Everyday Forces Friction Force. Types of Friction forces -Kinetic friction -Static Friction

Normal Force Force on an object perpendicular to the surface (Fn) It may equal the weight (Fg), as it does here. It does not always equal the weight (Fg), as in the second example. Fn = mg cos  Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

Chapter 4 Section 4 Everyday Forces Normal Force 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, Fn = mg cos q.

Chapter 4 Section 4 Everyday Forces Friction 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.

Static Friction Force that prevents motion Abbreviated Fs How does the applied force (F) compare to the frictional force (Fs)? Would Fs change if F was reduced? If so, how? If F is increased significantly, will Fs change? If so, how? Are there any limits on the value for Fs? These questions should help students understand that static friction balances the external force (F), so it increases and decreases as F increases and decreases. Eventually, F will be so large that the static frictional force (Fs) will no longer be able to balance it, and the net force will cause the object to slide. At this point, frictional forces become kinetic (see next slide).

Kinetic Friction Force between surfaces that opposes movement Abbreviated Fk Does not depend on the speed 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 Fk. Ask students if it requires more force to get an object moving when it is at rest or to keep it moving once it is already in motion. When pushing an object, we exert enough force to overcome static friction. At that point the object moves. The opposing force is now kinetic friction, which is less than static friction. Therefore, in order to maintain a constant speed and not accelerate, the force pushing the object is reduced.

Friction Forces in Free-Body Diagrams 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.

Calculating the Force of Friction (Ff) Ff is directly proportional to Fn (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 Point out to students that Ff is the general term for both static friction (Fs) and kinetic friction (Fk).

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

Typical Coefficients of Friction Values for  have no units and are approximate. Point out that static is greater than kinetic for each example. Also explain that the coefficient is generally less than 1 but there could be sticky surfaces where the frictional force was greater than the normal force. This would lead to coefficients greater than 1.

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 (Fn) the force of friction (Ff) Find the coefficient of friction. Answer: s = 0.32 This is a relatively simple example from the book (Sample Problem D). Ask students to follow the steps. It is easy to get the answer by skipping the free-body diagram, but they need this diagram to understand that normal force = weight, and the 75 N horizontal push is equal to the force of friction. More complicated problems (next slide) can’t be solved without a free- body diagram.

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.

Sample Problem, continued 1. Define Given: m = 20.0 kg mk = 0.500 Fapplied = 90.0 N at q = 30.0° Unknown: a = ? Diagram:

Sample Problem, continued Chapter 4 Sample Problem, continued 2. Plan Choose a convenient coordinate system, and find the x and y components of all forces. The diagram on the right shows the most convenient coordinate system, because the only force to resolve into components is Fapplied. Fapplied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward) Fapplied,x = (90.0 N)(cos 30.0º) = 77.9 N (to the right)

Sample Problem, continued Choose an equation or situation: A. Find the normal force, Fn, by applying the condition of equilibrium in the vertical direction: SFy = 0 B. Calculate the force of kinetic friction on the box: Fk = mkFn C. Apply Newton’s second law along the horizontal direction to find the acceleration of the box: SFx = max

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: Fg, Fn, and Fapplied,y. You know Fapplied,y and can use the box’s mass to find Fg. Fapplied,y = 45.0 N Fg = (20.0 kg)(9.81 m/s2) = 196 N Next, apply the equilibrium condition, SFy = 0, and solve for Fn. SFy = Fn + Fapplied,y – Fg = 0 Fn + 45.0 N – 196 N = 0 Fn = –45.0 N + 196 N = 151 N Tip: Remember to pay attention to the direction of forces. In this step, Fg is subtracted from Fn and Fapplied,y because Fg is directed downward.

Sample Problem, continued B. Use the normal force to find the force of kinetic friction. Fk = mkFn = (0.500)(151 N) = 75.5 N C. Use Newton’s second law to determine the horizontal acceleration. a = 0.12 m/s2 to the right

Sample Problem, continued Chapter 4 Section 4 Everyday Forces Sample Problem, continued 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.

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/s2 This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.