Problem 6/3  Page 428, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Bicycle moving down an incline.

Find: Deceleration that would cause the bicycle to tip about the front wheel.

Parameter Values:
q = 10 degrees = Incline angle.
M = Mass of bicycle plus bicylist (unknown)
B = 40" = Distance from front wheel contact point to rear wheel contact point.
H = 36" = Perpendicular height of mass center of bike plus bicyclist above inclined plane.
L = 25" = Distance along incline from front wheel contact point to mass center.

1. Mechanical System: Bike plus bicyclist as the brakes are applied.

2. Free Body Diagram: Shows bike and bicylist.  Forces include weight acting vertically downward.  The rear wheel is about to come off the ground so no force there.  Normal force perpendicular to incline at the front wheel.  Friction force parallel to incline and up the incline at the front wheel.  As the motion is down the incline, let the t axis point down the incline.  The n axis is perpendicular to the incline.

3. Equations
a) S F = ( M g sin(q) - F ) t + ( N - M g cos(q) ) n
S M_G = (-H n + L t ) X ( -F t + N n )
         = ( L N - H F ) k

b) m a_G = M ( a_t t + a_n n )
              = M a_t t 
Note that as the bike is traveling in a straight line (no curvature), then the normal acceleration must be zero.
IG a = 0
Note that bike is not yet tipping, so it is not rotating.

4. Solve
M g sin(q) - F = M a_t
N - M g cos(q) = 0
L N - H F = 0

From the second equation:
N = M g cos(q)
From the third equation:
F = (L/H) N = (L/H) M g cos(q)
From the first equation:
a_t = (1/M) ( M g sin(q) - F )
    = g { sin(q) - (L/H) M g cos(q) }
  = -0.510 g = -16.43 ft/s²