5Inclined PlanesA 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
6FNθInstead 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 haveΣ FY = 0FN - mg cos θ = 0FN = mg cos θIn the x-direction, we can immediately apply Newton’s Second Law to find the acceleration.Σ Fx = m amg sin θ = m ag sin θ = aThis is the formula we are trying to find.Fg = mg
7Example: 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=140Nq=18.5o
8Givens, Unknowns and Models m = 40.0 kgFapp = 140 Nq = 18.5oDx = 30 mv0 = 0v = ?This is constant force in the x direction (and constant acceleration) and equilibrium in y.FnFappqmg cos(q)mgmg sin(q)
13Discuss 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.
14Resistive ForcesResistive 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.
15FrictionFriction 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.FNFF
16Static 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.
17Static FrictionThe 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.
18Kinetic 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.FNFkFg
19Accelerating 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.FFThis tire will accelerate to the right because of the force of friction applied by the road.
20Rolling 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 FrictionThis tire will slow down if no other force is applied because rolling friction opposes the motion.
21Mini-lab- FrictionWhat 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.
22Lab ResultsFriction 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.
23Kinetic Friction and the Normal Force The force of kinetic friction is given by the formulaFN is the normal force between the surfaces.μK is the coefficient of kinetic friction, which depends on the nature of the surfaces in contact.FNFkFg
24Maximum Force of Static Friction The maximum force of static friction is given byμs depends on the surfaces in contact. It is usually larger than the corresponding μk.FAppFNFsFg
25Kinetic 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.
26Static Friction Example: Limiting Angle of Repose
32Drag ForcesThis 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”
33What 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.
34Terminal SpeedBecause 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.
43Homework 4c: 11-14A 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?
44Givens, Unknown and Model m = 3.00 kgq = 30.0oDx = mvi = 0Dt = 1.50 sax = ? Fk = ?mk = ? vf= ?Equilibrium in y, acceleration in x.qFnFKqmg cos(q)mgmg sin(q)
45Finding a: Method, Implement and Evaluate Solution FnFKqmg cos(q)mgmg sin(q)
46Finding FK: Method, Implement and Evaluate Solution FnFKqmg cos(q)mgmg sin(q)
47Finding μK: Method, Implement and Evaluate Solution FnFKqmg cos(q)mgmg sin(q)
49Practice 4D, #1A 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 Find the acceleration of the box.
54Evaluate 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.
55Chapter Review, # 39A 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.
56G.U.M. Note that both FN and Fg point down. Fapp = 85 N m = 4.00 kg q = 55oa=6 m/s2mK = ?Accelerates in x, equilibrium in y.FKFNFg