1. Chapter 6 Force and Motion II

2. Forces of Friction When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion This is due to the interactions between the object and its environment This is due to the interactions between the object and its environment This is resistance is called the force of friction This is resistance is called the force of friction

3. More About Friction Friction is proportional to the normal force Friction is proportional to the normal force The force of static friction is generally greater than the force of kinetic friction The force of static friction is generally greater than the force of kinetic friction The coefficient of friction (µ) depends on the surfaces in contact The coefficient of friction (µ) depends on the surfaces in contact The direction of the frictional force is opposite the direction of motion The direction of the frictional force is opposite the direction of motion The coefficients of friction are nearly independent of the area of contact The coefficients of friction are nearly independent of the area of contact

4. Static Friction, ƒ s Static friction acts to keep the object from moving Static friction acts to keep the object from moving If F increases, so does ƒ s If F increases, so does ƒ s If F decreases, so does ƒ s If F decreases, so does ƒ s ƒ s  µ n ƒ s  µ n

5. Kinetic Friction The force of kinetic friction acts when the object is in motion The force of kinetic friction acts when the object is in motion ƒ k = µ n ƒ k = µ n

6. Example: A Block of mass m 1 on a A Block of mass m 1 on a rough, horizontal surface rough, horizontal surface is connected to a ball of mass m 2 by a lightweight cord over a lightweight frictionless pulley. A force of magnitude F at an angle  with the horizontal is applied to the block as shown. The coefficient of Kinetic friction between block and surface is  k. Determine the acceleration of the objects.

7. Solution: Free Body diagrams: Newton’s 2 nd Law: X-axis: F cos  – T – f k = m 1 a (1) Y-axis: F sin  + n – m 1 g = 0(2) (2)  n = m 1 g – F sin  f k =  k n =  k (m 1 g – F sin  ) (1)  T = Fcos  -  k (m 1 g – F sin  ) - m 1 a But Also, T = m 2 a – m 2 g

8. Solution Cont’d: T = T  T = T  Fcos  -  k (m 1 g – F sin  ) - m 1 a = Fcos  -  k (m 1 g – F sin  ) - m 1 a = m 2 a – m 2 g Knowing the value of a one can easily deduce the value of T = m 2 a – m 2 g.

9. Connected Objects Apply Newton’s Laws separately to each object Apply Newton’s Laws separately to each object The acceleration of both objects will be the same The acceleration of both objects will be the same The tension is the same in each diagram The tension is the same in each diagram Solve the simultaneous equations Solve the simultaneous equations

10. More About Connected Objects Treating the system as one object allows an alternative method or a check Treating the system as one object allows an alternative method or a check Use only external forces Use only external forces Not the tension – it’s internal Not the tension – it’s internal The mass is the mass of the system The mass is the mass of the system Doesn’t tell you anything about any internal forces Doesn’t tell you anything about any internal forces

11. Newton’s Law applied to uniform circular motion: A particle moving with uniform speed v in a circular path of radius r, experiences a centripetal acceleration a c that has a magnitude: A particle moving with uniform speed v in a circular path of radius r, experiences a centripetal acceleration a c that has a magnitude: a c is directed toward the a c is directed toward the center of the circle and is always  to v.

12. Newton’s Law applied to uniform circular motion Cont’d: If we apply Newton’s law along the radial direction, If we apply Newton’s law along the radial direction, F acts toward the center F acts toward the center of the circle. of the circle.

13. Newton’s Law applied to uniform circular motion Cont’d Ex: The ball is put in a circular motion on a frictionless table. The ball is put in a circular motion on a frictionless table. If the string breaks at some point how the ball moves afterward?

14. Tangential ( a t ) and Radial ( a N ) Acceleration In the case when the speed changes along a circular path the total Acceleration vector is given by: In the case when the speed changes along a circular path the total Acceleration vector is given by: a = a t + a N Where, |a N | = v 2 /r (directed toward the center of the circle) |a t | = (directed tangent to the circle) dt d|v|

15. Tangential ( a t ) and Radial ( a N ) Acceleration Cont’d: atat aNaN a a = a t + a N |a N | = v 2 /r dt d|v| |a t | =

16. Example The above figure shows the path of a park ride that travels at constant speed through five circular arcs of radii R 0, 2R 0, and 3R 0. Rank the arcs according to the magnitude of the centripetal force on a rider traveling in the arcs, greatest first. 423.15

17. Application: A ball of mass m= 0.500 Kg is attached to the end of a cord 1.50 m long. The ball is whirled in a horizontal circle. A ball of mass m= 0.500 Kg is attached to the end of a cord 1.50 m long. The ball is whirled in a horizontal circle. If the ball can withstand a maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks. x

18. Solution: Applying Newton’s Law: Maximum tension correspond to max v: x mg N T 

19. Example: A small object of mass m is suspended from a string of length L. The object revolves with constant speed v in a horizontal circle of radius r, Find an expression for v as function of T, r and m? A small object of mass m is suspended from a string of length L. The object revolves with constant speed v in a horizontal circle of radius r, Find an expression for v as function of T, r and m?

20. Solution: Applying Newton’s Law: Applying Newton’s Law: Projection along the horizontal axis: Projection along the horizontal axis: Ac on x-axis Projection along the vertical axis: (1) (2) Dividing (1) by (2) 

21. Example: A 1500-kg car moving on a flat, horizontal road negotiates a curve, (See Figure). If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.500, find the maximum speed the car can have and still make the turn successfully. A 1500-kg car moving on a flat, horizontal road negotiates a curve, (See Figure). If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.500, find the maximum speed the car can have and still make the turn successfully.

22. Solution: Force of friction enables the car to remains in its circular path. Force of friction enables the car to remains in its circular path. Projection along the radius: But,

23. More Applications: A small sphere of mass m is attached to the end of a cord of length R and set into motion in a vertical circle about a fixed point O. Determine the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle  with the vertical. A small sphere of mass m is attached to the end of a cord of length R and set into motion in a vertical circle about a fixed point O. Determine the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle  with the vertical.

24. Solution: Note that along the path the speed is not constant. Note that along the path the speed is not constant. Projection along the radius:

25. Application Cont’d What speed would the ball have as it passes over the top of the circle if the tension in the cord goes to zero instantaneously? At the top  = 180° T = 0, (i.e. cord breaks)

26. Motion in the Presence of Resistive Forces: The medium through which the object moves exerts a resistive force R on it. The medium through which the object moves exerts a resistive force R on it. The medium can be either a liquid or a gas. The medium can be either a liquid or a gas. Some examples are the air resistance associated with moving vehicles (called also air Drag) and the viscous forces that act on a object moving through a liquid. Some examples are the air resistance associated with moving vehicles (called also air Drag) and the viscous forces that act on a object moving through a liquid.

27. Resistive Forces: The direction of the resistive force R is always opposite to the direction of motion. The direction of the resistive force R is always opposite to the direction of motion. The Magnitude of R is proportional to the speed of the object: The Magnitude of R is proportional to the speed of the object: R = -b v R = -b v v is the velocity of the object and b is a constant whose value depends on the properties of the medium and on the shape and dimensions of the object. v is the velocity of the object and b is a constant whose value depends on the properties of the medium and on the shape and dimensions of the object.

28. Terminal Speed When the Drag force ( upward air resistance for example) equals the driving force (downward force of gravity), the net force on the object is zero. When the Drag force ( upward air resistance for example) equals the driving force (downward force of gravity), the net force on the object is zero. The constant speed of the object is the terminal speed The constant speed of the object is the terminal speed

29. Application: Consider a small sphere of mass m released from rest in a liquid as shown in the adjacent figure, assuming that the buoyant force is negligible, describe the motion of the sphere. Consider a small sphere of mass m released from rest in a liquid as shown in the adjacent figure, assuming that the buoyant force is negligible, describe the motion of the sphere.

30. Application Cont’d: The forces acting on the sphere are: The forces acting on the sphere are: 1) Resistive Force: R =-b v R =-b v 2) The gravitational Force: F g = mg F g = mg

31. Application Cont’d: Applying Newton’s law: Applying Newton’s law: We can obtain the terminal speed by putting dv/dt = 0. We can obtain the terminal speed by putting dv/dt = 0. mg – b v T = 0 mg – b v T = 0 mgmg R y (+)

32. Application Cont’d: The Equation The Equation is called a differential equation. is called a differential equation.