g – μm1g + m2a
m1a + m2a = m2g – μm1g
(m 2  m1 )
a
g
(m1  m 2 )
Consider mass, m, pulled with a force Fpull up an
incline with an angle of θ. The coefficient of
friction between the surfaces is μ. Determine the
acceleration of the system.
For ANY object on
an incline, Fg can
be broken into two
components:
Draw the free body diagram:
F = ma
F = Fpull – f - F||
Fg
ma = Fpull – f - F||
F|| (Fg sinӨ) ▬
always parallel to
and down the
incline
FN (Fg cosӨ) ▬
always
perpendicular to
the incline
ma = Fpull – f – F||
F|| = Fg(sinθ) = mg(sinθ)
FN = Fg(cosθ) = mg(cosθ)
Determine friction from FN and µ:
f =μFN
f = μmg(cosθ)
Substitute and solve for a
ma = Fpull – μmg(cosθ) – mg(sinθ)
a
Fpull
m
 g(cos )  g(sin )
Consider a block of mass m pulled along a horizontal
surface of coefficient of friction, μ, with force, FT, at an
angle of θ. Determine the acceleration
Draw the free body diagram:
FN
Break FT into its components
FT
θ
FT
FT
θ
FT
f
Fg
y
x
a
FT = FT cosӨ
x
FT = FT sinӨ
y
F = FT – f
x
F = ma
Recall: The y-component of the
pull reduces the normal force
(the y-component of a push
would increase the normal
force)
ma = FT – f
x
FN = F g - F T
FN = mg – FT(sinθ)
ma = FT(cosθ) – μ(mg – FT(sinθ))
ma = FT(cosθ) – μmg + μFT(sinθ)
FT (cos    sin )
a
 g
m
y