1. FRICTION and Newton’s second law

2. The “Normal” Force, N When an object is pressed against a surface, the surface pushes back. (That’s Newton’s 3 rd Law) This “push back” from the surface is called the Normal Force, N The word “normal” in math terminology means “perpendicular” The surface pushes back in a direction that is perpendicular to the surface.

3. If the box is not accelerating up or down, then the net force is zero and the Normal force must balance the Weight Weight = mg Normal force, N

4. Weight = mg Normal force, N F A = 20 N f = 4 N m = 2 kg a = ? F net = ma 20 N – 4 N = ma a = F net / m a = 8 m/s 2 F A = 20 Nf = 4 N You pull on a box with an applied force of 20 N. The frictional force opposing the motion is 4 N. If the mass of the box is 2 kg, what is its acceleration? 1.Draw the free body diagram. 2.Write what you know and don’t know. 3.Write the equation, F net = ma 4.Calculate the value of the net Force and then the acceleration.

5. Friction, f A force that always opposes motion Depends on two things: the roughness of the surfaces and how hard they are pressed together. f =  N , mu- the “coefficient of friction” tells how rough the surfaces are. N, the Normal force tells how hard the surfaces are pressed together

6. Example: How large is the frictional force between 2 surfaces if the coefficient of friction is 0.2 and the Normal force is 80 N? f =  N f = 0.2 x 80 N f = 16 N

7. There are two kinds of friction: “static friction” (not moving) must be overcome to initiate motion. “kinetic friction” must be overcome while an object is moving Static friction > Kinetic friction

8. IF you pull just hard enough to make the object move, you overcome static friction: Your applied force = static friction force Your applied force =  static N Once an object is moving, if you pull in such a way that the object is moving with constant velocity, then Your applied force = kinetic friction force Your applied force =  kinetic N And… what is the normal force, N? N = mg Applied Force Friction =  N

9. A 75 kg firefighter is sliding down the pole at the fire station. If he is moving at a constant velocity, what is the friction force? F net = ma friction – weight = ma = 0 Therefore, friction = weight friction = mg = 75 kg x 10 m/s 2 friction = 750 N weight friction

10. Weight = mg Normal force, N F A = 20 N f = 4 N m = 2 kg a = ? F net = ma 20 N – 4 N = ma a = F net / m a = 8 m/s 2 F A = 20 Nf = 4 N You pull on a box with an applied force of 20 N. The frictional force opposing the motion is 4 N. If the mass of the box is 2 kg, what is its acceleration? 1.Draw the free body diagram. 2.Write what you know and don’t know. 3.Write the equation, F net = ma 4.Calculate the value of the net Force and then the acceleration.

11. Weight = mg Normal force, N F A = 30 Nm = 2 kg  = 0.4a = ? F net = maf =  N N = mg = 2 kg x 10 m/s 2 = 20 N f =  N = 0.4(20 N) = 8 N Horizontal : F net = F A - f F net = 30 N – 8 N = 22 N a = F net / m a = 11 m/s 2 FAFA f =  N = 30 N You pull on a box with an applied force of 30 N. The coefficient of friction is  0.4. If the mass of the box is 2 kg, what is its acceleration? 1.Draw the free body diagram. 2.Write what you know and don’t know. 3.Write the equations, F net = ma and f =  N 4.Calculate the Normal force and the friction force. 5.Calculate the value of the net Force and then the acceleration.

12. Pre-AP only…

13. Weight = mg Normal force, N F A = 25Nm = 2 kg  = 43  = 0.4 a = ?  F x = maf =  N  F x = -  N + F A cos  = ma (-0.4(36.65) + 25cos 43 ) / 2= a a = 1.81 m/s 2 FAFA f  N = mg + Fsin  

14. Weight = mg Normal force, N F A = 15 Nm = 2 kg  = 43  = 0.4a = ?  F x = maf =  N  F x = -  N + F A cos  = ma (-0.4(9.77) + 15cos 43) / 2 = a a = 3.53 m/s 2 FAFA f  N = mg – Fsin  

15. Weight = mg Normal force, N If the box is moving at constant velocity, there is no acceleration, Therefore the net force must be zero… so the horizontal forces must cancel each other.  N = Fcos  If you push hard enough to just get the box moving, the acceleration is zero in that case also, but the friction is static, not kinetic. FAFA f 

17. Friction Lab frictional force = (  Normal force f =  N Normal force, N = Wt 1 frictional force, f = Wt 2 Weight = mg N Wt 1 Frictional force, f

18. Pre-AP Only…

19. More examples… Xavier, who was stopped at a light, accelerated forward at 6 m/s 2 when the light changed. He had dice hanging from his rear view mirror. What angle did they make with the vertical during this acceleration? 1- draw free body diagram 2- write Newton’s 2 nd law for BOTH directions  F x = ma x Tsin  = ma x Tcos  = mg tan  = a/g  = 31.47 degrees   F y = ma y mg T Tcos  – mg = ma y = 0

20. An “Atwood’s Machine” Two masses of 5 kg and 2 kg are suspended from a massless, frictionless pulley, When released from rest, what is their acceleration? What is the Tension in the string? 1- Draw a free body diagram 2- Write Newton’s Second Law for EXTERNAL forces acting on the whole system.  F ext = m total a The Tension in the string is an “internal” force. The only “external” forces are the gravitational forces of weight, which oppose each other. So…  F ext = m 1 g – m 2 g = m total a Solving for the acceleration yields a = 4.2 m/s 2 Now, to find the Tension, we must “zoom in” on mass 2:  F= T – m 2 g = m 2 a T = m 2 a + m 2 g T = 28 N a +

21. Objects on Inclines- sliding down, no friction   mg The free-body diagram ALWAYS comes first: Draw the weight vector, mg Draw the Normal force vector. Are there any other forces??? Since the motion is parallel to the plane, ROTATE the axis from horizontal and vertical to “parallel” and “perpendicular”. Then draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Does the Normal force balance with the force of weight, mg? NO! What force must balance the Normal force? N Does all of the weight, mg, pull the box down the incline? Write Newton’s Second Law for the forces on box parallel to the plane with “down” being negative.  F = ma -mgsin  = ma N = mgcos  No! Only mgsin 

22. Inclines: pushed upward, no friction   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma F A - mgsin  = ma FAFA

23. Inclines: pushed downward, no friction   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma - F A - mgsin  = ma FAFA

25. Inclines: pushed upward, no friction   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector AND the applied force, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for all forces acting parallel to the incline.  F = ma F A cos  - mgsin  = ma FAFA 

26. Inclines: With friction, case 1: at rest or sliding down   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma f - mgsin  = ma  mgcos  - mgsin  = ma  gcos  - gsin  = a f What is friction? f =  N What is N ? N = mgcos  f =  mgcos 

27. Inclines: With friction, case 2: pushed downward   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma f - mgsin  – F A = ma  mgcos  - mgsin  – F A = ma f What is friction? f =  N f =  mgcos  FAFA

28. Inclines: With friction, case 3: pushed upward   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma F A - mgsin  – f = ma F A  - mgsin  -  mgcos  = ma f What is friction? f =  N f =  mgcos  FAFA

29. Friction along an incline An object placed along an incline will eventually slide down if the incline is elevated high enough. The angle at which it slides depends on how rough the incline surface is. To find the angle where it slides:  Since it doesn’t move: mgsin  =  mgcos  Therefore: Tan  max  =  max, the coefficient of static friction Or The angle  max  = tan -1  max

30. Spring Forces

31. Spring Force If a mass is suspended from a spring, two forces act on the mass: its weight and the spring force. The spring force for many springs is given by F s = kx, Where x is the distance the spring is stretched (or compressed) from its normal length and “k” is called the spring constant, which tells the stiffness of the spring. The stiffer the spring, the larger the spring constant k. This relationship is known as “Hooke’s Law”. mg FsFs

32. Spring Force If the mass is at rest, then the two forces are balanced, so that kx = mg These balanced forces provide a quick way to determine the spring constant of a spring, or anything springy, like a rubber band: Hang a known mass from the spring and measure how much it stretches, then solve for “k” !k = mg/x mg F s = kx