1. Newton’s Laws Overview
P H Y S I C S
   
Newton’s Laws Overview

2. Review of Newton’s Laws of Motion
Objects in motion stay in motion*
and objects at rest stay at rest
if there is zero net force (balanced)
ΣF = m·a
(the forces will be unbalanced)
Every force has an
equal and opposite force
* straight line/constant speed
1st
2nd
3rd

3. Inertia
Depends on mass
More mass  more resistance
Less mass  less resistance

4. Demo: NFL Hits

5. Equilibrium Equilibrium: Net force is zero (ΣF = 0) ΣFx = 0 ΣFy = 0 FN
FEngine
FAir Fg

6. Equilibrium ΣF = 0  Newton’s First Law applies
An object in equilibrium can be:
in motion (straight line/constant speed)
at rest FN Ff
Fengine
Fair Fg

7. Terminal Velocity
Once the forces of air resistance and gravity become balanced equilibrium is reached
No more
acceleration

8. Newton’s Second Law If there is a net force the object will accelerate
ΣF = m·a
Units: kg· 𝑚 𝑠 2 =𝑁
ΣF  net force (N)
m  mass (kg)
a  acceleration (m/s2)

9. Newton’s Second Law F = ma a = F/m m = F / a F = net force m = Mass
Equations:
F = ma
a = F/m
m = F / a
F = net force
m = Mass
a= Acceleration

10. Use one of the equations you just wrote down…

11. Acceleration 𝑎= Σ𝐹 𝑚 Increase acceleration by: Increasing force
Decreasing mass

12. Weight vs Mass Weight Force  Fg Fg = m ·g
Mass: Amount of matter (does not change)
Weight: Pull of gravity (changes)

13. Weight Force (Fg) g = 9.8 m/s2 g = 1.6 m/s2 g = 26 m/s2 m = 50 kg
Fg = 490 N
( 110 lb)
m = 50 kg
Fg = 80 N
( 18 lb)
m = 50 kg
Fg = 1300 N
( 292 lb)

14. In-Class Problem #1
A 2000 kg car has a push force of 5000 N from its engine. If it experiences a friction force of 3000 N determine it’s (a) acceleration, (b) weight and (c) the normal force acting on it. a = 1 m/s2 Fg = 19,600 N FN = 19,600 N

15. Review of Newton’s Laws of Motion
First Law
Second Law
a = 0 m/s2
Accelerates
at rest
in motion*
depends on net force
depends inversely on mass
stays
at rest
stays
in motion*
* Straight line/constant speed

16. Friction Force that resists motion due to imperfections in surfaces

17. Two Types Static (rest): Keeps object from moving
Kinetic (moving): Slows moving object

18. Friction Force Equation
Coefficient of Friction (μ): Ratio between friction force and normal force:
𝐹 𝑓 =μ· 𝐹 𝑁
𝐹 𝑓  friction force (N)
μ  coefficient of friction
𝐹 𝑁  normal force (N) μs
(static) μk
(kinetic)

19. Coefficient of Friction Table

20. In-Class Problem #2
A 30 kg desk is at rest on the floor. It takes 200 N of force to start it in motion. Determine the static coefficient of friction between the desk and the floor. μs = 0.68

21. In-Class Problem #3
Once the desk in the previous problem is set in motion the 200 N force continues to be applied. Determine the acceleration of the desk if the coefficient of kinetic friction is a = 1.57 m/s2