1. The Force of Friction

2. What Do We Already Know about forces? Newton’s Laws 1.An object at rest remains at rest until acted upon by an unbalanced force 2.F=ma 3.Every force has an equal and opposite force

3. Friction Opposes the Applied Force Static Friction: -The resistive force that keeps an object from moving -As long as an object does not move, the force of static friction is always equal to and opposite in direction to the applied force. F FsFsFsFs -When the applied force is as great as it can be without causing the object to move, the force of static friction reaches its maximum value, F s, max F s = -F applied

4. What happens when the applied force on an object exceeds F s,max ? The object begins to move with an acceleration in the direction of the applied force. There still is a frictional force, but the force is less than F s,max. The frictional force on the object is then called the force of kinetic friction (F k ) FkFk F a

5. Coefficient of Friction Frictional forces depend on the surface on which the object is in contact with. The coefficient of friction is the ratio of the force of friction to the normal force acting between two objects (μ ). μ k F n F k = μ k F n μ s F n F s,max = μ s F n

6. No applied force, car at rest FnFn FgFg Weak applied force, car remains at rest FaFa FsFs μ s F n F s is less than μ s F n FnFn FgFg Stronger applied force, car just about to move FaFa μ s F n F s = μ s F n FnFn FgFg FsFs Car moving at a constant speed μ k F n F k = μ k F n FnFn FgFg FsFs FaFa WHEN DOES THE CAR BEGIN TO MOVE AT A CONSTANT SPEED? a

7. What friction force is stronger? What is the frictional force on a 10kg rubber block on concrete just before it begins to slide? μ s =0.90 μ s =0.90 F n =(9.81m/s 2 )(10kg)=98.1N μ s F n =(0.90)(98.1N)=88.29N F s, max = μ s F n =(0.90)(98.1N)=88.29N What is the frictional force on a 10kg rubber block on concrete when it moves at a constant acceleration? μ k =0.68 F n =(9.81m/s 2 )(10kg)=98.1N μ k F n =(0.68)(98.1N)=66.71N F k = μ k F n =(0.68)(98.1N)=66.71N

8. Practice Problem A 24kg crate initially at rest on a horizontal floor requires a 75N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. 24kg F s,max FaFa FnFn FgFg F s,max = 75Nm=24kg μ s F n F s,max = μ s F n μ s = F s,max = _ 75N_____ = 0.32 F n 24kg x 9.81m/s 2

9. Practice Problem: Overcoming Friction A student moves a box of books by attaching a rope to the box and pulling with a force of 90.0N. The box of books has a mass of 20.0kg, and the coefficient of kinetic friction between the bottom of the box and the sidewalk is 0.50. Find the acceleration of the box. 20kg FaFa FgFg FaFa FkFk F g = mg = 20kg*9.81m/s 2 = 196.2N ΣF y = -F g + F n = -196.2N + F n = 0 F n = 196.2N F a = 90.0N ΣF x = -F k + F a = ma ΣF x = -F n μ k + 90.0N= ma ΣF x =-196.2N*0.50 + 90.0N=20kg*a a = -196.2N*0.50 + 90.0N = -9.41m/s 2 20kg