VIEWS: 3 PAGES: 20 POSTED ON: 3/27/2011 Public Domain
The Nature of Friction Friction (Ff) is a force that works against the motion of an object! Friction involves objects that are in contact with one another! Different Types of Friction • Static friction- the force you must overcome to get an object sliding. • Sliding friction- the force of friction that exists as one surface slides past another. Frictionstatic > Frictionsliding because of inertia • Rolling friction- wheels on the ground • Fluid friction- air resistance, water drag Characteristics of Sliding Friction 1) Friction acts parallel to the surface and opposite of the motion! 2) Friction depends on the type of materials in contact! 3) Sliding friction is always less than starting(static) friction! 4) Friction is independent of the surface area in contact! 5) Friction depends upon the Normal Force! Measuring Sliding Friction The amount of friction between two surfaces is represented by the Coefficient of Sliding Friction. Coefficient of Sliding Friction (µ) depends directly upon the amount of friction force (Ff) and inversely upon the force pressing the surfaces together (N). Ff µ= N Examples of Friction force diagrams: A box pulled horizontally across a level floor at constant speed: Balanced Forces? N Yes- constant speed! Fa Ff Fa = Ff Fw = N Fw A crate sliding down an incline at constant speed: N Balanced Forces? Ff Yes- constant speed! Fp Ff = Fp ø N = Fn ø Fn Fw A box weighing 450 N is pulled along a level floor at constant speed by a rope that makes an angle of 30.0˚ with the floor. If the force in the rope is 260 N, what is the coefficient of sliding friction? Fw= 450 N Fa = 260 N ø = 30.0˚ N Fa F ø Fv f Fh Fw Balanced Forces? Yes- constant speed! x direction: Fh = Ff y direction: Fw = N + Fv Fh = Ff µ = Ff cosø = Fh/Fa N Fh= cosø(Fa) = cos(30.0˚)(260 N) = 225N N = Fw - Fv sinø = Fv/Fa Fv = sinø(Fa) = sin(30.0˚)(260 N) = 130 N N = 450 N - 130 N = 320 N µ = Ff = 225 N = .703 N 320 N A force of 225 N is applied horizontally to a box of mass 40.0 kg and it produces an acceleration of 2.50 m/s2. What must be the coefficient of friction? Fa = 225 N N m = 40.0 kg Fw = mg = 392 N Ff Fa a = 2.50 m/s2 µ=? Fw Balanced Forces? No- Acceleration! x direction: Fa > Ff Fa - Ff = F y direction: Fw = N F = ma Ff = Fa - F = (40.0 kg)(2.50m/s2) µ = Ff = 225 N - 100 N = 100 N = 125 N N = 125 N = .319 392 N A rope is used to pull a 75.0 kg crate up a 35.0˚ incline at constant speed. If the coefficient of friction between the crate and the plane is .480, what must the tension in the rope be? N m = 75.0 kg Fa Fw = mg = 735 N Fp Ff ø = 35.0˚ ø µ = .480 ø Fa = ? Fn Fw Balanced Forces? Yes- constant speed! Fa = Ff + Fp N = Fn Fp = sinø(Fw)= sin(35.0)(735N) = 422N Fn = cosø(Fw)= cos(35.0)(735N) = 602 N Ff = µN = (.480)(602 N) = 289N Fa = 422 N + 289 N = 711 N If the rope breaks while the crate is at rest on the ramp, how far down the ramp will the crate be when it speed hits 5.25 m/s? N vi = 0 Ff vf = 5.25 m/s Fp ∆d = ? ø Balanced? No!- Acceleration! ø Fn Fw Fp > Ff N = Fn Fp - Ff = F F = 133 N ∆d = vf2 - vi2 a= m 75.0 kg 2a = 1.77 m/s2 ∆d = (5.25 m/s)2 2(1.77 m/s2) F = Fp - Ff = 422 N - 289 N = 133 N = 7.79 m A box of mass 125 kg is accelerated at 1.50 m/s2 across a level floor by a rope tied at a 28.5˚ angle to the horizontal. If the tension in the rope is found to be 555 N, what is µ? m = 125 kg N Fw = mg = 1230 N Fa a = 1.50 m/s2 Ff Fv ø = 28.5˚ ø Fa = 555 N Fh µ=? Fw Balanced? No- Acceleration! Fh > Ff Fw = N + Fv Fh - Ff = F Ff = Fh - F µ = Ff Ff = 488 N - 188 N = 300 N N F = ma Fh = cosø(Fa) = cos(28.5˚)(555 N) =(125 kg)(1.50 m/s2) = 188 N = 488 N N = Fw - Fv Fv = sinø(Fa) = sin(28.5˚)(555 N) = 265N N = 1230 N - 265 N = 965 N µ = Ff = 300 N = .311 N 965 N A 56.0 kg box is accelerated across a level floor by a force applied horizontally to the box. If the coefficient of friction is .340 and the box is accelerated at 3.00 m/s2, what force is being used to move the box? A 565 N crate is pulled up a 25.0˚ incline at constant speed. If the coefficient of sliding friction is measured as .250, what must be the tension in the rope used to pull the crate up the incline? A 3.00 kg wood box slides from rest down a 35.0˚ inclined plane. How long does it take the box to reach the bottom of the 4.75 m incline if the coefficient of friction is .350? A 60.0 kg crate is attached to a weight by a cord that passes over a frictionless pulley. A) If the coefficient of friction is .500, what weight will keep the crate moving up a 20.0˚ incline at constant speed? B) If the cord is cut when the crate is at rest at the top, how far will the crate have slid down the incline by the time its speed reaches 7.50 m/s?