1. 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)

2. Force Calculations If the object is moving its an = sign!!!!!!
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!!!!!!

3. 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

4. What is the tension in the rope in this scenario?
Consider a mass, m, hanging from a rope accelerating upwards with acceleration, a. FT
Draw the free body diagram:
F= FT – Fg
F = ma
ma = FT - Fg
ma = FT – mg
ma + mg = FT a Fg
What is the tension in the rope in this scenario?
FT = m(a + g)

5. What is the tension the rope in this scenario?
Consider a mass m, pulled along a frictionless
surface with acceleration, a.
Draw the free body diagram:
F = FT
F = ma
ma = FT FN FT Fg
What is the tension the rope in this scenario? a
FT = ma

6. What is the acceleration of the mass in this scenario?
Consider mass, m, pulled with a force of FT across a surface. The coefficient of friction between the surfaces is μ.
f = μFN
FN = Fg
Fg = mg
f = μmg
Draw the free body diagram: FN
F = FT - f
F = ma
ma = FT - f FT f Fg a
ma = FT - μmg
What is the acceleration of the mass in this scenario?

7. More involved dynamics problems
Multiple body
Incline plane
Push/Pull
Best reviewed with an
example of each to follow

8. Determine a in the following in terms of m1, m2, and g.
Draw two free body diagrams, one for each object: FN m1 FT FT m1 a m2 m2
Fg1 f
Fg2 a
Write a ΣF
expression for
each mass
mass 1
F = FT - f
F = m1a
m1a = FT - f
F = Fg2 - FT
F = m2a
m2a = Fg2 - FT
mass 2

9. Solve both for FT and set them equal
m1a = FT – μm1g
m2a = Fg2 – FT
f =μFN
FN = Fg1
Fg1 = m1g
f = μm1g
m2a = m2g - FT
FT = m2g – m2a
FT = m1a + μm1g
Solve for a
Solve both for FT and set them equal
(m1 + m2)a = (m2 – μm1)g
m1a + μm1g = m2g – m2a
+ m2a
– μm1g
– μm1g
+ m2a
m1a + m2a = m2g – μm1g

10. 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:
F|| (Fg sinӨ) ▬ always parallel to and down the incline
FN (Fg cosӨ) ▬ always perpendicular to the incline a
Draw the free body diagram: FN
F = ma
F = Fpull – f - F||
ma = Fpull – f - F||
Fpull
F|| f Fg

11. 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)
ma = Fpull – μmg(cosθ) – mg(sinθ)
Substitute and solve for a

12. FTx= FT cosӨ FTy = FT sinӨ
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 f
FTy θ Fg
FTx a
FTx= FT cosӨ FTy = FT sinӨ

13. ma = FT(cosθ) – μ(mg – FT(sinθ)) ma = FT(cosθ) – μmg + μFT(sinθ)
Recall: The y-component of the pull reduces the normal force (the y-component of a push would increase the normal force)
F = FTx – f
F = ma
ma = FTx – f
FN = Fg - FTy
FN = mg – FT(sinθ)
ma = FT(cosθ) – μ(mg – FT(sinθ))
ma = FT(cosθ) – μmg + μFT(sinθ)