February 25, Week 7 Today: Chapter 5, Applying Newtons Laws - PowerPoint PPT Presentation

February 25, Week 7 Today: Chapter 5, Applying Newton’s Laws Homework Assignment #5 - Due March 1. Mastering Physics: 10 problems from chapters 4 and 5. Written Questions: 5.74 Thursday Office Hours: 12:00-2:30, 4:00-5:00(???) Exam #2, Next Friday, March 8 Newton’s Laws February 27, 2013 - p. 1/8

Maximum Static Friction Experiments show that the static friction’s maximum value obeys a simple equation. Newton’s Laws February 27, 2013 - p. 2/8

Maximum Static Friction Experiments show that the static friction’s maximum value obeys a simple equation. f s,max = µ s n Newton’s Laws February 27, 2013 - p. 2/8

Maximum Static Friction Experiments show that the static friction’s maximum value obeys a simple equation. f s,max = µ s n Newton’s Laws February 27, 2013 - p. 2/8

Maximum Static Friction Experiments show that the static friction’s maximum value obeys a simple equation. f s,max = µ s n Example: A wooden block is placed on a wooden ramp which is initially hor- izontal. When the ramp is slowly raised, at what an- gle will the block begin to slide? Newton’s Laws February 27, 2013 - p. 2/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? 15 N Newton’s Laws February 27, 2013 - p. 3/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? (a) 0 . 5 N 15 N Newton’s Laws February 27, 2013 - p. 3/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? (a) 0 . 5 N (b) 15 N 15 N Newton’s Laws February 27, 2013 - p. 3/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? (a) 0 . 5 N (b) 15 N (c) 25 N 15 N Newton’s Laws February 27, 2013 - p. 3/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? (a) 0 . 5 N (b) 15 N (c) 25 N (d) 50 N 15 N Newton’s Laws February 27, 2013 - p. 3/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? (a) 0 . 5 N (b) 15 N (c) 25 N (d) 50 N (e) 65 N 15 N Newton’s Laws February 27, 2013 - p. 3/8

Static Friction Exercise A 50 N crate is placed on a horizontal surface. The coefficient of static friction between the crate and the table is 0 . 5 . A horizontal force of 15 N is applied to the crate. It does not move. How much static friction is acting on the crate? (b) 15 N (a) 0 . 5 N (c) 25 N (d) 50 N (e) 65 N Problem never specified that f s at max 15 N � F x = 0 ⇒ f s − 15 N = 0 Newton’s Laws February 27, 2013 - p. 3/8

Kinetic Friction Kinetic Friction - − → f k , sliding friction. Newton’s Laws February 27, 2013 - p. 4/8

Kinetic Friction Kinetic Friction - − → f k , sliding friction. Experiments show that the kinetic friction’s value is approximately constant and obeys a simple equation. f k = µ k n Newton’s Laws February 27, 2013 - p. 4/8

Kinetic Friction Kinetic Friction - − → f k , sliding friction. Experiments show that the kinetic friction’s value is approximately constant and obeys a simple equation. f k = µ k n Newton’s Laws February 27, 2013 - p. 4/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? α = 37 ◦ Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? α = 37 ◦ − → n − → f k → − w � → − w ⊥ Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? (a) a = g (sin α + µ k cos α ) = 7 . 45 m/s 2 α = 37 ◦ − → n − → f k → − w � − → w ⊥ Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? (a) a = g (sin α + µ k cos α ) = 7 . 45 m/s 2 α = 37 ◦ (b) a = g (sin α − µ k cos α ) = 4 . 33 m/s 2 − → n → − f k − → w � − → w ⊥ Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? (a) a = g (sin α + µ k cos α ) = 7 . 45 m/s 2 α = 37 ◦ (b) a = g (sin α − µ k cos α ) = 4 . 33 m/s 2 − → n (c) a = g (cos α + µ k cos α ) = 9 . 39 m/s 2 − → f k − → w � − → w ⊥ Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? (a) a = g (sin α + µ k cos α ) = 7 . 45 m/s 2 α = 37 ◦ (b) a = g (sin α − µ k cos α ) = 4 . 33 m/s 2 − → n (c) a = g (cos α + µ k cos α ) = 9 . 39 m/s 2 − → f k − → w (d) a = g (cos α − µ k cos α ) = 6 . 26 m/s 2 � → − w ⊥ Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? (a) a = g (sin α + µ k cos α ) = 7 . 45 m/s 2 α = 37 ◦ (b) a = g (sin α − µ k cos α ) = 4 . 33 m/s 2 − → n (c) a = g (cos α + µ k cos α ) = 9 . 39 m/s 2 − → f k − → w (d) a = g (cos α − µ k cos α ) = 6 . 26 m/s 2 � − → w ⊥ (e) a = g (sin α − µ k sin α ) = 4 . 72 m/s 2 Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? (a) a = g (sin α + µ k cos α ) = 7 . 45 m/s 2 α = 37 ◦ (b) a = g (sin α − µ k cos α ) = 4 . 33 m/s 2 − → n (c) a = g (cos α + µ k cos α ) = 9 . 39 m/s 2 − → f k − → w (d) a = g (cos α − µ k cos α ) = 6 . 26 m/s 2 � − → w ⊥ (e) a = g (sin α − µ k sin α ) = 4 . 72 m/s 2 Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise I A wooden block is sliding down a 37 ◦ wooden incline. What is its acceleration? α = 37 ◦ (b) a = g (sin α − µ k cos α ) = 4 . 33 m/s 2 − → n � F ⊥ = Ma ⊥ ⇒ n − w ⊥ = 0 − → f k − → ⇒ n = w ⊥ = Mg cos α w � − → w ⊥ � F � = Ma ⊥ ⇒ w � − f k = Ma ⇒ Mg sin α − µ k Mg cos α = Ma Newton’s Laws February 27, 2013 - p. 5/8

Kinetic Friction Exercise II A 5 . 0 kg wooden block is given a push and slides up a 37 ◦ wooden incline. After the pushing force is removed but the block is still going up the incline, which of the following is a correct statement about the block’s acceleration? Newton’s Laws February 27, 2013 - p. 6/8

Kinetic Friction Exercise II A 5 . 0 kg wooden block is given a push and slides up a 37 ◦ wooden incline. After the pushing force is removed but the block is still going up the incline, which of the following is a correct statement about the block’s acceleration? The acceleration’s magnitude is larger than 4 . 33 m/s 2 . (a) Newton’s Laws February 27, 2013 - p. 6/8

Kinetic Friction Exercise II A 5 . 0 kg wooden block is given a push and slides up a 37 ◦ wooden incline. After the pushing force is removed but the block is still going up the incline, which of the following is a correct statement about the block’s acceleration? The acceleration’s magnitude is larger than 4 . 33 m/s 2 . (a) The acceleration’s magnitude is smaller than 4 . 33 m/s 2 . (b) Newton’s Laws February 27, 2013 - p. 6/8

Kinetic Friction Exercise II A 5 . 0 kg wooden block is given a push and slides up a 37 ◦ wooden incline. After the pushing force is removed but the block is still going up the incline, which of the following is a correct statement about the block’s acceleration? The acceleration’s magnitude is larger than 4 . 33 m/s 2 . (a) The acceleration’s magnitude is smaller than 4 . 33 m/s 2 . (b) The acceleration’s magnitude is equal to 4 . 33 m/s 2 . (c) Newton’s Laws February 27, 2013 - p. 6/8