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**Newton’s Laws Overview**

P H Y S I C S Newton’s Laws Overview

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**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

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Inertia Depends on mass More mass more resistance Less mass less resistance

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Demo: NFL Hits

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**Equilibrium Equilibrium: Net force is zero (ΣF = 0) ΣFx = 0 ΣFy = 0 FN**

FEngine FAir Fg

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**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

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

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**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)

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**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

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**Use one of the equations you just wrote down…**

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**Acceleration 𝑎= Σ𝐹 𝑚 Increase acceleration by: Increasing force**

Decreasing mass

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**Weight vs Mass Weight Force Fg Fg = m ·g**

Mass: Amount of matter (does not change) Weight: Pull of gravity (changes)

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**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)

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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

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**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

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**Friction Force that resists motion due to imperfections in surfaces**

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**Two Types Static (rest): Keeps object from moving**

Kinetic (moving): Slows moving object

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**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)

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**Coefficient of Friction Table**

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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

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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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Unit 2 Forces & Motion. Forces Force- Ability to change motion(push or pull) Units of lb, N=kg. m/sec 2 If forces are balanced then the object won’t move.

Unit 2 Forces & Motion. Forces Force- Ability to change motion(push or pull) Units of lb, N=kg. m/sec 2 If forces are balanced then the object won’t move.

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