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Unit Four: Force and Motion

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  1. GE253 Physics Unit Four:Force and Motion John Elberfeld JElberfeld@itt-tech.edu 518 872 2082

  2. Schedule • Unit 1 – Measurements and Problem Solving • Unit 2 – Kinematics • Unit 3 – Motion in Two Dimensions • Unit 4 – Force and Motion • Unit 5 – Work and Energy • Unit 6 – Linear Momentum and Collisions • Unit 7 – Solids and Fluids • Unit 8 – Temperature and Kinetic Theory • Unit 9 – Sound • Unit 10 – Reflection and Refraction of Light • Unit 11 – Final

  3. Chapter 4 Objectives • Relate force and motion and explain what is meant by a net or unbalanced force. • State and explain Newton's first law of motion and describe inertia and its relationship to mass • State and explain Newton's second law of motion, apply it to physical situations, and distinguish between weight and mass.

  4. Chapter 4 Objectives • State and explain Newton's third law of motion and identify action-reaction force pairs. • Apply Newton's second law in analyzing various situations, use free-body diagrams, and understand the concept of translational equilibrium • Explain the causes of friction and how friction is described by using coefficients of friction.

  5. Reading Assignment • Read and study Chapters 4 in College Physics: Wilson and Buffa • Expect a quiz on any of the material covered in the course so far

  6. Written Assignments • Do all the exercises on the handout. • You must show all your work, and carry through the units in all calculations • Use the proper number of significant figures and, when reasonable, scientific notation

  7. Introduction • In the previous week, you explored the motion of objects when the velocity and acceleration of the object are known. • In real life, you have to know what actions we need to perform to cause and controlthis motion

  8. Introduction • For example, when pushing an object, you have to know what to push, where to push it, and how hard to push it. • You also want to understand what happens to an object when you push it. • In other words, you want to understand the relation between forces and the motion of objects that forces act upon.

  9. Force • A force is a push or a pull. • For example, when you try to lift a heavy object, you feel the force of gravity pulling the object down. • Springs and magnets also exert forces, as do your muscles. • Force is also associated with motion because when you exert force on an object, it may move. • Force is a vector quantity because it has direction. • When you exert force, it is in a particular direction.

  10. Net Force • This brings us to the concept of net force or total force. • If the forces acting on an object are equal and in opposite directions, they will cancel each other out. • In such a case, the total force or net force is zero.

  11. Net Forces

  12. Net Forces • If two forces of equal magnitude but opposite direction are applied to the same object, the vector resultant, or net force acting on the crate, in the x direction, is zero. • The forces are balanced • If the forces are not equal in magnitude, the resultant is not zero • A net force accelerates an object in the direction of the force

  13. Classes of Forces • There are two classes of forces: • The action-at-a-distance forces are gravity, magnetism, and electrostatic forces. • Contact forces include friction and the so-called normal forces. • For example, the force exerted by a table on an object resting on it is a contact force.

  14. Thought Experiment • Force is associated with motion, but the relationship between the two may not be clear at first. • The diagram shows an experiment performed by Galileo in which a ball rolls down and then up an inclined plane, always reaching the same level. • Therefore, if the surface is horizontal and perfectly frictionless, the ball will continue to move without stopping. • However, the ball will stop if another force (like friction) acts upon it.

  15. Thought Experiment • How far would the ball roll on a frictionless horizontal surface?

  16. Newton’s First Law • Based upon Galileo’s experiment and other observations, Newton derived the first law of motion. • An object at rest stays at rest unless acted upon by a net external force. • An object moving at a constant velocity will continue to move at that velocity unless acted on by a net external force.

  17. Common Sense • Newton’s law is in sharp contrast to the science of ancient Greece. • According to Aristotle, an ancient Greek scientist, the natural state of an object is to be at rest. • An object moves only when a force acts on it.

  18. Inertia • All objects have mass. • This means Newton’s first law can be considered a property of mass called inertia. • Inertia is the property of mass which determines its resistance to changes in velocity.

  19. Mass is in Kilograms • Mass is always measured in kilograms in physics • The mass in kilograms determines how much gravity affects an object (weight) as well as how hard it is to start or stop its motion (inertia)

  20. Practice • Every object on earth experiences the force of the earth’s gravity, and every object in the solar systems experiences the force of the sun’s gravity • All mass always has some type of force on it • An object at rest, or at constant velocity (no change in speed OR direction), has no net force on it – all forces add up to zero • If an object has a net force on it, the velocity will change (acceleration!)

  21. Practice • On a jet airliner that is taking off, you feel that you are being pushed back into the seat. Why? • You are at rest and will stay at rest unless a force acts upon you. When the plane takes off, you feel the force of the plane acting on you. • Is the opposite true when a car stops?

  22. Newton’s Second Law • When a net force acts on an object, the object accelerates. • a = Δv/Δt • The mass of an object affects its acceleration. • According to Newton’s second law • F = ma • where F and a are vectors. The unit of force is Newton (N), defined as • 1N = kg m / s2

  23. Second Law • Assuming no additional forces, like friction:

  24. F = ma

  25. Weight • The weight of an object (mass) is the force of gravity acting on the object. • Near the earth’s surface, the acceleration due to gravity is about (-) 9.8 m/s2 downward • So, you can calculate the weight of a 1-kg object as follows: • Fgravity = ma = W = mg = m (-9.8m/s2) • The weight of a 1 kg mass is: • F = ma = (1kg)(-9.8m/s2) = - 9.8kg m/s2 = -9.8N

  26. Gravity • Gravity is an example of an action-at-a-distance force. • The universal law of gravity, discovered by Newton, is that every mass in the universe is attracted to every other mass with a gravitational force proportional to the two masses and inversely proportional to the square of the distance between the two masses. • F = Gm1m2/d2

  27. Forces are the Same • F = Gm1m2/d2 • The force of m1 on m2 is the same as the force of m2 on m1 • The sun is so much more massive than the earth, the earth has little noticeable affect on the sun, but astronomers use the motion of stars to discover invisible planet

  28. On the Earth • Weight = mass x gravity • W = m g • W = m (- 9.8 m/s2 ) • A person pulls on the earth as much as the earth pulls on the person • The force caused by gravity (weight) causes falling objects to accelerate at at rate of 9.8 m/s2 downward • F = m a = W = m g • a = g = - 9.8 m/s2 on the earth’s surface

  29. Practice • A tractor pulls a loaded wagon with a mass of 275 kg on a level road with a constant force of 440 N, as shown below. • (a) What is the acceleration of the wagon? • (b) What would the acceleration of the wagon be if there was an opposing frictional force of 140 N? • (c) Suppose the wagon starts from rest (with friction). How far will the wagon travel in 4.0 s?

  30. Calculations • a) F = ma a = F/m = 440N/275 kg =1.60 N/kg 1.60 N/kg (kg m/s2 / N) = 1.60 m/s2 • b) F = Ftractor – FFriction F = 440N – 140 N = 300 N a = F/m = 300N/275kg = 1.09 m/s2 • c) x = x0 +v0t +at2/2 x = 0 + 0 +   (1.09 m/s2) *(4s)2/2 x = 8.72 m X0 X

  31. Practice • A student weighs 588 N on the surface of the earth. • What is her mass? • What would be her mass on the moon? • What would be her mass at the bottom of the ocean?

  32. Calculations • W = mg • m = W/g = 588 N / 9.8m/s2 • m = 60 kg (about 132 pounds) • 1 kg has a weight of 2.2 pounds on the surface of the earth • Her mass is 60 kg on the Moon, at the bottom of the ocean, on Mars… • Mass does NOT change with location

  33. Practice • In a college homecoming competition, 15 students lift a sports car. While holding the car off the ground, each student exerts an upward force of 420 N. • (a) What is the mass of the car in kilograms? • (b) What is the weight in pounds?

  34. Calculations • Total upward force = 15 x 420 N • F = 6,300 N • W = mg • m = W/g = 6300N / 9.8m/s2 • m = 643 kg • W = 643 kg (2.2 pounds/kg) on earth’s surface • W = 1,415 pounds

  35. Practice • In an emergency stop to avoid an accident, a shoulder-strap seat belt holds a 50-kg passenger firmly in place. • If the car were initially traveling at 80 km/h and came to a stop in 6 s along a straight, level road, what was the average force applied to the passenger by the seatbelt?   • Suppose the time is reduced to .5 s?

  36. Calculations • Find velocity in m/s • 80 km/hr (1000m/km)(1hr/3600s)= 22.2m/s • v = v0 + at • 0 = 22.2m/s + a 6s • a = -22.2m/s / 6s = -3.7 m/s2 • F = ma = 50 kg (-3.7m/s2) = -185 N • Minus indicates the force acts in a direction opposite of the initial velocity • a = -22.2m/s / .5 s = -44.4 m/s2 • F = ma = 50 kg (-44.4m/s2) = -2200 N • F = -493 pounds for a fast stop!

  37. Introduction • In the previous lessons, you learned what a force is becauseyou learned the connection between force and acceleration. • In this lesson, you will study why forces exist and where they come from. • The law of action and reaction will also be examined here.

  38. Newton’s Third Law • Newton’s second law says F = ma • Newton’s third law says the forces come from other objects • Your muscles (first object) push the crate (second object), but that crate pushes back just as hard

  39. Newton’s Third Law • Newton’s third law of motion: • For every action there is an equal and opposite reaction. • In other words, for every force, there is an equal and opposite force. • In notation form, the law is written as follows: • F12 = – F21

  40. Pairs of Forces • When an object exerts force on another object, the second object exerts an equal force at the sametime on the first object. • Objects interact in pairs; and the interacting forces come in equal and opposite pairs.

  41. Free Fall • When the briefcase is released, there is an unbalanced, or nonzero net force acting on the briefcase (its weight force,) and it accelerates downward at g or -9.8m/s2 for free fall

  42. Action-Reaction • Both skateboarders move toward the center point

  43. Paired Forces • The third law is illustrated with any kind of propulsion. • When you walk, for example, your muscles exert a horizontal force against the ground and the ground exerts an equal and opposite horizontal force against you, which propels you in the direction you are walking.

  44. Jet Propulsion • In this case, the action-reaction pair is the force of the hot gases on the plane propelling the plane forward and the force of the plane on the hot gases ejecting the gases backward.

  45. Finding Forces • The diagram shows two blocks m1 and m2 of masses 2.5 kg and 3.5 kg, respectively, resting on a frictionless surface. • A force of 12 N acts on m1, which tautens the massless string connecting the two masses. • As a result, the string exerts a force on m2. • Therefore, both masses accelerate. • What is the acceleration of the system, and what is the force that the string exerts on m2?

  46. Calculations • For the system of masses • F = ma • 12 N = (2.5kg + 3.5kg) a • a = 12 N / (6 kg) = 2.0 m/s2 • For the second mass • F = ma • F = 3.5 kg 2.0 m/s2 = 7 N

  47. Action and Reaction 7N -7N • m2 pulls back on m1 with the same force a m1 pulls forward on m2 • The net force on m2 = 7N and the net force on m1 = 12N – 7N = 5 N • If m1 were alone, only 5N would be needed to accelerate it at 2.0m/s2 F = ma = 5N = 2.5 kg 2 m/s 12N

  48. Force • A car traveling at 72.0 km/h along a straight, level road comes uniformly to a stop at a distance of 40.0 m. • If the car weighs 8800 N, what is the breaking force?

  49. Calculations • F = ma – need both m and a • v2-v02=2a(Δx) now this is very useful • 72km/hr (1000m/km)(1hr/3600s)=20m/s • 0 - (20m/s)2 = 2 a 40m • a = 5 m/s2 • W = 8800 N = m 9.8m/s2 • m =898kg • F = ma = 898kg 5m/s = 4490 N

  50. Normal Force • The diagram shows a block resting on a table. • The table is exerting an upward force on the block that is perpendicular to the surface of the table. • This is called the normal force because normal means perpendicular. • The weight of the block is equal to the normal force because the block is in equilibrium.