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More Newton’s Laws Applications Unit 3 Presentation 2 Revisit Newton’s 3rd Law “Every action force has an equal in magnitude but opposite in direction reaction force” The action and reaction forces act on DIFFERENT objects Newton’s 3rd Law Examples Action Force Reaction Force A gun is fired, propelling a bullet forward. The gun fired backwards, in the opposite direction of the bullet. A person pushes on a wall. The wall pushes back on the person. A person walks down a The ground exerts a street, exerting a frictional normal force against the force against the ground. person, propelling them forward. Newton’s 3rd Law Example A block-block system, like the one pictured below, is pushed by a force of 15 N from the left to the right. Calculate (a) The acceleration of the block-block system (b) The magnitude of the action-reaction force in between the two blocks First, break the two blocks into individual problems and draw two free-body diagrams: 5 kg 15 N 2 kg Normal Force AR Force 15 N 2 kg Action/ Reaction Force (same on both blocks) Normal Force AR Force 15 N 5 kg Gravity 2 kg block Gravity 5 kg block Newton’s 3rd Law Example (cntd) Now, consider the free-body diagrams. Remembering to keep the x and y variables separate, set up some equations. Do we need to consider the y direction forces? NO! The motion is only in the x direction in this example. Now, using Newton’s 2nd Law on each block in the x-direction only: m1 2kg m2 5kg a? Far ? 15 N Far m1a Far m2 a Notice, we have 2 equations with 2 unknowns. Lets use substitution and substitute the right-hand equation into the left-hand equation: 15 N ( m2 a ) m1a Next, solve for the AR Force: 15 (5a ) 2a 15 7 a Far 5(2.14) 2.14m / s 2 a Far 10.7 N Another Newton’s 3rd Law Example A block-block system, like the one pictured below, is pushed and accelerates at 3.2 m/s2 from the left to the right. Calculate: (a) The force on the 8 kg block (b) The magnitude of the action-reaction force in between the two blocks 3 kg First, break the two blocks into individual problems and draw two free-body diagrams: 8 kg 3.2 m/s2 F 8 kg Action/ Reaction Force (same on both blocks) Normal Force AR Force Normal Force AR Force F 3 kg Gravity 8 kg block Gravity 3 kg block Another Newton’s 3rd Law Example (cntd) Now, consider the free-body diagrams. Remembering to keep the x and y variables separate, set up some equations. Do we need to consider the y direction forces? NO! The motion is only in the x direction in this example. Now, using Newton’s 2nd Law on each block in the x-direction only: m1 8kg m2 3kg F Far m1 (3.2m / s ) 2 Far m2 (3.2m / s 2 ) Now, solve the right-hand equation to determine the magnitude of the action-reaction force: a 3.2m / s 2 Far 3kg(3.2m / s 2 ) Far ? Far 9.6 N F ? Next, solve for the Force: F (3kg 3.2m / s 2 ) 8kg(3.2m / s 2 ) F 9.6 N 25.6 N F 35.2 N The Force of Friction Friction is a force that always opposes motion Opposite in direction to the motion of the object Friction is directly related to the normal force: f f friction f f FN = Coefficient of Friction (pure number between 0 and 1) =0 No Friction (frictionless surface) =1 Highest frictional force possible Types of Friction Static Friction: The force of friction that opposes motion from beginning from rest. Also known as sticking friction. Kinetic Friction: The force of friction that opposes motion as it is occurring. This only occurs after motion has begun. Example: When pushing a large block, static friction prevents you from moving the block at all. When you finally start pushing it, kinetic friction takes over and slows the motion. Always…..static friction is stronger than or equal to kinetic friction in magnitude. Coefficients of Friction s =Coefficient of Static Friction k =Coefficient of Kinetic Friction Always…. s k Coefficients of Friction for Common Surfaces Materials s value k value Steel on Steel 0.74 0.57 Aluminum on Steel 0.61 0.47 Copper on Steel 0.53 0.36 Rubber on Concrete ~1.0 0.8 Wood on Wood 0.25 – 0.50 0.2 Glass on Glass 0.94 0.4 Waxed wood on wet snow 0.14 0.10 Waxed wood on dry snow 0.01 0.04 Metal on Metal (lubricated) 0.15 0.06 Ice on Ice 0.10 0.03 Teflon on Teflon 0.04 0.04 Synovial joints in Humans 0.01 0.003 Frictional Forces Example Calculate the static and kinetic frictional forces of a 5 kg block of aluminum on a steel table if a force were applied to move it to the right. First, find the coefficients from the table on the previous slide: s 0.61 k 0.47 Normal Force f (friction) F (applied) m 5kg g 9.8m / s 2 Gravity (Weight) Noting that there is no acceleration in the y direction and using Newton’s Second Law: Fn mg 0 Fn mg 5kg(9.8m / s 2 ) 49 N Frictional Forces Example (cntd) Now, lets calculate the static frictional force: f s s Fn 0.61* 49 N 29.9 N Now, lets calculate the kinetic frictional force: f s s Fn 0.47 * 49 N 23.0 N