Answered step by step
Verified Expert Solution
Question
1 Approved Answer
Bicycle Stopping Reactions: (Individual) See attached Bicycle drawings. The center of gravity and weight is shown for the rider and the bicycle itself, as
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 MR Then: 0 : R = = -R(39.75) + 160(10.00) + 40(11.50) - 49.69(60.50) - 12.42(31.00) = +160(10.00) + 40(11.50) - 49.69(60.50) - 12.42(31.00) 39.75 = -33.49lb 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) :. F = ... and check: , 39.75 = -33.4940-160 + 233.49 = 0 - What does it mean when a reaction is negative? (R = -33.49lb) = 233.49lb 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. MATC #17.00 160 lb 40 lb R 39.75 -11.50- F -10.00 160 32.2 40 32.2 a -#54.00 31.00 60.50 DRAWN CATE ORDINARY BICYCLE CALE Jimmy 1:16 DWG.No. 125-04-001 A
Step by Step Solution
There are 3 Steps involved in it
Step: 1
Get Instant Access to Expert-Tailored Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started