4-8 Applications Involving Friction, Inclines An object sliding down an incline has three forces acting on it: the normal force, gravity, and the frictional force. The normal force is always perpendicular to the surface. The friction force is parallel to it. The gravitational force points down. If the object is at rest, the forces are the same except that we use the static frictional force, and the sum of the forces is zero.

Use slide show mode (press F5) for animations. Inclined Planes Label the direction of N and mg. N θ mg Use slide show mode (press F5) for animations.

Inclined Planes Mark the direction of acceleration a. N a θ mg

Inclined Planes Choose the coordinate system with x in the same or opposite direction of acceleration and y perpendicular to x. y N x a θ mg

Inclined Planes Now some trigonometry y N x a θ 90- θ mg θ

Inclined Planes Replace the force of gravity with its components. x a mg sinθ mg cosθ θ θ mg

Inclined Planes mg sinθ mg cosθ mg Use Newton’s second law for both the x and y directions y N x a mg sinθ mg cosθ θ θ mg The force and acceleration in the x-direction have a negative sign because they point in the negative x-direction.

Inclined Planes mg sinθ mg cosθ mg Why is the component of mg along the x-axis –mgsinθ Why is the component of mg along the y-axis –mgcosθ mg mg sinθ θ mg cosθ N a x y

Inclined Planes mg sinθ mg mg cosθ Why is the component of mg along the x-axis: –mgsinθ Why is the component of mg along the y-axis: –mgcosθ y N a mg sinθ x θ θ mg mg cosθ

Inclined Planes mg sinθ mg mg cosθ Why is the component of mg along the x-axis: –mgsinθ Why is the component of mg along the y-axis: –mgcosθ y N a mg sinθ x θ θ mg mg cosθ

Inclined Planes mg sinθ mg mg cosθ Why is the component of mg along the x-axis: –mgsinθ Why is the component of mg along the y-axis: –mgcosθ θ y opposite N sinθ = a hypotenuse adjacent cosθ = mg sinθ hypotenuse x θ mg mg cosθ

Inclines q q Ff FN q q q Tips Rotate Axis Break weight into components Write equations of motion or equilibrium Solve mg q

Friction & Inclines A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85 N. Calculate the coefficient of friction if the cart is pushed at a constant speed. Fa Fn q Ff mg 0.117 q

Example A 5-kg block sits on a 30 degree incline. It is attached to string that is thread over a pulley mounted at the top of the incline. A 7.5-kg block hangs from the string. a) Calculate the tension in the string if the acceleration of the system is 1.2 m/s/s b) Calculate the coefficient of kinetic friction. T FN m2 m2gcos30 Ff 30 T m2g m1 30 m2gsin30 m1g

Example 64.5 N 0.80 N

Practice Problems 43. (a) A box sits at rest on a rough 30 degrees inclined plane. Draw a free body diagram. (b) How would the diagram change if the box were sliding down the plane? (c)How would it change if the box were sliding up the plane after an initial shove? 40. The coefficient of static friction between hard rubber and normal street pavement is 0.8. On how steep a hill (maximum angle) can you leave a car parked?

4-9 Problem Solving – A General Approach Read the problem carefully; then read it again. Draw a sketch, and then a free-body diagram. Choose a convenient coordinate system. List the known and unknown quantities; find relationships between the knowns and the unknowns. Estimate the answer. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. Keep track of dimensions. Make sure your answer is reasonable.

Summary of Chapter 4 Newton’s first law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed. Newton’s second law: Newton’s third law: Weight is the gravitational force on an object. The frictional force can be written: (kinetic friction) or (static friction) Free-body diagrams are essential for problem-solving

Homework: Chapter 4 Problems 41, 53, 55, 56

Kahoot 4-8