Asked by meder
Matt Ashell, world famous downhill skier, waxed his skis with the wrong kind of wax and ended up with a relatively high coefficient of friction of 0.40. If Matt, who weights 667.5 N is on a downhill slope of 38 degrees, will he slide without a push?
Answers
Answered by
Henry
mg = 667.5 N.
F1 = mg*sinA = 667.5*sin38 = 411 N = Force parallel to the hill.
F2 = mg*sinA= 667.5*cos38 = 526 N = Force perpendicular to the hill.
Fs = u*F2 = 0.4 * 526 = 210.4 N = Force of static friction.
Yes, because F1 is greater than the forceof s tatic friction, Fs.
F1 = mg*sinA = 667.5*sin38 = 411 N = Force parallel to the hill.
F2 = mg*sinA= 667.5*cos38 = 526 N = Force perpendicular to the hill.
Fs = u*F2 = 0.4 * 526 = 210.4 N = Force of static friction.
Yes, because F1 is greater than the forceof s tatic friction, Fs.
Answered by
Henry
Correction: F2 = mg*cosA.
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