1. Section 4–4: Everyday Forces Coach Kelsoe Physics Pages 135–143

2. Objectives Explain the difference between mass and weight. Find the direction and magnitude of normal forces. Describe air resistance as a form of friction. Use coefficients of friction to calculate frictional force.

3. Weight The gravitational force (F g ) exerted on an object by Earth is a vector quantity, directed toward the center of Earth. The magnitude of this force (F g ) is a scalar quantity called weight. It is considered a scalar because on Earth, the direction is always toward the center of Earth. Weight changes with the location of an object in the universe.

4. Weight Calculating weight at any location: F g = ma g a g = free-fall acceleration at that location Calculating weight on Earth’s surface: a g = g = 9.81 m/s 2 F g = mg = m(9.81 m/s 2 ) The reason for the distinguishing formulas is that 9.81 m/s 2 is the gravitational acceleration of Earth. Each mass has a different gravitational acceleration.

5. 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, F n = mg cos θ.

6. 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.

7. 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.

8. 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, and is symbolized by the Greek letter μ. Coefficient of kinetic friction: μ k = F k /F n Coefficient of static friction: μ s = F s,max /F n Where “F n ” = normal force

9. Coefficients of Friction Table

10. Sample Problem Coefficients of Friction 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. 24 kg 75 N

11. Sample Problem Solution Step 1: Define your variables. Given: m = 24 kg F s, max = F applied = 75 N Unknown: μ s = ? 24 kg 75 N

12. Sample Problem Solution Step 2: Calculate μ s = F s,max /F n = F s,max /mg μ s = 75 N/24 kg · 9.81 m/s 2 μ s = 0.32 24 kg 75 N

13. 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 friction between the box and the sidewalk is 0.500. Find the acceleration of the box.

14. Sample Problem Solution Step 1: Define your variables. Given: m = 20.0 kg μ k = 0.500 F applied = 90.0 N at θ = 30.0° Unknown: a = ? Diagram:

15. Sample Problem Solution Step 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 F applied. F applied,y = (90.0 N)(sin 30.0°) = 45.0 N (upward) F applied,x = (90.0 N)(cos 30.0°) = 77.9 N (to the right)

16. Sample Problem Solution Step 2: Plan Choose an equation or situation: A.Find the normal force, F n, by applying the condition of equilibrium in the vertical direction. ΣF y = 0 B.Calculate the force of kinetic friction on the box: F k = μ k F n C.Apply Newton’s second law along the horizontal direction to find the acceleration of the box: ΣF x = ma x

17. Sample Problem Solution Step 3: Calculate 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, ΣF y = 0, and solve for F n. ΣF 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 Note: Remember to pay attention to the direction of the forces. In this step, F g is given an opposite sign because its direction is opposite of the other 2 forces.

18. Sample Problem Solution Step 3: Calculate Use the normal force to find the force of kinetic friction. F k = μ k F n = (0.500)(151 N) = 75.5 N Use Newton’s second law to determine the horizontal acceleration. ΣF x = F applied – F k = ma x ax = F applied – F k = 77.9 N – 75.5 N = 2.4 N m20 kg 20 kg a = 0.12 m/s 2 to the right

19. Sample Problem Solution Step 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.

20. Air Resistance Air resistance is a form of friction. Whenever an object moves through a fluid medium, such as air or water, the fluid provides a resistance to the object’s motion. For a falling object, when the upward force of air resistance balances the downward gravitational force, the net force on the object is zero. The object continues to move downward with a constant maximum speed, called the terminal speed.

21. Fundamental Forces There are four fundamental forces: Electromagnetic force Gravitational force Strong nuclear force Weak nuclear force The four fundamental forces are all field forces.