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Chapter 4. Force and Motion. Chapter Goal: To establish a connection between force and motion. Student Learning Objectives. • To recognize what does and does not constitute a force. • To identify the specific forces acting on an object. • To draw an accurate free-body diagram of an object.

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Chapter 4. Force and Motion


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    1. Chapter 4. Force and Motion Chapter Goal: To establish a connection between force and motion.

    2. Student Learning Objectives • To recognize what does and does not constitute a force. • To identify the specific forces acting on an object. • To draw an accurate free-body diagram of an object. • To begin the process of understanding the connection between force and motion.

    3. A force is an interaction between two objects • A force is a push or a pull on an object. • If I push a book across a table, the book pushes me back (inanimate objects can exert force!) • A force is a vector. It has both a magnitude and a direction. • The force (interaction) has the same magnitude for both me and the book. However the direction of the force on me is opposite to the direction of the force on the book.

    4. A force is an interaction between two objects • A force on an object requires an agent. The agent is another object. • This is another way of confirming that a force is an interaction between objects. It takes two to tango! • A force is either a contact force or a long-range (noncontact) force. • Gravity is the only long-range force we will study this semester. It is an interaction between two objects, but they are not necessarily in contact. • All other forces (this semester) only exist when the two objects are in contact.

    5. A Short Catalog of Forces - Gravity • Gravity is a long-range attractive force between two objects. • In this class, our emphasis is on the interaction between the Earth and objects on or near its surface (the weight force). • Your text uses W for the weight force. • Near the surface of the earth, W = mg, where m is the object mass in kilograms and g= 9.8 m/s2is the acceleration due to gravity. Earth

    6. A Short Catalog of Forces: Tension Ropes, strings, cables and hooks exert a tensional force. These agents always pull, they never push. Although it seems strange, the sled exerts a tension force on the rope.

    7. A Short Catalog of Forces - Normal The normal force is a contact force between 2 surfaces that is perpendicular to the surfaces. Your book uses FNfor the normal force

    8. A Short Catalog of Forces - friction The frictional force is a contact force between 2 surfaces that is parallel to the surfaces

    9. A Short Catalog of Forces Other forces are usually shown as an with an identifying subscript.

    10. You’ve just kicked a rock, and it is now sliding across the ground about 2 meters in front of you. Which of these forces act on the rock at this instant? Choose all that apply • Gravity, acting downward • The normal force, acting upward • The force of the kick, acting in the  direction of motion • Friction, acting opposite the direction of  motion • All of the above

    11. Drawing force vectors

    12. Free-body diagrams truck not moving due to friction • Identify all forces acting on the object. • Draw a coordinate system. If motion is along an incline, the coordinates system should be tilted relative to true horizontal and true vertical. • Represent the object as a dot at the origin. • Draw vectors representing each of the identified forces and label with the appropriate symbol.

    13. Freebody diagrams Draw a free body diagram for the truck. There is no friction, but a cable attached to the wall keeps the truck from moving.

    14. Freebody diagrams Draw a free body diagram for the sled. Assume friction between the snow and the sled.

    15. 4.3 Newton’s Second Law of Motion Newton’s Second Law An object of mass m, subject to forces F1 , F2 , etc. will undergo an acceleration with a magnitude directly proportional to the net force and inversely proportional to the mass: The direction of the acceleration is the same as the direction of the net force.

    16. 4.3 Newton’s Second Law of Motion Newton’s First Law This is a special case of the 2nd Law. If the net force equals 0, there is no acceleration and the velocity of the object will not change. If it is at rest, it will stay at rest. • In this case the object is in equilibrium. • Static equilibrium – object is at rest for a finite period • Dynamic equilibrium –object is moving at constant velocity.

    17. Graphical Interpretation of Newton’s Second Law • Newton’s 2nd Law is the equation of a line with a 0 value for the y-intercept (in this case the a- intercept!). a F

    18. The following graphs plot acceleration vs force for different objects. Which object has the greatest mass? A. B. C. D.

    19. Problem-Solving Strategy for Newton’s Law Problems

    20. Problem-Solving Strategy: Equilibrium Problems If the net force is equal to zero, must the object be at rest? The next example shows the answer to this:

    21. Newton’s 1st Law: Towing a car up a hill A car with a weight (not mass) of 15,000 N is being towed up a 20° frictionless slope at constant velocity. The tow rope is parallel to the slope surface. What is the tension on the tow rope? Why is this a Newton’s 1st Law problem (ΣF = 0) and not a Newton’s 2nd Law problem (ΣF = ma)? Draw a free body diagram for this problem.

    22. Newton’s 1st Law: Towing a car up a hill A car with a weight (not mass) of 15,000 N is being towed up a 20° frictionless slope at constant velocity. The tow rope is parallel to the slope surface. What is the tension on the tow rope? What part of the story indicates that ΣF = 0?

    23. Newton’s 1st Law: Towing a car up a hill Once the pictorial representation is completed, use Newton’s 1st Law in component form: T and n have only one non-zero component, but FGhas non-zero components in both x and y directions. How do we write the components of FG in terms of θ ?

    24. Newton’s 1st Law: Towing a car up a hill Solving the 2nd equation tells us that n = FG cos θ, which is not necessary information for this problem.

    25. Problem-Solving Strategy for Newton’s 2nd Law Problems Use the problem-solving strategy outlined for Newton’s 1st Law problems to draw the free body diagram and determine known quantities. Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. Reverse # 2 and 3 if necessary.

    26. Newton’s 2nd Law – Speed of a Towed Car • Draw a free-body diagram. Find the acceleration using Newton’s second law in component form. • Draw a pictorial representation and Find the velocity using one of the kinematic equations.

    27. Speed of a towed car – pictorial representation of motion, motion diagram and free body diagram

    28. Speed of a towed car Use the free body diagram to write Newton’s 2nd Law in component form: ΣFy = may = 0 (no change in speed up or down) ΣFy = n – FG , or n=FG (not pertinent in this problem) ΣFx = max T – f = max

    29. Speed of a towed car

    30. Speed of a towed car

    31. Apparent Weight – What the scale says (even if there is no scale!) • The value the scale reads when scale and object are accelerating or being acted upon by some force other than gravity (e.g the buoyant force in water) • Not equal to the true weight (FG = mg) • Apparent weight is either a normal force (step on scale) or tension force (hanging scale) • An object inside an accelerating elevator has the same acceleration as the elevator. • Acceleration is not a force so don’t include directly on free body diagram.

    32. Typical FBD for an apparent weight problem n FG = mg, but n does not!

    33. Apparent Weight – What the scale would say, even if there is no scale It takes the elevator in a skyscraper 4.0 s to reach a constant speed of 10.0 m/s. A 60.0 kg passenger gets on at the ground floor. What is her apparent weight during those 4 seconds? Use the problem-solving strategy (repeated in next slide).

    34. Problem-Solving Strategy for Newton’s Law Problems

    35. Problem-Solving Strategy for Newton’s 2nd Law Problems Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. Reverse # 1 and 2 if necessary (Hint: it is!)

    36. Mass – a quantitative measure of inertia An object’s mass (m) is a measure of the amount of matter that it contains. The SI unit of mass is the kilogram (kg). The English unit of mass is the slug, which is virtually never used. The mass of an object does not change unless you add or subtract matter from it. The mass of an object is the same on any planet and in deep space. Scales do not directly measure mass, except for a balance scale, which compares one mass to another.

    37. Weight • The weight of an object on or above the earth is the gravitational force that the earth exerts on the object (FG or W). • Since weight is a force, the SI unit of weight is the Newton (N); the English unit is the pound. • The weight of an object changes on other planets since the gravitational force of other planets are different than that of the earth. • A scale does not directly measure weight; it measures the normal force or (hanging scale) the tension force. In most situations, this force measured by the scale is numerically equal to the weight force.

    38. Apparent weight or “what the scale says” • In accelerating reference frames (e.g. the ever-popular elevator with a scale), the scale reading will differ from the true weight. This is called apparent weight. • Metric scales assume measurement on earth and are calibrated to read in mass units (kg). American scales read in English force units (lb). It’s a culture thing.

    39. 4.7 The Gravitational Force Newton’s Law of Universal Gravitation • Every particle in the universe exerts an attractive force on every other particle. • A particle is a piece of matter, small enough in size to be regarded as having no volume. In practice we use the particle model even for larger bodies • The Law of Universal Gravitation is an example of an inverse square law. The force between the two bodies is inversely proportional to the square of the distance between them.

    40. For two particles that have masses m1 and m2and are separated by a distance r, the force has a magnitude given by: F1-2 and F2-1 are equal in magnitude but point in opposite directions

    41. If m1 has a mass of 12 kg and m2 has a mass of 25 kg and the two objects are 1.2 meters apart, what is the magnitude of the gravitational force between them?

    42. Ratio Reasoning with Inverse Square Laws Two masses, separated by a distance of r experience a gravitational attraction, F. Now the distance is increased by a factor of 3. By what factor does the magnitude of the force change (what is the ratio )? • 1/3 c. 9 • 3 d. 1/9

    43. The figure shows the moon being attracted by the earth. The earth is also attracted by the moon. Compared to FEonM (the force shown), FMonEis: • much smaller and in the same direction. • much smaller and in the opposite direction. • the same size and same direction. • the same size and • opposite direction. moon FEonM earth

    44. 4.7 The Gravitational Force Definition of Weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. SI Unit of Weight: newton (N)

    45. If the object is close (within 100 miles) to the earth, this becomes: Define the quantity g, for all the unchanging entities: Calculate g ME = 5.98 x 1024 kg RE = 6.38 x 106 m (don’t forget to square!)

    46. The Gravitational Force g = 9.8 m/s2 on Earth 1 kg of mass x g = 9.8 N, where N = kg m/s2

    47. Ratio Reasoning A space traveler weighs 540.0 N on earth. What will he weigh on another planet whose radius is twice that of earth and whose mass is 3 times that of earth?

    48. Friction

    49. Kinetic Friction Experiments show that the kinetic friction force is nearly constant and proportional to the magnitude of the normal force. where the proportionality constant μk is called the coefficient of kinetic friction.

    50. Static Friction The box is in static equilibrium, so the static friction must exactly balance the pushing force: