1. Forces and the Laws of MotionSection 4 Click below to watch the Visual Concept. Visual Concept Everyday Forces

2. Forces and the Laws of MotionSection 4 Normal Force Force on an object perpendicular to the surface (F n ) It may equal the weight (F g ), as it does here. It does not always equal the weight (F g ), as in the second example. F n = mg cos 

3. Forces and the Laws of MotionSection 4 Chapter 4 Normal Force Section 4 Everyday Forces The normal force acts on a surface in a direction perpendicular to the surface. The normal force is not always opposite in direction to the force due to gravity. –In the absence of other forces, the normal force is equal and opposite to the component of gravitational force that is perpendicular to the contact surface. –In this example, F n = mg cos .

4. Forces and the Laws of MotionSection 4 Click below to watch the Visual Concept. Visual Concept Chapter 4 Section 4 Everyday Forces Normal Force

5. Forces and the Laws of MotionSection 4 Friction Click below to watch the Visual Concept. Visual Concept

6. Forces and the Laws of MotionSection 4 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.

7. Forces and the Laws of MotionSection 4 Static Friction Force that prevents motion Abbreviated F s –How does the applied force (F) compare to the frictional force (F s )? –Would F s change if F was reduced? If so, how? –If F is increased significantly, will F s change? If so, how? –Are there any limits on the value for F s ?

8. Forces and the Laws of MotionSection 4 Kinetic Friction Using the picture, describe the motion you would observe. –The jug will accelerate. How could the person push the jug at a constant speed? –Reduce F so it equals F k. Force between surfaces that opposes movement Abbreviated F k Does not depend on the speed

9. Forces and the Laws of MotionSection 4 Chapter 4 Friction Forces in Free-Body Diagrams 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.

10. Forces and the Laws of MotionSection 4 Calculating the Force of Friction (F f ) F f is directly proportional to F n (normal force). Coefficient of friction (  ): –Determined by the nature of the two surfaces –  s is for static friction. –  k is for kinetic friction. –  s >  k

11. Forces and the Laws of MotionSection 4 Chapter 4 The Coefficient of Friction Section 4 Everyday Forces The quantity that expresses the dependence of frictional forces on the particular surfaces in contact is called the coefficient of friction, . Coefficient of kinetic friction: Coefficient of static friction:

12. Forces and the Laws of MotionSection 4 Typical Coefficients of Friction Values for  have no units and are approximate.

13. Forces and the Laws of MotionSection 4 Classroom Practice Problem A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. –Draw a free-body diagram and use it to find: the weight the normal force (F n ) the force of friction (F f ) –Find the coefficient of friction. Answer:  s = 0.32

14. Forces and the Laws of MotionSection 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 0.500. Find the acceleration of the box.

15. Forces and the Laws of MotionSection 4 Sample Problem, continued 1. Define Given: m = 20.0 kg  k = 0.500 F applied = 90.0 N at  = 30.0° Unknown: a = ? Diagram:

16. Forces and the Laws of MotionSection 4 Chapter 4 Sample Problem, continued 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)

17. Forces and the Laws of MotionSection 4 Sample Problem, continued 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

18. Forces and the Laws of MotionSection 4 Sample Problem, continued 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 + 45.0 N – 196 N = 0 F n = –45.0 N + 196 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.

19. Forces and the Laws of MotionSection 4 Sample Problem, continued 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

20. Forces and the Laws of MotionSection 4 Chapter 4 Sample Problem, continued Section 4 Everyday Forces 4. Evaluate The box accelerates in the direction of the net force, in accordance with Newton’s second law. The normal force is not equal in magnitude to the weight because the y component of the student’s pull on the rope helps support the box.

21. Forces and the Laws of MotionSection 4 Classroom Practice Problem 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 0.500. Find the magnitude of the acceleration of the box. –Start with a free-body diagram. –Determine the net force. –Find the acceleration. Answer: a = 0.12 m/s 2