Unit 5: Equilibrium Lesson 2: Friction

f Warm-Up: A 100 N block is at rest on a rough incline. Find the normal force and the friction force. N 100N 30º y X : 100sin30 = f f N f = 50 N Y : 100cos30 = N 10 min 30º N = 87 N x 100N

Definition of coefficient of friction (μ): Example: Find the minimum coefficient of friction which prevents this block from slipping. M= 5.0 kg , θ=30° f N mg 30° y N f mg 30°( x x: 10 min f = (5.0 x 9.8) sin30° = 24.5 N N = (5.0 x 9.8) cos30° = y: 42.4N 0.58 f = mg sinθ N = mg cosθ = tanθ

Brain Break!

2. Find the maximum value for mass B if the system is in equilibrium and mass A is 20 kg. The pulley is frictionless. a) no friction b) µ = 0.2 T N mg y x 37° ( f A . T mg B A B 37°( a) T = (20 x 9.8) sin37° = TB = TA 118N = mg = 118N = 12kg 15 min b) N = (20 x 9.8) cos37° = f = µN 156.5N = 0.2 x 156.5N = 31.3N X: T = f + mg sinθ = 31.3N + 118N = 149N mg = T = 15kg

Try This! 20N 40° ( a) A 20 N mass is held by a string as shown. Find the tension in the string. (Assume no friction.) b) If the tension is 5.0 N, find the coefficient of friction. T N mg y x 40°( f T = mg sinθ = 20 x sin40° = 12.9 N X: T + f = mg sinθ f = 20 sin40° - 5.0 = 7.9 N Y: N = mg cosθ = 20 cos40° = 15.32 N 20 min = 0.51

Lesson 3: Practice with Newton’s First Law Unit 5: Equilibrium Lesson 3: Practice with Newton’s First Law

Try This! A box 200 N is held in place by a 150 N horizontal force. Find the minimum coefficient of friction. Y N N f f x F = 150N 200N F ) 60° ) 30° ) 30° mg mg y: N = mg cos30° + F cos60° = 200 cos30° + 150 cos60° = 248.2 N x: f + 150 sin60° = 200 sin30° f = 200 sin30° - 150 sin60° = -29.9 N Friction is going the other way! 15 min = 0.12

Brain Break!

Practice Time! Work on the Newton’s First Law Problem Set. (You should be able to do all of the questions.) 40 min

Lesson 4: More Practice with Newton’s First Law Unit 5: Equilibrium Lesson 4: More Practice with Newton’s First Law

Practice Time! You have today’s class to finish working through the Newton’s First Law Problem Set. 60 min