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Chapter 6: Forces and Equilibrium  6.1 Mass, Weight and Gravity  6.2 Friction  6.3 Equilibrium of Forces and Hooke’s Law.

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Presentation on theme: "Chapter 6: Forces and Equilibrium  6.1 Mass, Weight and Gravity  6.2 Friction  6.3 Equilibrium of Forces and Hooke’s Law."— Presentation transcript:

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2 Chapter 6: Forces and Equilibrium  6.1 Mass, Weight and Gravity  6.2 Friction  6.3 Equilibrium of Forces and Hooke’s Law

3 Chapter 6 Objectives  Calculate the weight of an object using the strength of gravity (g) and mass.  Describe the difference between mass and weight.  Describe at least three processes that cause friction.  Calculate the force of friction on an object when given the coefficient of friction and normal force.  Calculate the acceleration of an object including the effect of friction.  Draw a free-body diagram and solve one-dimensional equilibrium force problems.  Calculate the force or deformation of a spring when given the spring constant and either of the other two variables.

4 Chapter 6 Vocabulary  ball bearings  coefficient of friction  coefficient of static friction  compressed  deformation  dimensions  lubricant  normal force  prototype  restoring force  rolling friction  sliding friction  engineering  engineering cycle  extended  free-body diagram  g forces  Hooke’s law  spring  spring constant  static friction  subscript  viscous friction  weightless

5 Inv 6.1 Mass versus Weight Investigation Key Question: How are mass and weight related on Earth?

6 6.1 Mass, Weight, and Gravity  Mass is a measure of matter.  Mass is constant.  Weight is a force.  Weight is not constant.

7 6.1 Mass, Weight, and Gravity  The weight of an object depends on the strength of gravity wherever the object is.  The mass always stays the same.

8 6.1 Calculating weight with mass and gravity  The weight of an object depends on its mass and the strength of gravity.  The formula gives the weight (F w ) in terms of the mass of an object, m, and the strength of gravity, g.

9 6.1 Two meanings for “g”  The symbol g stands for the acceleration of gravity in free fall, which is 9.8 m/s 2.  Another meaning for g is the strength of gravity, which is 9.8 N/kg.  Sometimes it is more natural to discuss gravity in N/kg instead of m/s 2 because objects may not be in motion but they still have weight.  The two meanings for g are equivalent since a force of 9.8 N acting on a mass of 1 kg produces an acceleration of 9.8 m/s 2.

10 6.1 Gravity, acceleration and weightlessness  An object is weightless when it experiences no net force from gravity.  If an elevator is accelerating downward at 9.8 m/sec 2, the scale in the elevator shows no force because it is falling away from your feet at the same rate you are falling.

11 6.1 Gravity, acceleration and weightlessness  Airplane pilots and race car drivers often describe forces they feel from acceleration as g forces.  These g forces are not really forces at all, but are created by inertia.  Remember, inertia is resistance to being accelerated.

12 6.1 Using weight in physics problems  Like other forces, weight is measured in newtons or pounds.  Very often, weight problems involve equilibrium where forces are balanced.  The other common type of weight problem involves other planets, or high altitudes, where the strength of gravity (g) is not the same as on Earth’s surface.

13 1.You are asked to find force. 2.You are given a mass of 10 kilograms. 3.The force of the weight is Fw = mg and g = 9.8 N/kg. 4.The word “supported” means the ball is hanging motionless at the end of the rope. That means the tension force in the rope is equal and opposite to the weight of the ball.  Fw = (10 kg) × (9.8 N/kg) = 98 N.  The tension force in the rope is 98 newtons. Calculating force required to hold up an object  A 10-kilogram ball is supported at the end of a rope. How much force (tension) is in the rope?

14 1.You are asked for the weight. 2.You are given the weight on Earth and the strength of gravity on Jupiter. 3.Use Fw = mg. 4.First, find the person’s mass from weight on Earth:  m = (490 N) ÷ (9.8 N/kg) = 50 kg.  Next, find the weight on Jupiter:  Fw = (50 kg) × (23 N/kg) = 1,150 N (259 lbs) Calculating weight on Jupiter How much would a person who weighs 490 N (110 lbs) on Earth weigh on Jupiter? Since Jupiter may not have a surface, on means at the top of the atmosphere. The value of g at the top of Jupiter’s atmosphere is 23 N/kg.

15 Chapter 6: Forces and Equilibrium  6.1 Mass, Weight and Gravity  6.2 Friction  6.3 Equilibrium of Forces and Hooke’s Law

16 Inv 6.2 Friction Investigation Key Question: What happens to the force of sliding friction as you add mass to a sled?

17 6.2 Friction  Friction results from relative motion between objects.  Friction is a resistive force.  Describing friction as resistive means that it always works against the motion that produces it.

18 6.2 Types of Friction  Static friction  Sliding friction  Rolling friction

19 6.2 Types of Friction  Air friction  Viscous friction

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21 6.2 A model for friction  No single model or formula can accurately describe the many processes that create friction.  Some of the factors that affect friction include the type of material, the degree of roughness, and the presence of dirt or oil.  Even friction between two identical surfaces changes as the surfaces are polished by sliding across each other.

22 6.2 A model for friction  The coefficient of friction is a ratio of the strength of sliding friction between two surfaces compared to the force holding the surfaces together, called the normal force.  The coefficient of friction is most often a number between zero and one.

23 6.2 Dry sliding friction  The symbol for coefficient of friction is the Greek letter μ.  A coefficient of one means the force of friction is equal to the normal force.  A coefficient of zero means there is no friction no matter how much force is applied to squeeze the surfaces together. F f =  F n Normal force (N) Coefficient of friction Friction force (N)

24 1.You are asked for the force of friction F f. 2.You are given weight F w, applied force F, and coefficient of sliding friction μ. 3.The normal force is the sum of forces pushing down on the floor, so use F f = μF n. 4.First, find the normal force: F n = 100 N + 10 N = 110 N  Use F f = μF n and substitute values: F f = (0.25)(110 N) = 27.5 N Calculate force of friction A 10-N force pushes down on a box that weighs 100 N. As the box is pushed horizontally, the coefficient of sliding friction is Determine the force of friction resisting the motion.

25 6.2 Calculating the force of friction  The normal force is the force perpendicular to two surfaces which are moving relative to each other.  In many problems, the normal force is the reaction in an action-reaction pair.

26 6.2 Static friction  It takes a certain minimum amount of force to make an object start sliding.  The maximum net force that can be applied before an object starts sliding is called the force of static friction.

27 6.2 Static Friction  The coefficient of static friction ( μ s ) relates the maximum force of static friction to the normal force.  It takes more force to break two surfaces loose than it does to keep them sliding once they are already moving. F f =  s F n Normal force (N) Coefficient of sliding friction Friction force (N)

28 6.2 Table of friction coefficients

29 1.You are asked for the force to overcome static friction F f 2.You are given the weight F w. Both surfaces are steel. 3.Use F f ≤ μ s F n 4.Substitute values: F f ≤ (0.74) (50 N) = ≤ 37 N Calculate the force of static friction A steel pot with a weight of 50 sits on a steel countertop. How much force does it take to start to slide the pot?

30 6.2 Friction and motion  When calculating the acceleration of an object, the F that appears in Newton’s second law represents the net force.  Since the net force includes all of the forces acting on an object, it also includes the force of friction.  The real world is never friction-free, so any useful physics must incorporate friction into practical models of motion.

31 1.You are asked for the acceleration a. 2.You are given the applied force F, the mass m, and the coefficient of rolling friction μ. 3.Use: a = F ÷ m, F f = μF n, F w = mg and g = 9.8 N/kg. Calculating the acceleration of a car including friction The engine applies a forward force of 1,000 newtons to a 500-kilogram car. Find the acceleration of the car if the coefficient of rolling friction is 0.07.

32  The normal force equals the weight of the car:  F n = mg = (500 kg)(9.8 N/kg) = 4,900 N.  The friction force is: F f = (0.07)(4,900 N) = 343 N.  The acceleration is the net force divided by the mass:  a = (1,000 N – 343 N) ÷ 500 kg = 657 N ÷ 500 kg  a = 1.31 m/s 2 Calculating the acceleration of a car including friction

33 6.2 Reducing the force of friction  Friction cannot be completely eliminated but it can be reduced.  A fluid used to reduce friction is called a lubricant.  In systems where there are axles, pulleys, and rotating objects, ball bearings are used to reduce friction.  Another method of reducing friction is to separate two surfaces with a cushion of air.

34 6.2 Using friction  There are many applications where friction is both useful and necessary.  Friction between brake pads and the rim slows down a bicycle.  All-weather tires have treads, patterns of deep grooves to channel water away from the road-tire contact point.  Friction keeps nails and screws in place.  Cleats greatly increase the friction between the sports shoe and the ground.

35 Chapter 6: Forces and Equilibrium  6.1 Mass, Weight and Gravity  6.2 Friction  6.3 Equilibrium of Forces and Hooke’s Law

36 Inv 6.3 Equilibrium of Forces and Hooke’s Law Investigation Key Question: How do you predict the force on a spring?

37 6.3 Equilibrium and Hooke's Law  When the net force acting on an object is zero, the forces on the object are balanced.  We call this condition equilibrium.

38 6.3 Equilibrium and Hooke's Law  A moving object continues to move with the same speed and direction.  Newton’s second law states that for an object to be in equilibrium, the net force, or the sum of the forces, has to be zero.

39 6.3 Equilibrium and Hooke's Law  Acceleration results from a net force that is not equal to zero.

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41  You are asked for acceleration.  You are given mass and force.  Use a = F ÷ m.  First add the forces to find the net force.  F = - 75N - 25N + 45N + 55N = 0 N, so a = 0 Calculating the net force from four forces Four people are pulling on the same 200 kg box with the forces shown. Calculate the acceleration of the box.

42 6.3 Free-body diagrams  To keep track of the number and direction of all the forces in a system, it is useful to draw a free-body diagram.  A free-body diagram makes it possible to focus on all forces and where they act

43 6.3 Free-body diagrams  Forces due to weight or acceleration may be assumed to act directly on an object, often at its center.  A reaction force is usually present at any point an object is in contact with another object or the floor.  If a force comes out negative, it means the opposes another force.

44 6.3 Applications of equilibrium  If an object is not moving, then you know it is in equilibrium and the net force must be zero.  You know the total upward force from the cables must equal the downward force of the sign’s weight because the sign is in equilibrium. What is the upward force in each cable?

45  You are asked for the force on one chain.  You are given 2 forces and the mass  Use: net force = zero, F w = mg and g = 9.8 N/kg.  Substitute values: F w = mg = (150 kg)(9.8 N/kg) = 1,470 N.  Let F be the force in the other chain, equilibrium requires:  F + (600 N) = 1,470 N F = 1,470 N – 600 N  So: F = 870 N. Using equilibrium to find an unknown force Two chains are used to lift a small boat. One of the chains has a force of 600 newtons. Find the force on the other chain if the mass of the boat is 150 kilograms.

46 6.3 Applications of equilibrium  Real objects can move in three directions: up-down, right-left, and front-back.  The three directions are called three dimensions and usually given the names x, y, and z.  When an object is in equilibrium, forces must balance separately in each of the x, y, and z dimensions.

47 6.3 The force from a spring  A spring is a device designed to expand or contract, and thereby make forces in a controlled way.  Springs are used in many devices to create force.  There are springs holding up the wheels in a car, springs to close doors, and a spring in a toaster that pops up the toast.

48 6.3 The force from a spring  The most common type of spring is a coil of metal or plastic that creates a force when it is extended (stretched) or compressed (squeezed).

49 6.3 The force from a spring  The force from a spring has two important characteristics:  The force always acts in a direction that tries to return the spring to its unstretched shape.  The strength of the force is proportional to the amount of extension or compression in the spring.

50 6.3 Restoring force and Hooke’s Law  The force created by an extended or compressed spring is called a “restoring force” because it always acts in a direction to restore the spring to its natural length.  The change a natural, unstretched length from extension or compression is called deformation.  The relationship between the restoring force and deformation of a spring is given by the spring constant (k).

51 6.3 Restoring force and Hooke’s Law  The relationship between force, spring constant, and deformation is called Hooke’s law.  The spring constant has units of newtons per meter, abbreviated N/m.

52 6.3 Hooke's Law  The negative sign indicates that positive deformation, or extension, creates a restoring force in the opposite direction. F = - k x Spring constant N/m Force (N) Deformation (m)

53  You are asked for force.  You are given k and x.  Use F = - kx  Substitute values: F = - (250 N/m)(0.01 m) F = N Calculate the force from a spring A spring with k = 250 N/m is extended by one centimeter. How much force does the spring exert?

54 6.3 More about action-reaction and normal forces  The restoring force from a wall is always exactly equal and opposite to the force you apply, because it is caused by the deformation resulting from the force you apply.

55  You are asked for the deformation, x.  You are given force, F and spring constant, k.  Use F = - kx, so x = - F ÷ k  Substitute values: x = - (500 N/m) ÷ (1 × 10 8 N/m)  x = - 5 × meters (a very small deformation) Calculate the restoring force The spring constant for a piece of solid wood is 1 × 10 8 N/m. Use Hooke’s law to calculate the deformation when a force of 500 N (112 lbs) is applied.

56  We are surrounded by structures.  To design a structure well, you first need to know what forces act and how, and where the forces are applied.  Engineering is the application of science to solving real-life problems, such as designing a bridge. The Design of Structures


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