Question

A 177-Ib snowboarder has speed v = 13 ft/sec when in the position shown on the halfpipe. Determine the normal force on his snowboard and the magnitude of his total acceleration at the instant depicted. Use a value MIx = 0.08 for the coefficient of kinetic friction between the snowboard and the surface. Neglect the weight of the snowboard and assume that the mass center of the snowboarder is 2.6 feet from the surface of the snow. 310 G 25 25 Answers N 218.472 Ib 25.56 ft/sec

Answer 1

weight, W of Snowboardex = 17726 Speed v of snowboarder = 13 ft/sec Mk (kinematic ) = D.08 masscenter Gt of snowboardex is 2-6 ft from surface of snow let į bc center of cwivatu xe 0=31 DON then AC = padius = 25 - 25 Ft A & GA= 2.6ft GC = 25-2-6 = 22.4 ft с Forces acting Fc 31 ४ w Die W weight from ci downwards (Fc) centrifugal force = my away from centre radially outward a friction force, F = MAN (tangentially opposite to motion A at A) Normal reaction N, vadially inward at A Nolan E = mv m = mass of snowboardex va tangential, velocity 8= distance toom center c' to Gil 8=CG= 22.4ft Resolving all forces along tangential & Nos mal directions Along Normal discction (AGC line) 61 E F = 0 = N-Wsino - F = 0 M N = Wsino + F wsinet Fc WCOSO W N m=w 9 = 32-19 ft/5² 1 А ܟܘ m = 111 32:19 = 5.4986 [slug] I mass of Snowboardes

v=3 ttls N- W sino + mv o 8= 22.6ft 2 = 117-5in (31') + 5-4936 x 13 22-6 >> N = 132.2795 lb ( unit is force) N = 132.2795 lb Along tangential direction: Lex Ce are unit vectors EFE = WcOSO - F = 177 cos(31) - MKN 177 C05 (31) –0-08) (132*2795) EFT = 14:1362 = = m-at 141-1362 5.4986 & Fr = at 25.6676 ft/gt m At this instant, tangential acceleration of Snowboardes is az = 25.6676 ft/52 13 o 22.4 Also, Snowboarded experience normal acceleration an = an = 7.5 4 4 6 ft 13 Magnitude of total acceleration, a = Jaz tan = |(25.6616) + (7-5446² = 26.75 3 4 ft /s² a = 26.7534 f1/5?
I hope you understand. Comment if you have any doubt.