Newtons First Second Laws Quiz Questions Newtons 1

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Newton’s First & Second Laws • • Quiz, Questions. Newton’s 1 st Law Newton’s

Newton’s First & Second Laws • • Quiz, Questions. Newton’s 1 st Law Newton’s 2 nd Law Examples of 2 nd Law Weight mg Examples of 2 nd Law with weight 2 -D force examples

Force and Newton’s 1 st Law • Force – Intuitive definition. – Contact, tension,

Force and Newton’s 1 st Law • Force – Intuitive definition. – Contact, tension, gravity, normal. – Vector, magnitude and direction, units. • Newton’s First Law – Every object continues in its state of rest, or uniform velocity in straight line, as long as no net force acts on it. • • • State of rest or uniform velocity (a= 0) Uniform velocity in straight line (a= 0) No force (F = 0) No NET force ( F = 0) Vector statement F = 0 ↔ a = 0

Mass and Newton’s 2 nd Law • Mass – Quantity of matter – Resistance

Mass and Newton’s 2 nd Law • Mass – Quantity of matter – Resistance to force, inertia – Scalar, units kilogram, approx. 1 L plastic soda bottle • Newton’s 2 nd Law – The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on it. • Force proportional to acceleration • Force inversely proportional to mass • Units of force kg-m/s 2 (Newtons - N) • Force direction same as acceleration direction • Net force on the left • Vector statement between F and a. X and y components • 1 st law is a subset of 2 nd law F = ma

Examples of Newton’s 2 nd • 10 N force on 2 kg mass (“forward”)

Examples of Newton’s 2 nd • 10 N force on 2 kg mass (“forward”) – Find acceleration • 10 N – 4 N force on 2 kg mass (“forward”) – Find acceleration – Find velocity and position at 2 s • 2 kg mass accelerates at 4 m/s 2 (“reverse”) – Find force • 2 kg mass accelerates to 4 m/s in 4 m (“reverse”) – Find net force – Find pulling force if friction = 3 N

Acceleration “g” and weight “mg” • All free-falling objects* accelerate at g, then gravity

Acceleration “g” and weight “mg” • All free-falling objects* accelerate at g, then gravity exerts force mg – Mass m accelerating g requires force mg (“reverse”) – 1 kg mass weighs 1 kg * 9. 8 m/s 2 = 9. 8 Newtons (N). – Force mg divided by mass m causes acceleration g (“forward”) – Weight - draw force vector on object pointing down. – If only weight, ΣF = mg and a = g – If countered by suspension, ΣF = 0 and a = 0. – Suspension force pushes/pulls up, perfectly balances. – Remove suspension force, object falls. – Suspension force = tension or “normal” force.

Vertical Forces - Example 4 -6, 4 -7 • 98 N ? N 40

Vertical Forces - Example 4 -6, 4 -7 • 98 N ? N 40 N 98 N ? N 98 N 100 N ? N 98 N

Elevator – Example 4 -8 • FT ma 696 N

Elevator – Example 4 -8 • FT ma 696 N

Force Vectors - Example 4 -9 • Crate acted on by – horizontal forces

Force Vectors - Example 4 -9 • Crate acted on by – horizontal forces – vertical forces. • Canal Boat – Add forces as vectors – Get net force components – Get net force and direction – Get acceleration components – Get acceleration – Problem solving strategy

2 -D Example 4 -11 • Box with x and y problem • Zero

2 -D Example 4 -11 • Box with x and y problem • Zero acceleration in y • Constant acceleration in x • Calculate forces in y – Fp sin(30) = 20 N, FN= 78 N • Calculate acceleration in x – Fp cos(30) = 34. 6 N – a = 34. 6 N / 10 kg = 3. 46 m/s 2 – Can then calculate v and x