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Chapter 5

Chapter 5. Force and Motion. The Concept of Force. A force describes the action of a body on another body. A force is capable of changing an object’s state of motion, or shape. The Concept of Force. We distinguish two types of forces:

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Chapter 5

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  1. Chapter 5 Force and Motion

  2. The Concept of Force A force describes the action of a body on another body. A force is capable of changing an object’s state of motion, or shape.

  3. The Concept of Force • We distinguish two types of forces: • A contact force, such as a push or pull, friction, tension from a rope or string, and so on. • A force that acts at a distance, such as gravity, the magnetic force, or the electric force.

  4. Contact forces - Involve physical contact between objects.

  5. Field forces: • (acting at a distance) • No physical contact between objects • Forces act through empty space gravity magnetic electric

  6. The Concept of Net Force Net force—the vector sum of all forces acting on the body. Perfectly equivalent to the group of forces, causing the same effect. ΣF = F1+F2+...Fi+...Fn Fi F2 ΣFx = F1x+F2x+...Fix+...Fnx ΣFy = F1y+F2y+...Fiy+...Fny Fn F1

  7. Example Net Force Two tugboats pull a tanker with 10000 N each at 30o. What is the net force? ΣF=173000N

  8. Newton’s First Law:The Law of Inertia Galileo’s Principle of Inertia

  9. Inertia: property to maintain actual state of motion or rest, or oppose to any change in motion. Measured by mass (kg)

  10. Inertial frame of reference: • A frame (system) that is not accelerating. • Newton’s laws hold only true in non-accelerating (inertial) frames of reference! • Are the following inertial frames of reference: • A cruising car? • A braking car? • The earth? • Accelerating car?

  11. Is Austin a good IRF? • Is Austin accelerating? • YES! • Austin is on the Earth. • The Earth is rotating. • What is the centripetal acceleration of Austin? • T = 1 day = 8.64 x 104 sec, • R ~ RE = 6.4 x 106 meters . • Plug this in: aU = .034 m/s2 ( ~ 1/300 g) • Close enough to 0 that we will ignore it. • Austin is a pretty good IRF.

  12. Force, Mass, and Newton’s Second Law Fundamental Principle

  13. Newton’s second law (very important) a (acceleration) ΣF (applied) m (mass) - A greater force causes, more acceleration to an object (effect) • The greater the mass of an object, the less it accelerates under the action of an applied force.

  14. Newton’s second law (very important) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

  15. Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m? v = 0 F m a i

  16. Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ? v F m a i d

  17. Example: Pushing a Box on Ice... Start with F= ma. a= F / m. Using Galileo’s formula: v2 - v02 = 2ad So v2 = 2Fd / m v F m a i d

  18. Example: Pushing a Box on Ice... Plug in F = 50 N, d = 10 m, m = 100 kg: Find v = 3.2 m/s. v F m a i d

  19. A force F acting on a mass m1results in an acceleration a1.The same force acting on a different mass m2 results in an acceleration a2=2a1. m1 m2 F a1 F a2 = 2a1 • If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? m1 m2 F a = ? (a)2/3 a1(b)3/2 a1(c)3/4 a1

  20. Unit of force: • The unit of force is the Newton (N) • One Newton: The force required to accelerate a 1 kg mass to 1m/s2. • 1N = 1kg·m/s2 • US Customary unit of force is a pound (lb): • 1 N = 0.225 lb

  21. The Force Due to Gravity: Weight Gravity=Weight Near Earth

  22. The force of gravity and weight • Objects are attracted to the Earth. • This attractive force is the force of gravity Fw. • The magnitude of this force is called the weight of the object. • The weight of an object is, thus mg. The weight of an object can very with location (less weight on the moon than on earth, since g is smaller). The mass of an object does not vary.

  23. You’re stranded away from your space ship. Fortunately you have a propulsion unit that provides a constant force F for 3 s. After 3 s you moved 2.25 m. If your mass is 68 kg, find F. 1. The constant force F provides the required acceleration: F = ma. 2. Find acceleration from law of motion: x = at2/2, a = 2x/t2 a = 2 * 2.25 m/(3s)2 = 0.5 m/s2 3. F = 68 kg * 0.5 m/s2 = 34 N

  24. Newton’s third law “For every action there is an equal and opposite reaction.” If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1: Action and reaction forces always act on different objects.

  25. Normal Force N or FN • Exerted on an object lying on a surface. Always perpendicular to the surface. N mg N=-mg

  26. Tension T or FT • Exerted on an object hanging on a string or rope. Same if the string or rope is not interrupted. T T = -mg mg

  27. Tension Fig. 5-9 (a) The cord, pulled taut, is under tension. If its mass is negligible, the cord pulls on the body and the hand with force T, even if the cord runs around a massless, frictionless pulley as in (b) and (c). When a cord is attached to a body and pulled taut, the cord pulls on the body with a force T directed away from the body and along the cord.

  28. Forces in Nature 4 Fundamental Forces

  29. Gravitational Force

  30. Electromagnetic Force

  31. H bomb Strong Nuclear b/w hadrons (protons, neutrons).

  32. Muon (green) from a cosmic ray interacts with an electron (red) knocked of an atom Weak Nuclear b/w leptons (electrons, muons) and b/w hadrons

  33. Hooke’s Law

  34. Molecular forces like spring forces

  35. A 110 kg basketball player hangs on a rim following a slam dunk, the rim being deflected down by 15 cm. Calculate the force constant of the rim.

  36. Problem Solving:Free-Body Diagrams Force Representation of Interactions

  37. Dog Sled Find acceleration of sled. Show forces on sled only! F: force exerted by the dog on the sled. w: weight of sled Fn: normal force

  38. Dogsled Race F = 150 N at 25o, weight of sled = 80 kg, find acceleration, normal force. 1. Draw FBD 2. Apply 2nd Law in Ox Direction: 3. Apply 2nd Law in Oy Direction:

  39. Equilibrium Example A picture weighing 8N is supported by two wires with tensions T1 and T2. Find the two tensions 1. Draw FBD 2. Apply ΣF = 0 in this case: T1 + T2 + w = 0 T1x + T2x + wx = 0 3. Break down into components: T1y + T2y + wy = 0 T1cos30 - T2cos60 + 0 = 0 T1sin30 + T2sin60 –8 N = 0 4. Solve for T2 in terms of T1 in the first equation: T2 =T1(cos30)/cos60 = √3T1 5. Plug in into second equation and solve for T1: T1sin30 + √3T1 sin60 –8 = 0 T1 = 4 N 6. Find T2 T2 = √3 x 4N = 6.93 N

  40. As Mr. Avram’s jet plane takes of with his physics class, you note that your 40 g yo-yo is deflected at an angle of 22o with the vertical. a) What is the acceleration of the plane and b) the tension in the string? Plane Example 1. Draw a FBD. Tsinθ + 0 = ma, or Tsinθ = ma * 2. Apply ΣFx = max, and develop using trig.: Tx + wx = max Tcosθ -mg = 0, or Tcosθ = mg ** 3. Apply ΣFy = may, and develop using trig.: Ty + wy = mav (Tsinθ)/(Tcosθ) = ma/mg, or tanθ = a/g or a = gtanθ , a = 9.8 x tan 22 = 3.96 m/s2 4. Divide * by **: T = mg/cosθ = (0.04Kg x 9.8 m/s2)/cos 22 = 0.423 N 5. Find T:

  41. Suppose your mass is 80 Kg, and you’re standing on a scale placed on the floor of an elevator. • What is the reading of the scale when: • the elevator is rising with an upward acceleration of a. • the elevator is descending with a downward acceleration of a. • The elevator is rising at at 20m/s, and decelerating at 8 m/s2.

  42. Inclined Plane A crate of mass m is placed on a frictionless plane of incline q. y O x x

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