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Newton’s Laws Overview

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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

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