Upcoming SlideShare
Loading in …5
×

# 08 newton's law of motion

6,792 views

Published on

Published in: Technology, Business
0 Comments
2 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

No Downloads
Views
Total views
6,792
On SlideShare
0
From Embeds
0
Number of Embeds
10
Actions
Shares
0
Downloads
43
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide
• Figure 5-1. 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.
• Figure 5-2. Figure 5-3. Caption: Example 5–1. 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.
• Figure 5-4. Answer: Friction, of course! The normal force points out from the wall, and the frictional force points up.
• Figure 5-5. 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 .
• Figure 5-6. Answer: Pulling decreases the normal force, while pushing increases it. Better pull.
• Figure 5-7. 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).
• Figure 5-8. 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.
• Figure 5-9. 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
• [CORRECT 5 ANSWER]
• ### 08 newton's law of motion

1. 1. Newton’s Laws of Motion Topic 4 (cont.)
2. 2. Lecture Outline <ul><li>Newton’s Law Involving Friction </li></ul>
3. 3. <ul><li>Friction is always present when two solid surfaces slide along each other. </li></ul><ul><ul><li>Even the smoothest looking surface is quite rough on a microscopic scale </li></ul></ul><ul><li>Even for a rolling object, there is a friction called rolling friction </li></ul><ul><li>In this topic we are focusing on sliding friction, kinetic friction </li></ul><ul><li>(kinetic – ‘moving’ in Greek) </li></ul>Newton’s Laws Involving Friction
4. 4. <ul><li>Kinetic friction act on the opposite direction of the object’s velocity </li></ul><ul><li>Magnitude of the kinetic friction depend on </li></ul><ul><ul><li>Normal force between the two surface </li></ul></ul><ul><ul><li>Nature of the two sliding surface – coefficient of static friction </li></ul></ul><ul><li>Approximation of the frictional force: </li></ul><ul><li>F fr = μ k F N . </li></ul><ul><li>Here, F N is the normal force, and μ k is the coefficient of kinetic friction, which is different for each pair of surfaces. </li></ul>
5. 5. <ul><li>Static friction applies when two surfaces are at rest with respect to each other (such as a book sitting on a table). </li></ul><ul><li>F fr ≤ μ s F N </li></ul><ul><li> μ s is the coefficient of static friction </li></ul><ul><li>The static frictional force is as big as it needs to be to prevent slipping, up to a maximum value. </li></ul><ul><li>Usually it is easier to keep an object sliding than it is to get it started. </li></ul>
6. 6. Note that, in general, μ s > μ k .
7. 7. Example 5-1: 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.
8. 8. Conceptual Example 5-2: 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?
9. 9. Example 5-3: 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.
10. 10. Conceptual Example 5-4: 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.
11. 11. Example 5-5: 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 moves down, box A moves to the right.
12. 12. Example 5-6: The skier. This skier is descending a 30° slope, at constant speed. What can you say about the coefficient of kinetic friction?
13. 13. Example 5-7: A ramp, a pulley, and two boxes. <ul><li>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. </li></ul><ul><li>If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest. </li></ul><ul><li>If the coefficient of kinetic friction is 0.30, and m B = 10.0 kg, determine the acceleration of the system. </li></ul>
14. 14. <ul><li>A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because: </li></ul>ConcepTest 5.1 Friction 1) the force from the rushing air pushed it off 2) the force of friction pushed it off 3) no net force acted on the box 4) truck went into reverse by accident 5) none of the above
15. 15. <ul><li>A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because: </li></ul>Generally, the reason that the box in the truck bed would move with the truck is due to friction between the box and the bed. If there is no friction, there is no force to push the box along, and it remains at rest. The truck accelerated away, essentially leaving the box behind!! ConcepTest 5.1 Friction 1) the force from the rushing air pushed it off 2) the force of friction pushed it off 3) no net force acted on the box 4) truck went into reverse by accident 5) none of the above
16. 16. <ul><li>Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? </li></ul>1)  k >  s so sliding friction is better 2)  k >  s so static friction is better 3)  s >  k so sliding friction is better 4)  s >  k so static friction is better 5) none of the above ConcepTest 5.2 Antilock Brakes
17. 17. <ul><li>Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? </li></ul>1)  k >  s so sliding friction is better 2)  k >  s so static friction is better 3)  s >  k so sliding friction is better 4)  s >  k so static friction is better 5) none of the above Static friction is greater than sliding friction , so by keeping the wheels from skidding, the static friction force will help slow the car down more efficiently than the sliding friction that occurs during a skid. ConcepTest 5.2 Antilock Brakes
18. 18. <ul><li>Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? </li></ul>ConcepTest 5.3 Going Sledding 1) pushing her from behind 2) pulling her from the front 3) both are equivalent 4) it is impossible to move the sled 5) tell her to get out and walk 1 2
19. 19. <ul><li>Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? </li></ul>ConcepTest 5.3 Going Sledding In Case 1, the force F is pushing down (in addition to mg ), so the normal force is larger . In Case 2, the force F is pulling up , against gravity, so the normal force is lessened . Recall that the frictional force is proportional to the normal force. 1) pushing her from behind 2) pulling her from the front 3) both are equivalent 4) it is impossible to move the sled 5) tell her to get out and walk 1 2
20. 20. <ul><li>A box of weight 100 N is at rest on a floor where  s = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N . Which way does the box move? </li></ul>ConcepTest 5.4 Will it Budge? 1) moves to the left 2) moves to the right 3) moves up 4) moves down 5) the box does not move T m Static friction (  s  = 0.4  ) 
21. 21. <ul><li>A box of weight 100 N is at rest on a floor where  s = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N . Which way does the box move? </li></ul>The static friction force has a maximum of  s N = 40 N . The tension in the rope is only 30 N . So the pulling force is not big enough to overcome friction. ConcepTest 5.4 Will it Budge? 1) moves to the left 2) moves to the right 3) moves up 4) moves down 5) the box does not move Follow-up: What happens if the tension is 35 N ? What about 45 N ? T m Static friction (  s  = 0.4  ) 
22. 22. <ul><li>A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? </li></ul>1) component of the gravity force parallel to the plane increased 2) coeff. of static friction decreased 3) normal force exerted by the board decreased 4) both #1 and #3 5) all of #1, #2 and #3 ConcepTest 5.5a Sliding Down I Net Force Normal Weight
23. 23. <ul><li>A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? </li></ul>1) component of the gravity force parallel to the plane increased 2) coeff. of static friction decreased 3) normal force exerted by the board decreased 4) both #1 and #3 5) all of #1, #2 and #3 <ul><li>As the angle increases, the component of weight parallel to the plane increases and the component perpendicular to the plane decreases (and so does the Normal force). Since friction depends on Normal force, we see that the friction force gets smaller and the force pulling the box down the plane gets bigger . </li></ul>ConcepTest 5.5a Sliding Down I Net Force Normal Weight
24. 24. <ul><li>A mass m is placed on an inclined plane (  > 0 ) and slides down the plane with constant speed . If a similar block (same  ) of mass 2m were placed on the same incline, it would: </li></ul>1) not move at all 2) slide a bit, slow down, then stop 3) accelerate down the incline 4) slide down at constant speed 5) slide up at constant speed ConcepTest 5.5b Sliding Down II m
25. 25. <ul><li>A mass m is placed on an inclined plane (  > 0 ) and slides down the plane with constant speed . If a similar block (same  ) of mass 2m were placed on the same incline, it would: </li></ul>The component of gravity acting down the plane is double for 2m . However, the normal force (and hence the friction force) is also double (the same factor!). This means the two forces still cancel to give a net force of zero. 1) not move at all 2) slide a bit, slow down, then stop 3) accelerate down the incline 4) slide down at constant speed 5) slide up at constant speed ConcepTest 5.5b Sliding Down II  W N f  W x W y