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

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## Unit Four: Force and Motion

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**GE253 Physics**Unit Four:Force and Motion John Elberfeld JElberfeld@itt-tech.edu 518 872 2082**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**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.**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.**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**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**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**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.**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.**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.**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**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.**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.**Thought Experiment**• How far would the ball roll on a frictionless horizontal surface?**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.**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.**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.**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)**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!)**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?**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**Second Law**• Assuming no additional forces, like friction:**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**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**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**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**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?**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**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?**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**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?**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**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?**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!**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.**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**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**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.**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**Action-Reaction**• Both skateboarders move toward the center point**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.**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.**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?**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**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**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?**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**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.

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