Applications of Newton’s Laws Mark Lesmeister Dawson High School Physics

Acknowledgements © 2013 Mark Lesmeister/Pearland ISD This work is licensed under the Creative Commons Attribution- ShareAlike 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA. Selected graphics and problems from OpenStax College. (2012, June 12). College Physics. Retrieved from the Connexions Web site: http://cnx.org/content/col11406/1.7/

This graphic shows three things we will talk about this week: Inclined planes (the banked track), friction (it’s how the car moves) and air resistance (the shape of the car is designed to reduce it.)

Inclined Planes Part 1

Inclined Planes A box slides down two smooth rails with no friction. Find the acceleration of the box. Show incline. Ask what it is used for. Ask how that works. Demonstrate using student and incline plane that less force is required to bring something up inclined plane. Say we will analyze this. Bring up inclined plane picture. Ask what forces act on object. (Gravity, normal force, and applied force.) Show free body diagram by bringing up forces. Point out that analyzing these forces with x and y directions as usual would be complicated. Ask why we use x and y. (Perpendicular) So, do they have to be horizontal and vertical? Elucidate that we can just as well use perpendicular and parallel directions. Show how two forces already are in those directions. Thus, only Force of gravity will be broken into components. Show how angle between gravity and perp. Is same as angle of incline, using plumb bob. Then, derive each component. Then, see if scale is correct at 30 degrees. Then do the example problem. q

FN θ Instead of using axes that are horizontal and vertical, we use a system with the x axis parallel to the plan, and the y axis perpendicular to the plane. This way the acceleration will be entirely along the x-axis, in the positive x axis as I have drawn it here. Also, now all the forces, including friction when we start to include it, will lie along an axis, with the exception of gravity. But, gravity will always have the same two components. Once we figure out what those components are, we can use them in every incline plane problem. Here, the two forces acting on the block are gravity and the normal force. There is no friction. So, the free body diagram is as shown above. The two components of gravity are shown above with dotted lines. From geometry, it is easy to show that the angle of incline is the same as the angle θ shown above. The angle we want to use is always the angle between the force of gravity and the perpendicular, or y, axis. The components of gravity are then Fgx = mg sin θ and Fgy = mg cos θ. As long as we use this angle, the components of gravity will always be the same. In the y-direction, there is no acceleration, since the block does not move perpendicular to the incline. The y direction is in equilibrium, so we have Σ FY = 0 FN - mg cos θ = 0 FN = mg cos θ In the x-direction, we can immediately apply Newton’s Second Law to find the acceleration. Σ Fx = m a mg sin θ = m a g sin θ = a This is the formula we are trying to find. Fg = mg

Example: Forces on an Incline A 40 kg wagon is towed up a hill at an 18.5o incline. The tow rope exerts a force of 140 N. The wagon starts from rest. How fast is the wagon going after 30 m? Fapp=140N q=18.5o

Givens, Unknowns and Models m = 40.0 kg Fapp = 140 N q = 18.5o Dx = 30 m v0 = 0 v = ? This is constant force in the x direction (and constant acceleration) and equilibrium in y. Fn Fapp q mg cos(q) mg mg sin(q)

Method Fn Fapp q mg cos(q) mg mg sin(q)

Implement

Evaluate the Solution Fapp=140N q=18.5o This is about 11 mi/hr, a reasonable speed. It is positive, so the wagon moves uphill, as expected.

Friction Part 2

Discuss this question and be prepared to answer. What force pushes a car forward? Gravity pulls down. The normal force pushes upward. The car’s engine pushes on the wheels, but that is not an outside force. Point out that the only thing pushing on the car in the direction of motion is the frictional forces of the road against the tire.

Resistive Forces Resistive forces, such as friction and drag forces, are forces which oppose the relative motion of two surfaces, or a surface and a fluid. Friction occurs when two solid surfaces interact. Drag forces occur in fluids, i.e. liquids and gases. Note that these are not fundamental forces. Friction results from the same electric forces that bond molecules together.

Friction Friction is really just one component of the force between two surfaces. Friction is the force of interaction parallel to the surfaces. The perpendicular component is the normal force. We generally treat friction and the normal force as separate forces. FN FF

Static and Kinetic Friction The amount of friction between an object and a surface depends on whether the object is moving relative to the surface. Static friction applies when the object is at rest relative to the surface. Kinetic friction applies when the object is moving relative to the surface. Demonstrate difference between static and kinetic friction using force meter and Logger Pro. Then show above.

Static Friction The force is equal to the applied force, until it exceeds a certain maximum. Once the maximum force of static friction is exceeded, the object breaks free and begins to slide. Tell actuary joke.

Kinetic Friction FN Fk Fg Kinetic friction applies when the object is moving across the surface. The force of kinetic friction is typically less than the corresponding maximum force of static friction. In other words, once things are moving they are easier to keep moving. FN Fk Fg

Accelerating a car. The force the road applies to accelerate a car (speed up, slow down or change direction) is a frictional force. If the wheels do not slip, the friction can be treated as static friction. If the wheels slip, kinetic friction is more appropriate. FF This tire will accelerate to the right because of the force of friction applied by the road.

Rolling Friction Frolling Friction If the car moves along at constant velocity, ideally no forces are needed to accelerate it. However, real tires deform, and the road and tire must be continually peeled apart as the tire rolls. A small force is needed to overcome the rolling friction to keep the car moving with constant velocity. Frolling Friction This tire will slow down if no other force is applied because rolling friction opposes the motion.

Mini-lab- Friction What factors affect the amount of the frictional force between two surfaces? Using a force sensor and blocks of wood, investigate your answers to the above question. Every group should investigate the relationship between normal force and friction, and at least one other factor.

Lab Results Friction generally doesn’t depend on the surface area between two objects. Objects with less surface area have more normal force / unit area. Friction depends directly on the normal force between two objects. The types of surfaces in contact make a difference in friction observed.

Kinetic Friction and the Normal Force The force of kinetic friction is given by the formula FN is the normal force between the surfaces. μK is the coefficient of kinetic friction, which depends on the nature of the surfaces in contact. FN Fk Fg

Maximum Force of Static Friction The maximum force of static friction is given by μs depends on the surfaces in contact. It is usually larger than the corresponding μk. FApp FN Fs Fg

Kinetic Friction Example: Eins, Zwei, Soffa A bartender slides a 0.45 kg beer stein horizontally along a bar with an initial speed of 3.5 m/s. (The stein contains root beer of course!) The stein comes to rest near the customer after sliding 2.8 meters. Find the coefficient of kinetic friction.

Static Friction Example: Limiting Angle of Repose

FN FS Fg = mg

Practice Problems: Friction OpenStax College Physics textbook, “Friction”, Exercises 5 and 6.

Practice Problems: Friction

Drag Forces Part 3

Drag Forces This chute is called a drag chute, but not directly for same reason as a car is called drag racer. Source: Wikipedia commons, “B-52 landing with drogue chute”

What factors influence drag? Shape of the object. Cars and planes are given aerodynamic shape to reduce drag. Properties of the fluid. Denser fluids increase drag. Speed of the object relative to the fluid. The simplest case is when Fd is proportional to v.

Terminal Speed Because the drag force increases with increasing speed, a falling object will reach a speed where the drag force balances the force of gravity. This speed is called terminal speed.

Terminal Speed At terminal speed

Inclined Planes with Friction

Practice 4D, #3 A 75 kg box slides down a 25.0o ramp with an acceleration of 3.60 m/s2. Find mk between the box and the ramp. q=25o

Givens, Unknowns and Models m = 75 kg q = 25o ax= -3.60 m/s2 mk = ? Constant force in x direction, equilibrium in y. q=25o FN Fk q mg cos(q) mg mg sin(q)

Method Fn Fk Use x equation to find Fk. Use y equation to find FN. mg cos(q) Use x equation to find Fk. Use y equation to find FN. Use Fk and FN to find m mg mg sin(q)

Implement FN Fk q mg cos(q) mg mg sin(q)

FN Fk q mg cos(q) mg mg sin(q)

Fn Fk q mg

Homework 4c: 11-14 A 3.00 kg block starts from rest at the top of a 30.0o incline and accelerates uniformly down the incline, moving 2.00 m in 1.50 s. What is the magnitude of the acceleration of the block? What is the coefficient of kinetic friction? What is the frictional force acting on the block? What is the speed of the block at the end?

Givens, Unknown and Model m = 3.00 kg q = 30.0o Dx = -2.00 m vi = 0 Dt = 1.50 s ax = ? Fk = ? mk = ? vf= ? Equilibrium in y, acceleration in x. q Fn FK q mg cos(q) mg mg sin(q)

Finding a: Method, Implement and Evaluate Solution Fn FK q mg cos(q) mg mg sin(q)

Finding FK: Method, Implement and Evaluate Solution Fn FK q mg cos(q) mg mg sin(q)

Finding μK: Method, Implement and Evaluate Solution Fn FK q mg cos(q) mg mg sin(q)

Forces Applied At an Angle Part 4

Practice 4D, #1 A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 185 N at and angle of 25.0o above the horizontal. The box has a mass of 35.0 kg, and μK between the box and the floor is 0.27. Find the acceleration of the box.

Identify G.U.M. m= 35 kg q = 25o Fapp = 185 N mK= 0.27 ax = ? FN m= 35 kg q = 25o Fapp = 185 N mK= 0.27 ax = ? Fapp FK Fg Y- direction- Equilibrium X-direction- Constant Force

Method FN Fapp FK Fg

Method FN Fapp FK Fg

Implement

Evaluate the solution. Our answer is about 1/3 of g. This is a reasonable acceleration for an object to have over a short period of time.

Chapter Review, # 39 A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0o with the horizontal. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between the block and the ceiling.

G.U.M. Note that both FN and Fg point down. Fapp = 85 N m = 4.00 kg q = 55o a=6 m/s2 mK = ? Accelerates in x, equilibrium in y. FK FN Fg

Method

Implement and Evaluate the Solution