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Foundation of Technical Education Al-Dour Technical Institute Mechanical Department 1st Stage Training Package In Friction , Dry Friction For Students of first class Mechanical Department/ Production By Dr. Salah F. Al-Samarraaee 2013 Overview Friction forces is very important subject to be studied In order to have a full knowledge about the contact resistance exerted by one body upon asecond when the second body moves or tends to move past the first body. For this reason I have been designed this modular unit for this knowledge to be understood. Objectives :After studying the first modular unit , the student will be able to:1- Concept of friction force. 2- Angle of friction force. Flow Chart:- Activities And the exchanges In The Package Pre Test IF β₯ π Pre Test Objectives Overview No No IF β₯ π Yes Yes Next Package Pre Test : For the (40N) horizontal body shown ,Find:1- The force acting on the thread which make the body move from rest ( motionless)is( µs= 0.5). 2- The force needed to keep the body moving continuously in a uniform velocity and in a straight line if ( µk = 0.2). ( sin 37=0.6 cos 37 =0.8 ) 40 N 37 Ν¦ Notes - 10 degree for the above question. - Check your answers in key answer. The Text 9. Friction Friction is the tangential forces generated between contacting Impending surfaces. motion F Static friction (no motion) πΉπππ₯=ππ .π Kinetic friction ( motion) πΉπ=ππ .N Figure (12) P 9.1 Static Friction:- the region in the fig. above up to point of slippage or impending motion is called the range of static friction, and in this range the value of static friction force is determined by the equation of equilibrium. This friction force may have any value from zero up to and including the maximum value which proportional to the normal force N. thus we may write:- πΉπππ₯ = ππ π where ππ is the proportionality constant, called the coefficient of static friction. ( ππ = β«)Ω ΨΉΨ§Ω Ω Ψ§Ψ§ΩΨΨͺΩΨ§Ω Ψ§ΩΨ³ΩΩΩΩβ¬ Be aware that the last equation describes only the limiting or maximum value of the static friction force and not any lesser value, thus the equation applies only to where motion is impending ( β«) ΩΨ΄ΩΩβ¬ with the friction force at its peak value. For a condition of static equilibrium where motion is not impending, the static friction force is πΉ < ππ π 9.2 Kinetic Friction:- After slippage occurs, a condition of kinetic friction accompanies the ensuing motion. Kinetic friction force is usually somewhat less than the maximum static friction force. The kinetic friction force is also proportional to the normal force, thus πΉπ = ππ π Where ππ is the coefficient of kinetic friction. Also we can write ππ < ππ 9.3 Friction on Horizontal surface Fπ = π sin β tan β = ππ = πΉπ π πΉπ P π β = tan β N figure 13 ( ππ =β«)Ω ΨΉΨ§Ω Ω Ψ§Ψ§ΩΨΨͺΩΨ§Ω Ψ§ΩΨ³ΩΩΩΩβ¬ β«)Ψ§ΩΩ ΨͺΨ§ΩΩ Ψ³ΩΩβ¬ (Fs = β« ( )ΩΩΨ© Ψ§Ψ§ΩΨΨͺΩΨ§Ω Ψ£ΩΨ³ΩΩΩΩβ¬N= β«Ψ§ΩΩΩΨ© Ψ§ΩΨΉΩ ΩΨ―ΩΨ© ΨΉΩΩ Ψ§ΩΨ³Ψ·ΨΩΩβ¬ 9.4 Friction on Incline Surface There are two components for the weight (mg) as shown. Before moving we have :- F R β πΉπ₯ = 0 πΉ = ππ sin ΞΈ β πΉπ¦ = 0 y π = ππ cos ΞΈ N π= πΉ π = ππ π ππΞΈ πππππ ΞΈ F = tan ΞΈ ππ sin ΞΈ ΞΈ ΞΈ ππ cos ΞΈ mg Figure 14 Post Test If the force 100 Newtons acts as shown on the 300 N body. Make a conclusion if the body is in equilibrium or not and find the friction force? ππ = 0.25 100 Newtons 3 4 ππ = 0.2 Key answer Pre test :β πΉπ₯ = 0 π × cos 37 β πΉ = 0 πΉπ = π × cos 37 β β β β β β(1) β πΉπ¦ = 0 π + π π ππ 37 β 40 = 0 β β β β β β β β(2) πΉπ = ππ π = 0.5 π β β β β β β β β β (3) Sub. (3) in (1) π = 1.6 π 0.8P=0.5N Sub. In (2) 1.6 π + 0.6 π = 40 2.2 P=40 P=18.181818 Newtons Post test :Sol: To find the friction force needed For equilibrium, assume the friction force in a direction downward the incline surface, then the F.B.D will N be as shown: F 3 100 β × 300 β πΉ = 0 5 100 Newtons 3 4 300 × 300 4 5 β«πΉ = βͺ100 β 180β¬β¬ β«π πππ‘π€ππβͺπΉ = β80β¬β¬ β«Ψ§ΩΨΨΈ Ψ§Ψ₯ΩΨ΄Ψ§Ψ±Ψ© Ψ§ΩΨ³Ψ§ΩΨ¨Ψ© ΨͺΨΉΩΩ Ψ£Ω Ψ§ΨͺΨ¬Ψ§Ω πΉ ΨΉΩΨ³ Ψ§ΩΩΨ±ΨΆβ¬ β«βͺβ πΉπ¦ = 0β¬β¬ β«βͺ4β¬β¬ β«βͺ× 300 = 0β¬β¬ β«βͺ5β¬β¬ β«βͺπββ¬β¬ β«βͺN=240 Newtonsβ¬β¬ β«Ψ§ΩΨΨΈ Ψ£Ω ΩΩΨ© Ψ§Ψ§ΩΨΨͺΩΨ§Ω Ψ§ΩΩ Ψ·ΩΩΨ¨Ψ© ΩΨΩΨΈ Ψ§Ψ§ΩΨͺΨ²Ψ§Ω ΩΩπ πππ‘π€ππβͺ 80β¬Ψ¨ΩΩΩ Ψ§ Ψ£ΩΨ΅Ω ΩΩΨ©β¬ β«ΩΨ§ΩΨΨͺΩΨ§Ω ΩΩ βͺ-:β¬β¬ β«π πππ‘π€ππ βͺπΉπππ₯ = ππ π = 0.25 × 240 = 60β¬β¬ β«Ψ₯Ψ°Ω Ψ§ΩΩΩΨ© Ψ§ΩΩ Ψ·ΩΩΨ¨Ψ© ΩΨΩΨΈ Ψ§Ψ§ΩΨͺΨ²Ψ§Ω (βͺ 80β¬ΩΩΩΨͺΩ) Ψ£ΩΨ¨Ψ± Ω Ω ΩΩΨ© Ψ§Ψ§ΩΨΨͺΩΨ§Ω Ψ§ΩΩΨ΅ΩΩ ( βͺ60β¬β¬ β«ΩΩΩΨͺΩ)β¬ β«Ψ£Ω Ψ£Ω Ψ§ΩΨ¬Ψ³Ω ΩΩΨ²ΩΩ ΩΨΩ Ψ§Ψ£ΩΨ³ΩΩ ΩΨͺΩΩΩ ΩΩΩ Ψ© Ψ§Ψ§ΩΨΨͺΩΨ§Ω Ψ§ΩΨΩΩΩΩΨ© ΩΩ βͺ-:β¬β¬ β«π πππ‘π€ππ βͺπΉπ = ππ π = 0.2 × 240 = 48β¬β¬ β«βͺReferences:β¬β¬β«"βͺ1- Bed ford & fowler "Engineering Mechanicsβ¬β¬ β«βͺthβ¬β¬ β«βͺStatic & dynamics 4 Edition 2005.β¬β¬ β«βͺ2- J.L Meriam "Engineering Mechanics- Statics" Secondβ¬β¬ β«βͺEdition 1980.β¬β¬ β«βͺ3- Jan Kiusslaas "Engineering Mechanics Statics" Secondβ¬β¬ β«βͺEdition 2001.β¬β¬
