Classical Physics I. PHY131 Lecture 7 Friction Forces and Newton s Laws. Lecture 7 1

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1 Classical Phsics I PHY131 Lecture 7 Friction Forces and Newton s Laws Lecture 7 1

2 Newton s Laws: 1 & 2: F Net = ma Recap LHS: All the forces acting ON the object of mass m RHS: the resulting acceleration, from which the motion can be deduced 3: F AB = F BA ( Action is Reaction ) Valid onl in Inertial Reference Frames! Use Free Bod Diagram to find the Net Force F Net : diagram with all the (vector) forces acting ON THE BODY OF MASS m Put the ais convention ou will be using on it as well Make separate diagrams for separate parts (if an) of the full problem Lecture 7 2

3 Frictional Forces Generall, friction alwas opposes (relative) motion between the materials in question (block and incline, ball and air, fish and water, etc ). is directed anti-parallel to the relative motion (if an). In case of fluids and gases we speak of Drag ; we ll discuss this mostl later in case of rubbing surfaces we speak of Friction ; toda s topic Friction has atomic causes, essentiall electric in nature, and is intimatel related to the detailed surface properties and smoothness of the rubbing surfaces Note, that smooth often means more rather than less friction! Empirical: Moving a Crate: Initiall, a certain force is needed to start the crate moving; to un-glue the crate, one has to overcome Static Friction, which is tpicall larger than the force needed to keep it moving at constant velocit Kinetic Friction In case of no motion, the static frictional force counteracts eactl the (increasing) eternal force tring to move the crate, until the eternal force overcomes static friction Static friction has thus no fied value, but onl a maimum! The (maimum) value of the friction forces is in most cases closel proportional to the Normal force N that presses the rubbing surfaces crate bottom and floor together: μn, where μ for static friction is tpicall larger than μ = μ k for kinetic (= moving) friction Lecture 7 3

4 Friction 1 st Eample block of mass M = 2.0 kg sits on a horizontal table. The coefficients of static and kinetic friction are μ s = 0.30 and μ k = 0.25 respectivel. Calculate the minimum horizontal force F necessar to start the block moving. Note: normal forces, like static friction, are reaction forces, and tpicall calculable onl from other forces N = W N(-i) M F W = Mg(-j) the minimum force needed is to just overcome static friction: W + N = ma j = 0; N = W F + s = 0; s Ni= i = 5.9i N Calculate the acceleration of the block afterwards: F + k = ma = (μ s μ k )mg i = 0.98 i N; a = 0.48 m/s 2 Lecture 7 4

5 Friction 1 st Eample For what angle is F minimal? N(-i) M N = W Fsinθ(j) Fsinθ F Fcosθ W = Mg(-j) the weight W of the block is now partiall counteracted b the upward component of F. Because the vertical acceleration is zero, we have: W + N + Fsinθ = 0, and N = W Fsinθ Thus, the minimum force Fcosθ i needed to just overcome static friction: Fcosθ = s N (W Fsinθ) F W/(cosθ + μ s sinθ) = F(θ) The minimum occurs when df/dθ = 0: df/dθ W( sinθ + μ s cosθ)/(cosθ + μ s sinθ) 2 which equals zero for tanθ, i.e. θ = 16.7 ; The force F is then: F W/(cosθ + μ s sinθ) = sinθw = 5.6 N Lecture 7 5

6 Block on Slope Single block of mass M = 3.0 kg, is initiall at rest on a slope of angle θ. The angle is slowl increased until the block starts sliding at θ = 25 Q1: calculate the coefficient of static friction = μn N Wsinθ Wcosθ θ W=mg The block is at rest in the direction perpendicular to the slope, and thus F Net, = F = N Wcosθ = 0. When the block starts sliding: s = Wsinθ. Thus, for the parallel forces: F Net,// = F = s Wsinθ = 0. s N Wcosθ = Wsinθ Thus: μ s = tan(25 ) = 0.47 (independent of mass M!) Lecture 7 6

7 Block on Slope Q2: With θ kept at 25 the block slides down the slope (with constant acceleration) and slides a distance of 1.60 m in 2.0 seconds. Calculate the coefficient of kinetic friction N = μn when the block began sliding: F Net = Wsinθ k = Wsinθ μ k N = Wsinθ μ k Wcosθ = mg(sinθ μ k cosθ) = ma μ k = tanθ a/(gcosθ) Wcosθ W=mg Wsinθ θ Thus, we need the constant acceleration a from the rest of the info: D = ½ at 2 a = 2D/t 2 = 0.80 m/s 2 Then: μ k = tanθ a/(gcosθ) = 0.38 Lecture 7 7

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