Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Everyday Forces Chapter 4 Objectives Describe air resistance as a form of friction. Use coefficients of friction to calculate frictional force.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Friction Section 4 Everyday Forces Static friction is a force that resists the initiation of sliding motion between two surfaces that are in contact and at rest. Kinetic friction is the force that opposes the movement of two surfaces that are in contact and are sliding over each other. Kinetic friction is always less than the maximum static friction.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Two Kinds of Friction Static friction ◦Must be overcome in order to budge an object ◦Present only when there is no relative motion between the bodies, e.g., the box & table top Kinetic friction ◦Present only when objects are moving with respect to each other (skidding) Applied force fs Applied force fk ΣF is to the right. a is positive Objects are still. ΣF = 0.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Friction Section 4 Everyday Forces

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Static Friction Force fs   s N static frictional force coefficient of static friction normal force fs, max =  s N maximum force of static friction fs, max is the force you must exceed in order to budge a resting object.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Static friction force varies f s, max is a constant in a given problem, but f s varies. f s matches F A until F A exceeds f s, max. Example: In the picture below, if  s for a wooden crate on a tile floor is 0.6, f s, max = 0.6 (10 ) (9.8) = 58.8 N. 10 kg F applied = 27 Nfs = 27 N 10 kg F applied = 43 Nfs = 43 N 10 kg F applied = 66 N The box finally budges when F applied surpasses fs, max. Then kinetic acts on the box. fk

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Friction Strength The magnitude of the friction force is proportional to: how hard the two bodies are pressed together (the normal force, N ). the materials from which the bodies are made (the coefficient of friction,  ). Attributes that have little or no effect: sliding speed contact area

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Kinetic Friction Once object budges, forget about  s. Use  k instead. f k is a constant so long as the materials involved don’t change. There is no “maximum f k.” fk =  k N kinetic frictional force coefficient of kinetic friction normal force

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Coefficients of Friction Static coefficient …  s. Kinetic coefficient …  k. Both depend on the materials in contact. ◦Small for steel on ice or scrambled egg on Teflon frying pan ◦Large for rubber on concrete or cardboard box on carpeting The bigger the coefficient of friction, the bigger the frictional force.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Section 4 Everyday Forces

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu  values Typically, 0 <  k <  s < 1. This is why it’s harder to budge an object than to keep it moving. If  k > 1, it would be easier to lift an object and carry it than to slide across the floor. Dimensionless (’s have no units).

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Example: A 24.0 kg crate initially at rest on a horizontal floor requires a 75. N horizontally force to set it in motion. Find the coefficient of static friction between the crate and the floor.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Friction Example 1 Barrel o’ Monkeys 14.7 kg You push a giant barrel o’ monkeys setting on a table with a constant force of 63 N. If  k = 0.35 and  s =0.58, when will the barrel have moved 15 m? answer: Never, since this force won’t even budge it! 63 < 0.58 (14.7) (9.8)  83.6 N

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Friction Example 2 Barrel o’ Monkeys 14.7 kg Same as the last problem except with a bigger FA: You push the barrel o’ monkeys with a constant force of 281 N.  k = 0.35 and  s =0.58, same as before. When will the barrel have moved 15 m? step 1: fs, max = 0.58 (14.7) (9.8)  83.6 N step 2: FA= 281N > fs, max. Thus, it budges this time. step 3: Forget fs and calculate fk: fk = 0.35 (14.7) (9.8) = N (continued on next slide)

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Friction Example 2 (continued) step 4: Free body diagram while sliding: mg N FA fk step 5: ΣF = FA – fk = = N To avoid compounding of error, do not round until the end of the problem? step 6: a = ΣF / m = / 14.7 = m/s 2 step 7: Kinematics:  x = +15 m, v 0 = 0, a = m/s 2, t = ?  x = v 0 t + ½ a t 2 t = 2  x / a  1.38 s

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Friction Forces in Free-Body Diagrams Section 4 Everyday Forces In free-body diagrams, the force of friction is always parallel to the surface of contact. The force of kinetic friction is always opposite the direction of motion. To determine the direction of the force of static friction, use the principle of equilibrium. For an object in equilibrium, the frictional force must point in the direction that results in a net force of zero.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Sample Problem Overcoming Friction A student attaches a rope to a 20.0 kg box of books.He pulls with a force of 90.0 N at an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is Find the acceleration of the box. Section 4 Everyday Forces

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Sample Problem, continued Section 4 Everyday Forces 1. Define Given: m = 20.0 kg  k = F applied = 90.0 N at  = 30.0° Unknown: a = ? Diagram:

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Sample Problem, continued Section 4 Everyday Forces The diagram on the right shows the most convenient coordinate system, because the only force to resolve into components is F applied. 2. Plan Choose a convenient coordinate system, and find the x and y components of all forces. F applied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward) F applied,x = (90.0 N)(cos 30.0º) = 77.9 N (to the right)

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Sample Problem, continued Section 4 Everyday Forces Choose an equation or situation: A. Find the normal force, F n, by applying the condition of equilibrium in the vertical direction:  F y = 0 B. Calculate the force of kinetic friction on the box: F k =  k F n C. Apply Newton’s second law along the horizontal direction to find the acceleration of the box:  F x = ma x

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Sample Problem, continued Section 4 Everyday Forces 3. Calculate A. To apply the condition of equilibrium in the vertical direction, you need to account for all of the forces in the y direction: F g, F n, and F applied,y. You know F applied,y and can use the box’s mass to find F g. F applied,y = 45.0 N F g = (20.0 kg)(9.81 m/s 2 ) = 196 N Next, apply the equilibrium condition,  F y = 0, and solve for F n.  F y = F n + F applied,y – F g = 0 F n N – 196 N = 0 F n = –45.0 N N = 151 N Tip: Remember to pay attention to the direction of forces. In this step, F g is subtracted from F n and F applied,y because F g is directed downward.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 4 Sample Problem, continued Section 4 Everyday Forces B. Use the normal force to find the force of kinetic friction. F k =  k F n = (0.500)(151 N) = 75.5 N C. Use Newton’s second law to determine the horizontal acceleration. a = 0.12 m/s 2 to the right

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Practice problems Pg 139 #1, 2 Pg 141 #1, #4 Pg 147 #