Download presentation

Presentation is loading. Please wait.

Published byCaiden Polley Modified over 2 years ago

1
Friction

2
Friction - the force needed to drag one object across another. (at a constant velocity) Depends on: How hard the surfaces are held together What type of surface it is (i.e. rough, smooth) Not supposed to depend on: Surface area (pressure) Speed (low speeds)

3
F Fr = μF N Force of Friction in N Coefficient of Friction. 0 < μ < 1 Normal Force - Force exerted by a surface to maintain its integrity Usually the weight (level surfaces)

5
K inetic Friction - Force needed to k eep it going at a constant velocity. F Fr = μ k F N Always in opposition to velocity St atic Friction - Force needed to st art motion. F Fr < μ s F N Keeps the object from moving if it can. Only relevant when object is stationary. Always in opposition to applied force. Calculated value is a maximum

6
A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. a. Calculate the limit of static friction, and the kinetic friction. 20.1105 ≈ 20. N, 11.772 ≈ 12 N 5.0 kg

7
A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. (20.1105 N, 11.772 N) b. The block is at rest, and you exert a force of 15 N to the right to try to make it move. Draw and label all the forces acting on the block 5.0 kg

8
A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. (20.1105 N, 11.772 N) c. If the block is initially at rest, and we exert a force of 35 N to the right, calculate the block’s acceleration. 4.6456 ≈ 4.6 m/s/s 5.0 kg

9
A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. (20.1105 N, 11.772 N) d. Kyle and Sally both exert a force on the block, and it accelerates from rest to a 5.2 m/s over a distance of 10. m to the right. If Kyle exerts a force of 25 N to the right, what force in what direction does Sally exert? -6.468 ≈ -6.5 N 5.0 kg

10
Whiteboards: Friction 11 | 2 | 3 | 4 | 5 | 623456 TOC

11
What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete? F = ma, F Fr = μ k F N F N = weight = mg = (12 kg)(9.8 N/kg) = 117.6 N F Fr = μ k F N = (.8)(117.6 N) = 94.08 N = 90 N 90 N

12
What is the force needed to start a 150 kg cart sliding across wet concrete from rest if the wheels are locked up? F = ma, F Fr < μ s F N F N = weight = mg = (150 kg)(9.8 N/kg) = 1470 N F Fr = μ k F N = (.7)(1470) = 1029 N = 1000 N 1000 N

13
μ s =.62, μ k =.48 What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding? F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 N F = ma = (8.5 kg)a, a = 3.77 = 3.8 ms -2 3.8 m/s/s 72 N 8.5 kg v F Fr

14
μ s =.62, μ k =.48 A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop? F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)(22 kg)(9.8 N/kg) = 103.488 N F = ma = (22 kg)a, a = -4.704 ms -2 v = u + at, v = 0, u = 12, a = -4.704 ms -2, t = 2.55 s = 2.6 s 2.6 s 22 kg v=12m/s F Fr

15
μ s =.62, μ k =.48 A 6.5 kg box accelerates and moves to the right at 3.2 m/s/s, what force must be applied? F Fr = μ k F N, m = 6.5 kg F Fr = (.48)(6.5 kg)(9.8 N/kg) = 30.576 N F = ma = (6.5 kg)(3.2 ms -2 ), F = 51.376 = 51 N 51 N F = ? F Fr 6.5 kg v a = 3.2 ms -2

16
μ s =.62, μ k =.48 A 22 kg block is sliding on a level surface initially at 12 m/s stops in 2.1 seconds. What external force is acting on it besides friction?? v = u + at, v = 0, u = 12 m/s, t = 2.1 s, a = -5.7143 ms -2 F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)(22 kg)(9.8 N/kg) = 103.488 N F = ma = (22 kg)(-5.7143 ms -2 ), F = -22 N (left) -22 N (to the left) 22 kg v=12m/s F Fr F = ?

17
μ s =.62, μ k =.48 A force of 35 N in the direction of motion accelerates a block at 1.2 m/s/s in the same direction What is the mass of the block? F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)m(9.8 ms -2 ) = m(4.704 ms -2 ) F = ma = m(1.2 ms -2 ) 35 N = m(4.704 ms -2 ) + m(1.2 ms -2 ) = m(4.704 ms -2 + 1.2 ms -2 ) m = (35 N)/(5.904 ms -2 ) = 5.928 kg = 5.9 kg 5.9 kg 35 N F Fr m v a = 1.2 ms -2

Similar presentations

OK

Review for Dynamics test Page 1Page 1 - Net force with weight Page 2Page 2 - Friction on a level surface Page 3 Page 3 - Inclined Plane Page 4 Page 4.

Review for Dynamics test Page 1Page 1 - Net force with weight Page 2Page 2 - Friction on a level surface Page 3 Page 3 - Inclined Plane Page 4 Page 4.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on self awareness theory Ppt on articles of association definition Ppt on history of indian cinema Ppt on environment and sustainable development class 11 Ppt on education problems in india Ppt on no plastic bags Reliability of the bible ppt on how to treat Ppt on history generation and classification of computer Ppt on antimicrobial activity of ginger Ppt on english grammar articles