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Chapter 4- Forces and Motion

Chapter 4- Forces and Motion. Erupting Volcano!!. Think about the following questions: What is this object? Where is it? Why does it look like that?. IO is a moon of Jupiter Competing forces between Jupiter and the other Galilean moons cause the center of Io compress and melt. Consequently Io

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Chapter 4- Forces and Motion

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  1. Chapter 4- Forces and Motion

  2. Erupting Volcano!! Think about the following questions:What is this object? Where is it? Why does it look like that? IO is a moon of Jupiter Competing forces between Jupiter and the other Galilean moons cause the center of Io compress and melt. Consequently Io is the most volcanically active body in the solar system.

  3. Other examples of forces

  4. What is a force? • IPC definition: A push or a pull exerted on some object • Better definition: Force represents the interaction of an object with its environment • The Unit for Force is a Newton

  5. Two major types of forces • Contact Forces: Result from physical contact between two objects • Examples: Pushing a cart, Pulling suitcase • Field Forces: Forces that do not involve physical contact • Examples: Gravity, Electric/Magnetic Force

  6. Force is a vector! (yay more vectors ) • The effect of a force depends on magnitude and direction

  7. Force Diagrams (p. 126) • Force Diagram: A diagram that shows all the forces acting in a situation

  8. Free Body Diagrams p.127 • Free Body Diagrams (FBDs) isolate an object and show only the forces acting on it • FBDs are essential! They are not optional! You need to draw them to get most problems correct!

  9. How to draw a free body diagram Situation: A tow truck is pulling a car (p. 127) We want to draw a FBD for the car only.

  10. Step 1: Draw a shape representing the car (keep it simple) Step 2: Starting at the center of the object, Draw and label all the external forces acting on the object Force of Tow Truck on Car= 5800 N Steps for drawing your FBD

  11. Force of Tow Truck on Car= 5800 N Gravitational force (Weight of car)= 14700 N Add force of gravity

  12. Normal Force = 13690 N Force of Tow Truck on Car= 5800 N Gravitational force (Weight of car)= 14700 N Add force of the road on the car(Called the Normal Force)

  13. Normal Force = 13690 N Force of Tow Truck on Car= 5800 N Force of Friction= 775 N Gravitational force (Weight of car)= 14700 N Finally add the force of friction acting on the car

  14. Fkick Fg A Free Body Diagram of a Football Being Kicked

  15. FN= 500 N Ff= 30 N Fapp= 185 N Fg= 500 N A person is pushed forward with a force of 185 N. The weight of the person is 500 N, the floor exerts a force of 500 N up. The friction force is 30 N.

  16. Forces you will need

  17. Sample Problem p. 128 #3 • Draw a free body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.

  18. Newton’s 1st Law of Motion • The Law of Inertia • An object at rest remains at rest, and an object in motion continues in motion with constant velocity (constant speed in straight line) unless the object experiences a net external force • The tendency of an object not to accelerate is called inertia

  19. Acceleration • The net external force (Fnet) is the vector sum of all the forces acting on an object • If an object accelerates (changes speed or direction) then a net external force must be acting upon it

  20. Equilibrium • If an object is at rest (v=0) or moving at constant velocity, then according to Newton’s First Law,Fnet=0 • WhenFnet =0, the object is said tobe inequilibrium

  21. How do we use this information?Sample Problem p. 133 #2 • A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N and downward with a force of 236 N. • A. Find the net external force in the x direction • B. Find the net external force in the y direction • C. Find the magnitude and direction of the net external force on the crate.

  22. Step 1: Draw a FBD Fup = 565 N Fright = 82 N Fleft = 115 N Fdown = 236 N

  23. R = 331 N at 84.3 North of West 329 N 33 N Find the vector sum of forces • A. 82 N + (-115 N )= -33 N • B. 565 N + (-236 N) = 329 N • C. Find the resultant of the two vectors from part a and b.

  24. Newton’s 1st Law • Review Newton’s 1st Law: • When Fnet=0, an object is in equilibrium and will stay at rest or stay in motion • In other words, if the net external force acting on an object is zero, then the acceleration of that object is zero

  25. Newton’s 2nd Law (p.137) • The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass

  26. Example p. 138 # 4 • A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline?

  27. Solving the problem • To calculate Fnet, we need m and a • M=2.0 kg • What is a? • Vi= 0 m/s, t=0.50 s, • displacement=85 cm=.85 m • Welcome back kinematic equations! 

  28. Newtons’ 3rd Law • Forces always exist in pairs • For every action there is an equal and opposite reaction

  29. Action- Reaction Pairs Some action-reaction pairs:

  30. Although the forces are the same, the accelerations will not be unless the objects have the same mass.

  31. Everyday Forces • Weight= Fg = mg • Normal Force= FN= Is always perpendicular to the surface. • Friction Force= Ff • Opposes applied force • There are two types of friction: static and kinetic

  32. Static Friction Force of Static Friction (Fs) is a resistive force that keeps objects stationary As long as an object is at rest: Fs = -Fapp

  33. Kinetic Friction • Kinetic Friction (Fk) is the frictional force on an object in motion • Fnet= Fapp- Fk

  34. Coefficients of Friction • The coefficient of friction (μ) is the ratio of the frictional force to the normal force • Coefficient of kinetic Friction • Coefficient of Static Friction

  35. Sample Problem p. 145 #2 • A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the char is in motion, a 327 N horizontal force keeps it moving at a constant velocity. • A. Find coefficient of static friction • B. Find coefficient of kinetic friction

  36. Coefficient of Static Friction • In order to get the chair moving, it was necessary to apply 365 N of force to overcome static friction. Therefore Fs = 365 N. • The normal force is equal to the weight of the chair (9.81 x 25= 245 N)

  37. Fk= 327 N Fapplied= 327 N Coefficient of Kinetic Friction • The problem states that the chair is moving with constant velocity, which means Fnet=0. Therefore, Fapp must equal -Fk.

  38. Solve for Coefficient of Kinetic Friction

  39. Forces at an angle • A woman is pulling a box to the right at an angle of 30 above the horizontal. The box is moving at a constant velocity. Draw a free body diagram for the situation.

  40. FN= Normal Force Fapp= Applied Force F app,y F app,x Ff= Friction Force Fg=Weight FBD

  41. What is Fnet? • Since the suitcase is moving with constant velocity, Fnet=0. • That means the forces in the x direction have to cancel out and the forces in y direction have to cancel out • Fk = Fapp,x • FN + Fapp,y = Fg • NOTICE THAT NORMAL FORCE DOES NOT EQUAL WEIGHT IN THIS SITUATION

  42. Let’s do an example. P. 154 #42 • A 925 N crate is being pushed across a level floor by a force F of 325 N at an angle of 25 above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.

  43. What do we need to know? So we need mass and Fnet. We have weight (925 N). So what is mass? How to find Fnet? Find vector sum of forces acting on crate.

  44. FN= Normal Force Fapp= 325 N Fapp,y Fapp,x Ff= Friction Force Fg=Weight=925 N FBD

  45. Finding Fnet,y • Is box accelerating in y direction? • No. Therefore Fnet in y direction is 0 • So FN + Fapp,y = Fg • So FN = Fg- Fapp,y= 925 N- 325sin(25) • FN= 787.65 N

  46. Finding Fnet,x • Is box accelerating in x direction? • Yes. Therefore Fnet,x is not 0 • Fnet,x= Fapp,x – Ff • Fapp,x = Fappcos(25)=294.6 N • Use coefficient of friction to find Ff • Ff=μFN=(0.25)(787N)=197 N

  47. Finish the Problem • Fnet,x = 294 N – 197 N= 97 N • So now we know that the Fnet on the box is 97 N since Fnet,y is 0

  48. Another example. P. 154 #54 part a • A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal. • If μk between the floor and the box is 0.57, how long does it take to move the box 4.00 m starting from rest?

  49. DRAW FBD FN Fapp,x Ff Fapp,y Fg=319 N Fapp= 485 N

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