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Website   friction
 

Website friction

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    Website   friction Website friction Presentation Transcript

    • FRICTIONAL FORCES • There are two types of frictional forces: Type of friction Definition EquationStatic Friction A force that acts parallel to the two surfaces & Fmax = μsFN(used for stationary objects) keeps an object from moving.Kinetic Friction A force that acts opposite to the direction of an(used for moving objects) Ff = μkFN object’s motion. Fmax is the maximum force that can be applied to an Ff is the frictional force object before it begins to move The greek symbol “μ” is pronounced “mu”
    • NEW FREE-BODY DIAGRAM• Now we include the force of friction! FN Ff Fa mg
    • HOW TO CALCULATE FRICTIONAL FORCES • When an object is moving at a CONSTANT SPEED, we can find the force of friction due to the coefficient of kinetic friction. Constant speed means that acceleration = 0 FN So Fnet = 0 SO…. Fa – Ff = 0Ff Fa Ff = Fa mg • When a force is applied that causes an object to JUST BEGIN TO MOVE, we can find the force of friction due to the coefficient of static friction Once again…. Ff = Fa
    • Example: Net force = 0A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N.a) Calculate the Force of Friction Fa = Ff = 50Na) Calculate the Force Normal mg Fn (10 )(9.8) 98 Na) Calculate the coefficient of kinetic friction Ff = mk FN FN Fa 50 = m k (98) Ff 50 mg m k = = 0.51 98
    • Example: Net force = 0 Suppose the same box is now pulled at an angle of 30 degrees above the horizontal. a) Calculate the Force of Friction Fax Fa cos 50 cos30 43.3N Ff Fax 43.3N a) Calculate the Force Normal FN mg! FN Fay mg FN Fa FN mg Fay (10)(9.8) 50 sin 30 Fay FN 73NFf 30 Fax mg
    • EXAMPLE: NET FORCE = MAA 50 N applied force drags an 8.16 kg log to the right across a horizontal surface. What is the acceleration of the log if the force of friction is 40.0 N? Fn a FNET = ma 50 N40 N Fa - Ff = ma mg 50 - 40 = 8.16a 10 = 8.16a a= 1.23 m/s/s
    • EXAMPLE: NET FORCE = MAA sled is being accelerated to the right at a rate of 1.5 m/s/s by a rope at a 33 degree angle above the + x . Calculate the acceleration of the sled if the Frictional Force is 26.8 N, the mass of the sled is 66 kg and the tension in the rope is 150 N. a FN Tsin Tcos FNET = ma T cosq - Ff = ma T cosq - ma = Ff Ff 150 cos33- (66)(1.5) = Ff mg Ff = 1.5 m/s/s