Bicycle Stopping Reactions: (Individual) See attached Bicycle drawings. The center of gravity and weight is shown for the rider and the bicycle itself, as well as relevant dimensions. Be sure to find reactions when NOT braking first and then reactions when braking. Assuming the bicycles are moving at a constant velocity in a straight line, find the reactions at the front and rear wheels. Assume that the rider weighs 160lb in each case: The "ordinary" weighs 40lb The "safety" weighs 21lb The "recumbent" weighs 36lb Use M = 0; take moments about where the front wheel touches the roadway and then about where the rear wheel touches the roadway. Then use Fy to check your answers. OVERWRITE THE ATTACHED AUTOCAD TO MAKE YOUR FBD FOR EACH CASE. Note that the first one already is an FBD If each bike is travelling at 18mph and needs to stop in 40ft, find the deceleration. Use equations from the previous assignment. Yes, the value is the same in each case, so only calculate once! Once something is being accelerated/decelerated, there is a force involved: E.g. F=ma, where m = W/g and g = 32.2 ft/s (acceleration due to gravity) and 'a' is acceleration calculated above. = If a 10ft/sec (it doesn't!) for the rider: FR = (160/32.2)(10) = 49.69lb and for the bike: FB = (40/32.2)(10) = 12.42lb Then: - -R(39.75) + 160(10.00) + 40(11.50) - 49.69(60.50) 12.42(31.00) = 0 - +160(10.00) + 40(11.50) 49.69(60.50) - 12.42(31.00) R = = -33.49lb 39.75 EMR F(39.75) - 160(29.75) - 40(28.25) - 49.69 (60.50) - 12.42(31.00) = 0 +160(29.75) + 40(28.25) + 49.69(60.50) + 12.42(31.00) = 233.49lb F 39.75 ... and check: - - -33.49 40 160 + 233.49=0 What does it mean when a reaction is negative? (R = -33.49lb) MATC 31.25 1/20.00 17.00 RECUMBENT BICYCLE 36.25 38.00 52.50 $16.30 DWG.No. 125-04-003 A Statics Find Reactions Now solve for reactions at the wheels for the "Safety" and the "Recumbent" using the deceleration that you calculated for the 40' stopping distance. State if anyone is going over the handlebars.