| eHttps://loncapal.fsu.edufe Syllabus-CollegePhysicsA : |PHYSICS 2053C Syllabus
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| eHttps://loncapal.fsu.edufe Syllabus-CollegePhysicsA : |PHYSICS 2053C Syllabus Fal collegePhysics-OP.pdf × + d3bxy9euwde 147.cloudfront iald ollegePhysics-OP pdf Find on page table 5.1 × 1of4 Options Magnitute of Kinetic Frkction The magnitude of kineticrictonf is given by (5.5) where k is the wellicient of kirvelk: friction. As seen in thecoet icients of kinetic friction are less than their static counterparts. That values of Table 5.1 are stated to only one or, at most, two digits is an indication of the approximate description of triction gven by the above two Table 5.1 Coetticients of Static and Kinetic Friction Static friction uKinetic friction Rubber on dry concrete Rubber on wet concrete Wool on wood Waxed wood on wet snoW Metal on wood Steel on steel (dry) Steel on steel (olled) Tellon on steel Bone lubricated by synovial fluid Shoes on wood Shoes on ice Ice on ice Steel on ice 1.0 0.5 0.14 05 0.6 0.05 0.04 0.7 0,5 0.3 0.1 0.3 0.3 0.9 D.1 0.1 0.4 0.7 0,05 0.03 176 Chapor 5 l Further Appictores of Newars Low. Fiction Dr and Easoary The equations given carlier include the dependence of friction on materials and the nomal force, The directon of friction is alsays opposite that o notion, parallel to the surface between objects, and perpendicular lo the normal force. For example, i the crate you try to push with a torce paralel to the floor has a mass of 100 kg then the normal force would be equal 10 weight, W(100 kg)9.80 ms2980N, perpendicutar to the tlor. f the coefficlent of static triction is 0.45, you would have to exert a lorce parallel to the floor greater than ,N-(0.45)(980 N) 440 N to move the date 4:05 PM Type here to search 9/30/2017Explanation / Answer
A. Given, force to be exerted = F = 150 N
So , Normal reaction, N
but F = mu*N
so, mu = 0.3 ( coefficient of kinetic friction, metyal vs metal, dry)
N = F/mu = 150/0.3 = 500 N
B. If the part was oiled, the normal; reaction would still remain the same
F = mu*N
mu = 0.03 ( for oiled parts)
so, F = mu*(N) = 0.03*500 = 15 N
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