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Forces

Forces. Unit 2: Forces Chapter 6: Systems in Motion. 6.1 Motion in Two Dimension 6.2 Circular Motion 6.3 Centripetal Force, Gravitation, and Satellites 6.4 Center of Mass. Key Question: Which launch angle will give a marble the best range?. 6.1 Investigation: Launch Angle and Range.

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Forces

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  1. Forces

  2. Unit 2: Forces Chapter 6: Systems in Motion • 6.1 Motion in Two Dimension • 6.2 Circular Motion • 6.3 Centripetal Force, Gravitation, and Satellites • 6.4 Center of Mass

  3. Key Question: Which launch angle will give a marble the best range? 6.1 Investigation: Launch Angle and Range • Objectives: • Use the Marble Launcher to find the launch angle that produces the maximum range for a projectile.

  4. Motion in two dimensions • Real objects do not move in straight lines alone; their motion includes turns and curves. • To describe a curve you need at least two dimensions (x and y).

  5. Displacement • Distance is scalar, but displacement is a vector. • A displacement vector shows a change in position.

  6. Displacement • A displacement vector’s direction is often given using words. • Directional words include left, right, up, down, and compass directions.

  7. Solving displacement problems • Displacement vectors can be added just like force vectors. • To add displacements graphically, draw them to scale with each subsequent vector drawn at the end of the previous vector. • The resultant vectorrepresents the displacement for the entire trip.

  8. Adding vectors A mouse walks 5 meters north and 12 meters west. Use a scaled drawing and a protractor to find the mouse’s displacement. Use the Pythagorean theorem to check your work. • Looking for:… the displacement. • Given:… distances and direction (5 m, N) and (12 m, W) • Relationships:Pythagorean theorem a2+ b2= c2 • Solution:Make a drawing with a scale of 1 cm = 2 meters. Measure the angle from the x axis. Pythagorean theorem: (5)2 + (12)2 = c2 169 = c2 13 = c The mouse walks 13 meters at 157°.

  9. Velocity vectors • Velocity is speed with direction, so velocity is a vector. • As objects move in curved paths, their velocity vectors change because the direction of motion changes. • The symbol v is used to represent a velocity vector.

  10. Velocity vectors • Suppose a ball is launched at 5 m/s at an angle of 37° • At the moment after launch, the velocity vector for the ball written as a magnitude‑angle pair is v = (5 m/s, 37°).

  11. Velocity vectors • In x-y components, the same velocity vector is written as v = (4, 3) m/s. • Both representations tell you exactly how fast and in what direction the ball is moving at that moment.

  12. Using velocity vectors A train moves at a speed of 100 km/h heading east. What is its velocity vector in x-y form? • Looking for:… the velocity vector. • Given:… speed (100 km/h) and direction (east). • Relationships: x-velocity is east and y-velocity is north. • Solution:v = (100,0) km/h

  13. Projectile motion • Any object moving through air and affected only by the force of gravity is called a projectile. • Flying objects such as airplanes and birds are not projectiles, because they are affected by forces generated from their own power.

  14. Projectile motion • The path a projectile follows is called its trajectory. • The trajectory of a projectile is a special type of arch- or bowl-shaped curve called a parabola.

  15. Trajectory and range • The rangeof a projectile is the horizontal distance it travels in the air before touching the ground. • A projectile’s range depends on the speed and angle at which it is launched.

  16. Two dimensional motion • Projectile motion is two-dimensional because there is both horizontal and vertical motion. • Both speed and direction change as a projectile moves through the air.

  17. A ball rolling off a table • The horizontal and vertical components of a projectile’s velocity are independent of each other.

  18. Horizontal velocity • The ball’s horizontal motion looks exactly like the its motion if it was it rolling along the ground at 5 m/s.

  19. Vertical velocity • The vertical (y) velocity increases due to the acceleration of gravity.

  20. NOTE: These equations are suitable only for situations where the projectile starts with zero vertical velocity, such as a ball rolling off a table.

  21. Range of a Projectile The range, or horizontal distance, traveled by a projectile depends on the launch speed and the launch angle.

  22. Range of a Projectile The range of a projectile is calculated from the horizontal velocity and the time of flight. • The air time and height are greatest when a ball is hit at an angle of 90°, but air time and height are zero when a ball is hit at an angle of 0°.

  23. Projectile motion A stunt driver steers a car off a cliff at a speed of 20.0 m/s. The car lands in a lake below 2.00 s later. Find the horizontal distance the car travels and the height of the cliff. • Looking for:… vertical and horizontal distances. • Given:… the time (2.00 s) and initial horizontal speed (20.0 m/s). • Relationships:Use : dx= vxt dy = 4.9t2 • Solution:dx= (20 m/s)(2 s) = 40 m. dy= (4.9 m/s2)(2 s)2 = (4.9 m/s2)(4 s2) = 19.6 m.

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