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Friction is a critical force that exists between any two sliding surfaces and must be accounted for in realistic scenarios. There are two main types of friction: static friction (when there’s no motion) and kinetic friction (when motion occurs). The size of the friction force depends on various factors, including the materials and their conditions (smooth or rough, wet or dry). This guide provides key formulas and experiments demonstrating the relationship between friction force and normal force, emphasizing the coefficients of static and kinetic friction.
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Friction • Friction: We must account for it to be realistic! • Exists between any 2 sliding surfaces. • Two types of friction: Static (no motion) friction Kinetic (motion) friction • The size of the friction force: Depends on the microscopic details of 2 sliding surfaces. • The materials they are made of • Are the surfaces smooth or rough? • Are they wet or dry? • Etc., etc., etc.
Kinetic Friction: Experiments determine the relation used to compute friction forces. • Friction force fk is proportional to the magnitude of the normal force n between 2 sliding surfaces. • DIRECTIONSof fk & n areeach other!!fk n • Write relation as fk k n (magnitudes) k Coefficient of kinetic friction kdepends on the surfaces & their conditions kis dimensionless & < 1 n a fk FA(applied) mg
Static Friction:Experiments are used again. • The friction force fs exists || 2 surfaces, even if there is no motion. Consider the applied force FA: ∑F = ma = 0 & also v = 0 There must be a friction forcefs to oppose FA FA – fs = 0 orfs = FA n fk FA(applied) mg
Experiments find that the maximum static friction forcefs(max) is proportional to the magnitude (size) of the normal force n between the 2 surfaces. • DIRECTIONSof fk & n areeach other!! fk n • Write the relation as fs(max) = sn (magnitudes) s Coefficient of static friction • Depends on the surfaces & their conditions • Dimensionless & < 1 • Always find s > k Static friction force:fs sn
Coefficients of Friction μs > μk fs(max, static) > fk(kinetic)
Example n f
Example ∑F = ma n n fs fs y direction: ∑Fy = 0;n- mg + Fy = 0 n = mg - Fy; fs(max) = μsn x direction:∑Fx = ma y direction: ∑Fy = 0; n– mg – Fy = 0 n = mg + Fy ; fs (max) = μsn x direction: ∑Fx = ma
Example n a fk ∑F = ma For EACH mass separately! x & y components plus fk = μkn a
Example 5.11 Place a block, mass m, on an inclined plane with static friction coefficient μs. Increase incline angle θ until the block just starts to slide. Calculate the critical angle θcat which the sliding starts. Solution:Newton’s 2nd Law (static) y direction: ∑Fy = 0 n – mg cosθ = 0; n = mg cosθ(1) x direction: ∑Fx = 0 mg sinθ – fs = 0; fs= mg sinθ(2) Also: fs = μsn(3) Put (1) into (3): fs = μsn mg cosθ(4) Equate (2) & (4) & solve for μsμs = tanθc