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PROJE
 CT
WORKBY:
     SOUVIK 1
      XI ‘A’
Newton’s
     Laws:
Friction, Circular M otion &
          Drag F orces


                               2
1. Applications of Newton’s
   Laws Involving Friction
            Friction is always
              present when two
              solid surfaces slide
              along each other.

            T microscopic
             he
             details are not
             yet fully
             understood.
                                 3
Sliding friction is called kinetic
  friction.
Approximation of the frictional
 force:
             Ffr = μkFN .
Here, FN is the normal force, and
 μk is the coefficient of kinetic
 friction, which is different for    4
Static friction applies when two
  surfaces are at rest with respect to
  each other (such as a book sitting
  on a table).
T static frictional force is as big as
 he
 it needs to be to prevent slipping,
 up to a maximum value.
              Ffr ≤ μsFN .
Usually it is easier to keep an          5

 object sliding than it is to get it
Sliding friction (Animation)




                               6
Note that, in general, μs > μk.




                                  7
2. Uniform Circular Motion—
          Kinematics
Uniform circular motion: motion in a circle
 of constant radius at constant speed
Instantaneous velocity is always tangent
  to the circle.




                                              8
Looking at the change in velocity in the limit
  that the time interval becomes
  infinitesimally small, we see that
                           .




                                            9
T acceleration is called the
 his
 centripetal, or radial, acceleration, and
 it points toward the center of the
 circle.




                                        10
A centrifuge works
 by spinning very
 fast. T means
        his
 there must be a
 very large
 centripetal force.
 T object at A
   he
 would go in a
 straight line but for
 this force; as it is,
                   11

 it winds up at B   .
3. Dynamics of Uniform
       Circular Motion
For an object to be in uniform circular motion,
  there must be a net force acting on it.

                        We already know the
                         acceleration, so can
                         immediately write the
                         force:


                                                  12
W can see that
  e
 the force must
 be inward by
 thinking about
 a ball on a
 string. Strings
 only pull; they
 never push.
                   13
There is no centrifugal force pointing
  outward; what happens is that the natural
  tendency of the object to move in a straight
  line must be overcome.
If the centripetal force vanishes, the object
   flies off at a tangent to the circle.




                                                14
4. Highway Curves:
     Banked and Unbanked
When a car goes around a curve, there must
 be a net force toward the center of the
 circle of which the curve is an arc. If the
 road is flat, that force is supplied by
 friction.




                                          15
If the frictional
  force is
  insufficient, the
  car will tend to
  move more
  nearly in a
  straight line, as
  the skid marks      16

  show.
As long as the tires do not slip, the friction
 is static. If the tires do start to slip, the
 friction is kinetic, which is bad in two
 ways:
1. T kinetic frictional force is smaller than
    he
  the static.
2. T static frictional force can point
    he
  toward the center of the circle, but the
  kinetic frictional force opposes the
  direction of motion, making it very
  difficult to regain control of the car and   17

  continue around the curve.
Circular Motion - Car on a Banked Track (Animation)




                                              18
Banking the curve can help
  keep cars from skidding. In
  fact, for every banked curve,
  there is one speed at which
  the entire centripetal force is
  supplied by the
     horizontal component of
        the normal force, and
        no friction is required.
        T occurs when:
         his


                               19
5. Nonuniform Circular
        Motion

       If an object is moving in
         a circular path but at
         varying speeds, it
         must have a
         tangential component
         to its acceleration as
         well as the radial one.
                              20
This concept can be used for an
 object moving along any curved
 path, as any small segment of the
 path will be approximately circular.




                                        21
6. Velocity-Dependent
Forces: Drag and Terminal
         Velocity
When an object moves through a fluid, it
 experiences a drag force that depends on
 the velocity of the object.
For small velocities, the force is
  approximately proportional to the
  velocity; for higher speeds, the force is
  approximately proportional to the square
  of the velocity.
                                              22
If the drag force on a falling
   object is proportional to
   its velocity, the object
   gradually slows until the
   drag force and the
   gravitational force are
   equal. Then it falls with
   constant velocity, called
   the terminal velocity.



                                 23
Terminal Velocity Animation




                          24
REVIEW QUESTIONS
      &
       EXAMPLES
 ON EACH TOPICS
               25
A.   The force on the object acts directly
     toward the center of the circle.
B.   The force on the object acts directly
     away from the center of the circle.
C.   The net force on the object acts in a
     direction tangent to the circle at the
     position of the object.
D.   The net force cannot be in any of the
     three directions.

AN OBJECT IS SPEEDING UP AS IT GOES
AROUND A CIRCLE. WHICH STATEMENT IS
NECESSARILY TRUE ABOUT THE NET FORCE
ACTING ON THE OBJECT?
                                              26
ANTI-LOCK BRAKING SYSTEMS ON CARS ARE
DESIGNED TO KEEP THE TIRES ROLLING ON THE ROAD
RATHER THAN SLIDING. WHY IS THIS ADVANTAGEOUS
TO STOPPING AND CONTROLLING A VEHICLE?


A.   It isn’t advantageous. The car will take longer to stop.
B.   The car manufacturer can charge more money for the
     car.
C.   The coefficient of static friction of the tires on the road
     is higher than the coefficient of kinetic friction.
D.   The coefficient of static friction of the tires on the road
     is lower than the coefficient of kinetic friction.
E.   The tires won’t sqeal when you stop.                     27
1. Applications of Newton’s Laws Involving Friction

Example 1: Friction: static and kinetic.
Our 10.0-kg mystery box rests on a horizontal floor. The
  coefficient of static friction is 0.40 and the coefficient of
  kinetic friction is 0.30. Determine the force of friction acting
  on the box if a horizontal external applied force is exerted
  on it of magnitude:
(a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.




                                                                     28
1. Applications of Newton’s Laws Involving Friction

Conceptual Example 2:                A box against a
 wall.
You can hold a box against a rough
 wall and prevent it from slipping
 down by pressing hard
 horizontally. How does the
 application of a horizontal force
 keep an object from moving
 vertically?




                                                        29
1. Applications of Newton’s Laws Involving Friction



Example 3: Pulling against friction.
A 10.0-kg box is pulled along a horizontal
  surface by a force of 40.0 N applied at a
  30.0° angle above horizontal. The
  coefficient of kinetic friction is 0.30.
  Calculate the acceleration.




                                                      30
1. Applications of Newton’s Laws Involving Friction



Conceptual Example 4: To push or to pull a
 sled?
Your little sister wants a
  ride on her sled. If you
  are on flat ground, will
  you exert less force if
  you push her or pull
  her? Assume the same
  angle θ in each case.


                                                        31
1. Applications of Newton’s Laws Involving Friction


Example 5: Two boxes and a pulley.
Two boxes are connected by a cord
  running over a pulley. The coefficient
  of kinetic friction between box A and
  the table is 0.20. We ignore the mass
  of the cord and pulley and any
  friction in the pulley, which means we
  can assume that a force applied to
  one end of the cord will have the
  same magnitude at the other end. We
  wish to find the acceleration, a, of the
  system, which will have the same
  magnitude for both boxes assuming
  the cord doesn’t stretch. As box B
                                                        32
  moves down, box A moves to the
  right.
1. Applications of Newton’s Laws Involving Friction


 Example 6: The skier.
 This skier is descending a
   30° slope, at constant
   speed. What can you
   say about the coefficient
   of kinetic friction?




                                                      33
1. Applications of Newton’s Laws Involving Friction



Example 7: A ramp, a pulley, and two boxes.
Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the
  horizontal. It is connected by a lightweight cord, which
  passes over a massless and frictionless pulley, to a second
  box B, which hangs freely as shown. (a) If the coefficient of
  static friction is 0.40, determine what range of values for mass
  B will keep the system at rest. (b) If the coefficient of kinetic
  friction is 0.30, and mB = 10.0 kg, determine the acceleration of
  the system.




                                                                 34
2. Uniform Circular Motion—Kinematics

Example 8: Acceleration of a revolving ball.
A 150-g ball at the end of a string is
  revolving uniformly in a horizontal circle
  of radius 0.600 m. The ball makes 2.00
  revolutions in a second. What is its
  centripetal acceleration?




                                               35
2. Uniform Circular Motion—Kinematics




Example 9: Moon’s centripetal acceleration.
The Moon’s nearly circular orbit about the
  Earth has a radius of about 384,000 km and
  a period T of 27.3 days. Determine the
  acceleration of the Moon toward the Earth.




                                              36
2. Uniform Circular Motion—Kinematics




Example 10: Ultracentrifuge.
The rotor of an ultracentrifuge
  rotates at 50,000 rpm (revolutions
  per minute). A particle at the top of
  a test tube is 6.00 cm from the
  rotation axis. Calculate its
  centripetal acceleration, in “g’s.”


                                          37
3. Dynamics of Uniform Circular Motion


Example 11: Force on revolving ball
 (horizontal).
Estimate the force a person must exert on a
  string attached to a 0.150-kg ball to make
  the ball revolve in a horizontal circle of
  radius 0.600 m. The ball makes 2.00
  revolutions per second. Ignore the
  string’s mass.



                                               38
3. Dynamics of Uniform Circular Motion

Example 12: Revolving ball (vertical circle).

A 0.150-kg ball on the end of a
  1.10-m-long cord
  (negligible mass) is swung
  in a vertical circle. (a)
  Determine the minimum
  speed the ball must have at
  the top of its arc so that the
  ball continues moving in a
  circle. (b) Calculate the
  tension in the cord at the
  bottom of the arc,
  assuming the ball is
                                                39
  moving at twice the speed
  of part (a).
3. Dynamics of Uniform Circular Motion

Example 13: Conical pendulum.

A small ball of mass m,
  suspended by a cord of
  length l, revolves in a circle
  of radius r = l sin θ, where
  θ is the angle the string
  makes with the vertical. (a)
  In what direction is the
  acceleration of the ball, and
  what causes the
  acceleration? (b) Calculate
  the speed and period (time
  required for one revolution)
  of the ball in terms of l, θ, g,               40

  and m.
4. Highway Curves: Banked and Unbanked


Example 14: Skidding on a curve.
A 1000-kg car rounds a curve
  on a flat road of radius 50
  m at a speed of 15 m/s (54
  km/h). Will the car follow
  the curve, or will it skid?
  Assume: (a) the pavement
  is dry and the coefficient of
  static friction is μs = 0.60;
  (b) the pavement is icy and
  μs = 0.25.
                                             41
4. Highway Curves: Banked and Unbanked



Example 15: Banking angle.
(a) For a car traveling with speed v
  around a curve of radius r, determine a
  formula for the angle at which a road
  should be banked so that no friction is
  required. (b) What is this angle for an
  expressway off-ramp curve of radius 50
  m at a design speed of 50 km/h?


                                            42
43

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Physics_XI_Newton's_Laws_By Souvik

  • 1. PROJE CT WORKBY: SOUVIK 1 XI ‘A’
  • 2. Newton’s Laws: Friction, Circular M otion & Drag F orces 2
  • 3. 1. Applications of Newton’s Laws Involving Friction Friction is always present when two solid surfaces slide along each other. T microscopic he details are not yet fully understood. 3
  • 4. Sliding friction is called kinetic friction. Approximation of the frictional force: Ffr = μkFN . Here, FN is the normal force, and μk is the coefficient of kinetic friction, which is different for 4
  • 5. Static friction applies when two surfaces are at rest with respect to each other (such as a book sitting on a table). T static frictional force is as big as he it needs to be to prevent slipping, up to a maximum value. Ffr ≤ μsFN . Usually it is easier to keep an 5 object sliding than it is to get it
  • 7. Note that, in general, μs > μk. 7
  • 8. 2. Uniform Circular Motion— Kinematics Uniform circular motion: motion in a circle of constant radius at constant speed Instantaneous velocity is always tangent to the circle. 8
  • 9. Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that . 9
  • 10. T acceleration is called the his centripetal, or radial, acceleration, and it points toward the center of the circle. 10
  • 11. A centrifuge works by spinning very fast. T means his there must be a very large centripetal force. T object at A he would go in a straight line but for this force; as it is, 11 it winds up at B .
  • 12. 3. Dynamics of Uniform Circular Motion For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can immediately write the force: 12
  • 13. W can see that e the force must be inward by thinking about a ball on a string. Strings only pull; they never push. 13
  • 14. There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off at a tangent to the circle. 14
  • 15. 4. Highway Curves: Banked and Unbanked When a car goes around a curve, there must be a net force toward the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction. 15
  • 16. If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks 16 show.
  • 17. As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways: 1. T kinetic frictional force is smaller than he the static. 2. T static frictional force can point he toward the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and 17 continue around the curve.
  • 18. Circular Motion - Car on a Banked Track (Animation) 18
  • 19. Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed at which the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required. T occurs when: his 19
  • 20. 5. Nonuniform Circular Motion If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one. 20
  • 21. This concept can be used for an object moving along any curved path, as any small segment of the path will be approximately circular. 21
  • 22. 6. Velocity-Dependent Forces: Drag and Terminal Velocity When an object moves through a fluid, it experiences a drag force that depends on the velocity of the object. For small velocities, the force is approximately proportional to the velocity; for higher speeds, the force is approximately proportional to the square of the velocity. 22
  • 23. If the drag force on a falling object is proportional to its velocity, the object gradually slows until the drag force and the gravitational force are equal. Then it falls with constant velocity, called the terminal velocity. 23
  • 25. REVIEW QUESTIONS & EXAMPLES ON EACH TOPICS 25
  • 26. A. The force on the object acts directly toward the center of the circle. B. The force on the object acts directly away from the center of the circle. C. The net force on the object acts in a direction tangent to the circle at the position of the object. D. The net force cannot be in any of the three directions. AN OBJECT IS SPEEDING UP AS IT GOES AROUND A CIRCLE. WHICH STATEMENT IS NECESSARILY TRUE ABOUT THE NET FORCE ACTING ON THE OBJECT? 26
  • 27. ANTI-LOCK BRAKING SYSTEMS ON CARS ARE DESIGNED TO KEEP THE TIRES ROLLING ON THE ROAD RATHER THAN SLIDING. WHY IS THIS ADVANTAGEOUS TO STOPPING AND CONTROLLING A VEHICLE? A. It isn’t advantageous. The car will take longer to stop. B. The car manufacturer can charge more money for the car. C. The coefficient of static friction of the tires on the road is higher than the coefficient of kinetic friction. D. The coefficient of static friction of the tires on the road is lower than the coefficient of kinetic friction. E. The tires won’t sqeal when you stop. 27
  • 28. 1. Applications of Newton’s Laws Involving Friction Example 1: Friction: static and kinetic. Our 10.0-kg mystery box rests on a horizontal floor. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is 0.30. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude: (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N. 28
  • 29. 1. Applications of Newton’s Laws Involving Friction Conceptual Example 2: A box against a wall. You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving vertically? 29
  • 30. 1. Applications of Newton’s Laws Involving Friction Example 3: Pulling against friction. A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration. 30
  • 31. 1. Applications of Newton’s Laws Involving Friction Conceptual Example 4: To push or to pull a sled? Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the same angle θ in each case. 31
  • 32. 1. Applications of Newton’s Laws Involving Friction Example 5: Two boxes and a pulley. Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B 32 moves down, box A moves to the right.
  • 33. 1. Applications of Newton’s Laws Involving Friction Example 6: The skier. This skier is descending a 30° slope, at constant speed. What can you say about the coefficient of kinetic friction? 33
  • 34. 1. Applications of Newton’s Laws Involving Friction Example 7: A ramp, a pulley, and two boxes. Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. (a) If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest. (b) If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system. 34
  • 35. 2. Uniform Circular Motion—Kinematics Example 8: Acceleration of a revolving ball. A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration? 35
  • 36. 2. Uniform Circular Motion—Kinematics Example 9: Moon’s centripetal acceleration. The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km and a period T of 27.3 days. Determine the acceleration of the Moon toward the Earth. 36
  • 37. 2. Uniform Circular Motion—Kinematics Example 10: Ultracentrifuge. The rotor of an ultracentrifuge rotates at 50,000 rpm (revolutions per minute). A particle at the top of a test tube is 6.00 cm from the rotation axis. Calculate its centripetal acceleration, in “g’s.” 37
  • 38. 3. Dynamics of Uniform Circular Motion Example 11: Force on revolving ball (horizontal). Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass. 38
  • 39. 3. Dynamics of Uniform Circular Motion Example 12: Revolving ball (vertical circle). A 0.150-kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is 39 moving at twice the speed of part (a).
  • 40. 3. Dynamics of Uniform Circular Motion Example 13: Conical pendulum. A small ball of mass m, suspended by a cord of length l, revolves in a circle of radius r = l sin θ, where θ is the angle the string makes with the vertical. (a) In what direction is the acceleration of the ball, and what causes the acceleration? (b) Calculate the speed and period (time required for one revolution) of the ball in terms of l, θ, g, 40 and m.
  • 41. 4. Highway Curves: Banked and Unbanked Example 14: Skidding on a curve. A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s (54 km/h). Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is μs = 0.60; (b) the pavement is icy and μs = 0.25. 41
  • 42. 4. Highway Curves: Banked and Unbanked Example 15: Banking angle. (a) For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required. (b) What is this angle for an expressway off-ramp curve of radius 50 m at a design speed of 50 km/h? 42
  • 43. 43

Editor's Notes

  1. Chapter Opener. Caption: Newton’s laws are fundamental in physics. These photos show two situations of using Newton’s laws which involve some new elements in addition to those discussed in the previous Chapter. The downhill skier illustrates friction on an incline, although at this moment she is not touching the snow, and so is retarded only by air resistance which is a velocity-dependent force (an optional topic in this Chapter). The people on the rotating amusement park ride below illustrate the dynamics of circular motion.
  2. Figure 1. Caption: An object moving to the right on a table or floor. The two surfaces in contact are rough, at least on a microscopic scale.
  3. Figure 2. Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.
  4. Figure 3. Caption: Determining the change in velocity, Δ v, for a particle moving in a circle. The length Δ l is the distance along the arc, from A to B.
  5. Figure 4. Caption: For uniform circular motion, a is always perpendicular to v.
  6. Figure 5. Caption: Two positions of a rotating test tube in a centrifuge (top view). At A, the green dot represents a macromolecule or other particle being sedimented. It would tend to follow the dashed line, heading toward the bottom of the tube, but the fluid resists this motion by exerting a force on the particle as shown at point B.
  7. Figure 6. Caption: A force is required to keep an object moving in a circle. If the speed is constant, the force is directed toward the circle’s center.
  8. Figure 7. Caption: Swinging a ball on the end of a string.
  9. Figure 8. Caption: If centrifugal force existed, the revolving ball would fly outward as in (a) when released. In fact, it flies off tangentially as in (b). For example, in (c) sparks fly in straight lines tangentially from the edge of a rotating grinding wheel.
  10. Figure 9. Caption: The road exerts an inward force on a car (friction against the tires) to make it move in a circle. The car exerts an inward force on the passenger.
  11. Figure 10. Caption: Race car heading into a curve. From the tire marks we see that most cars experienced a sufficient friction force to give them the needed centripetal acceleration for rounding the curve safely. But, we also see tire tracks of cars on which there was not sufficient force—and which unfortunately followed more nearly straight-line paths.
  12. Figure 11. Caption: Normal force on a car rounding a banked curve, resolved into its horizontal and vertical components. The centripetal acceleration is horizontal ( not parallel to the sloping road). The friction force on the tires, not shown, could point up or down along the slope, depending on the car’s speed. The friction force will be zero for one particular speed.
  13. Figure 12. Caption: The speed of an object moving in a circle changes if the force on it has a tangential component, F tan . Part (a) shows the force F and its vector components; part (b) shows the acceleration vector and its vector components.
  14. Figure 13. Caption: Object following a curved path (solid line). At point P the path has a radius of curvature r . The object has velocity v, tangential acceleration a tan (the object is here increasing in speed), and radial (centripetal) acceleration a R (magnitude a R = v 2 /r) which points toward the center of curvature C.
  15. Figure 14. Caption: (a) Forces acting on an object falling downward. (b) Graph of the velocity of an object falling due to gravity when the air resistance drag force is F D = -bv. Initially, v = 0 and dv/dt = g, and but as time goes on dv/dt (= slope of curve) decreases because of F D . Eventually, v approaches a maximum value, v T , the terminal velocity, which occurs when F D has magnitude equal to mg .
  16. Figure 15. Figure 16. Caption: Example 1. Magnitude of the force of friction as a function of the external force applied to an object initially at rest. As the applied force is increased in magnitude, the force of static friction increases linearly to just match it, until the applied force equals μ s F N . If the applied force increases further, the object will begin to move, and the friction force drops to a roughly constant value characteristic of kinetic friction. Solution: Since there is no vertical motion, the normal force equals the weight, 98.0 N. This gives a maximum static frictional force of 29 N. Therefore, for parts a-d, the frictional force equals the applied force. In (e), the frictional force is 29 N and the box accelerates at 1.1 m/s 2 to the right.
  17. Figure 17. Answer: Friction, of course! The normal force points out from the wall, and the frictional force points up.
  18. Figure 18. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of F P ), which is 78 N. The frictional force is 23.4 N, and the horizontal component of F P is 34.6 N, so the acceleration is 1.1 m/s 2 .
  19. Figure 19. Answer: Pulling decreases the normal force, while pushing increases it. Better pull.
  20. Figure 20. Solution: For box A, the normal force equals the weight; we therefore know the magnitude of the frictional force, but not of the tension in the cord. Box B has no horizontal forces; the vertical forces on it are its weight (which we know) and the tension in the cord (which we don’t). Solving the two free-body equations for a gives 1.4 m/s 2 (and the tension as 17 N).
  21. Figure 21. Caption: Example 5–6. A skier descending a slope; F G = mg is the force of gravity (weight) on the skier. Solution: Since the speed is constant, there is no net force in any direction. This allows us to find the normal force and then the frictional force; the coefficient of kinetic friction is 0.58.
  22. Figure 22. Caption: Example 5–7. Note choice of x and y axes. Solution: Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns. The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down). 0.78 m/s 2
  23. Figure 23. Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path. Answer: a = v 2 /r; we can find v from the radius and frequency. v = 7.54 m/s, so a = 94.7 m/s 2 .
  24. Answer: Again, find v from r and the period; first convert r and T into meters and seconds, respectively. Then a = 2.72 x 10 -3 m/s 2 = 2.78 x 10 -4 g.
  25. Answer: Again, find v and then a. Convert minutes to seconds and centimeters to meters. Divide final answer by 9.8 m/s 2 . a = 167,000 g’s.
  26. Figure 24. Answer: Ignoring the weight of the ball, so F T is essentially horizontal, we find F T = 14 N.
  27. Figure 25. Caption: Example 5–12. Freebody diagrams for positions 1 and 2. Solution: See the freebody diagrams. The minimum speed occurs when the tension is zero at the top; v = 3.283 m/s. Substitute to find F T2 = 7.35 N.
  28. Figure 26. Caption: Example 5–13. Conical pendulum. Answer: a. The acceleration is towards the center of the circle, and it comes from the horizontal component of the tension in the cord. b. See text.
  29. Figure 27. Caption: Example 5–14. Forces on a car rounding a curve on a flat road. (a) Front view, (b) top view. Solution: The normal force equals the weight, and the centripetal force is provided by the frictional force (if sufficient). The required centripetal force is 4500 N. The maximum frictional force is 5880 N, so the car follows the curve. The maximum frictional force is 2450 N, so the car will skid.
  30. Answer: a. Set F N = mg in previous equation. Find tan θ = v 2 /rg. b. Tan θ = 0.40, so θ = 22°.