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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 4 Forces and Newton’s Laws of Motion. Rescheduling. Quiz  Cancelled. Lab  Thursday’s lab will meet. CAPA  Due Thursday at noon. The Quiz That Wasn’t.

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 4 Forces and Newton’s Laws of Motion Dr. Larry Dennis, FSU Department of Physics

  2. Rescheduling Quiz  Cancelled. Lab  Thursday’s lab will meet. CAPA  Due Thursday at noon.

  3. The Quiz That Wasn’t. • This quiz will be about a car driving along a level road. The car has a mass of 2000 kg. • 1)(2 points) If the car is driving along the level road, what do you know about the component of the car’s velocity in the vertical direction? • a)The magnitude of the vertical component of the car’s velocity is greater than zero, and its direction is up. • b)The magnitude of the vertical component of the car’s velocity is greater than zero, and its direction is down. • c)The vertical component of the car’s velocity is zero.

  4. The Quiz That Wasn’t. • 2. (2 points) If the car is driving along the level road, what do you know about the component of the car’s acceleration in the vertical direction. • The magnitude of the vertical component of the car’s acceleration is greater than zero and its direction is up. • The magnitude of the vertical component of the car’s acceleration is greater than zero and its direction is down. • The vertical component of the car’s acceleration is zero.

  5. The Quiz That Wasn’t. 3) (3 points) If the car travels 600 km in a constant direction at a constant speed in 6 hrs, what is the speed at which the car travels? Make sure you clearly specify what equation you are using.

  6. The Quiz That Wasn’t. 4) (3 points) If the car is driving along the level road at a constant speed and in a constant direction, what do you know about the horizontal component of the car's acceleration? • The magnitude of the horizontal component of the acceleration is zero. • b) The magnitude of the horizontal component of the acceleration is greater than zero, and its direction is in the same direction as the motion of the car. • c) The magnitude of the horizontal component of the acceleration is greater than zero, and its direction is opposite to the direction of the motion of the car.

  7. The Quiz That Wasn’t. 5) (2 points) If the car is driving along the level road with an increasing speed and in a constant direction, what do you know about the horizontal component of the car's acceleration? a) The magnitude of the horizontal component of the acceleration is zero. b) The magnitude of the horizontal component of the acceleration is greater than zero, and its direction is in the same direction as the motion of the car. c) The magnitude of the horizontal component of the acceleration is greater than zero, and its direction is opposite to the direction of the motion of the car.

  8. The Quiz That Wasn’t. 6) (3 points) If the car's speed increases from 25 m/s to 50 m/s in 3.5 s while continuing to travel in a constant direction, what is the magnitude of the horizontal component of the car's average acceleration? Make sure you clearly specify what equation you are using.

  9. The Quiz That Wasn’t. 7) (2 points) If the car is driving along the level road with a decreasing speed and in a constant direction, what do you know about the horizontal component of the car's acceleration? • The magnitude of the horizontal component of the acceleration is zero. • The magnitude of the horizontal component of the acceleration is greater than zero, and its direction is in the same direction as the motion of the car. • The magnitude of the horizontal component of the acceleration is greater than zero, and its direction is opposite to the direction of the motion of the car.

  10. The Quiz That Wasn’t. 8) (3 points) If the car's speed decreases from 40 m/s to 20 m/s in 2.5 s while continuing to travel in a constant direction, what is the magnitude of the horizontal component of the car's average acceleration? Make sure you clearly specify what equation you are using.

  11. Forces • A push or pull that often gives rise to motion. • Newton’s Laws: • If the force on an object is zero, then it’s velocity is constant. • The acceleration = net force / mass. • Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. F = m a

  12. Lift W =mg Vector Nature of Forces • Forces have: • Magnitude • Direction F = m a Vector Scalar Vector

  13. N = Normal Force Fboy f = Friction W= Force of gravity Example: Force and Acceleration • A boy pulls a 10 kg cart across the floor. If he pulls with a force of 10 N, the cart moves with constant speed. If he pulls with a force of 15 N what is the acceleration of the cart?

  14. N Fboy f W Example: Force and Acceleration • A boy pulls a 10 kg cart across the floor. If he pulls with a force of 10 N, the cart moves with constant speed. If he pulls with a force of 15 N what is the acceleration of the cart? The normal force and weight are equal and opposite in direction (ay = 0). N – W = 0 Friction opposes the force of the boy pulling on the cart. Fboy – f = max

  15. N Fboy f W Example: Force and Acceleration • A boy pulls a 10 kg cart across the floor. If he pulls with a force of 10 N, the cart moves with constant speed. If he pulls with a force of 15 N what is the acceleration of the cart? Fboy – f = max When Fboy = 10 N, ax = 0 Fboy – f = max = 0 f = Fboy = 10 N

  16. N Fboy f W Example: Force and Acceleration • A boy pulls a 10 kg cart across the floor. If he pulls with a force of 10 N, the cart moves with constant speed. If he pulls with a force of 15 N what is the acceleration of the cart? Fboy – f = max The force of friction does not increase when the boy increases his force. Fboy – f = max  ax = (Fboy – f)/m ax = (15 N – 10 N)/10 kg = 0.5 m/s2

  17. N F  =53o W = mg Force and Acceleration A boy on his sled is being pulled across the ice on a sled, by his sister. She pulls on the rope at an angle of 53o with a force of 47 N. Assume the ice is frictionless and that the boy an the sled have a mass of 42 kg. • Determine the acceleration of the sled. • What is the normal force of the ground on the sled? • Step #1: • Draw a free body diagram. • Identify all the forces. • Show relevant angles.

  18. N F  =53o W = mg Force and Acceleration A boy on his sled is being pulled across the ice on a sled, by his sister. She pulls on the rope at an angle of 53o with a force of 47 N. Assume the ice is frictionless and that the boy an the sled have a mass of 42 kg. Step #2: Use Newton’s Laws for forces in the horizontal and vertical directions. Horizontal forces. F cos  = max Vertical forces (acceleration = 0). F sin  + N – mg = 0

  19. N F  =53o W = mg Force and Acceleration A boy on his sled is being pulled across the ice on a sled, by his sister. She pulls on the rope at an angle of 53o with a force of 47 N. Assume the ice is frictionless and that the boy an the sled have a mass of 42 kg. • Determine the acceleration of the sled. Step #3: Solve for the requested values. F cos  = max ax = F cos /m = 47 N cos53o/ 42 kg = 47 * 0.6/42 = 0.67 m/s2

  20. N F  =53o W = mg Force and Acceleration A boy on his sled is being pulled across the ice on a sled, by his sister. She pulls on the rope at an angle of 53o with a force of 47 N. Assume the ice is frictionless and that the boy an the sled have a mass of 42 kg. • What is the normal force of the ground on the sled? Step #3 (continued): Solve for the requested values. F sin  + N – mg = 0 N = mg - F sin  = 42 kg * 10 m/s2 – 47 N sin 53o = 382 N

  21. Next Time • Finish Chapter 4. • Sample problems. • Please see me with any questions or comments. See you Friday.

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