An SUV has a height h and a wheelbase of length b. Its center of mass is midway between the wheels and at a distance αh above the ground, where
0 < α < 1.
The SUV enters a turn at a dangerously high speed, v. The radius of the turn is
R (R >> b),
and the road is flat. The coefficient of static friction between the road and the properly inflated tires is
μs.
After entering the turn, the SUV will either skid out of the turn or begin to tip.
(a) The SUV will skid out of the turn if the friction force reaches its maximum value,
F → μsN.
Determine the speed,
vskid,
for which this will occur. Assume no tipping occurs. (Use any variable or symbol stated above along with the following as necessary: g for acceleration due to gravity.)
vskid =
(b) The torque keeping the SUV from tipping acts on the outside wheel. The highest value this force can have is equal to the entire normal force. Determine the speed,
vtip,
at which this will occur. Assume no skidding occurs. (Use any variable or symbol stated above along with the following as necessary: g for acceleration due to gravity.)
vtip =
(c) It is safer if the SUV skids out before it tips. This will occur as long as
vskid < vtip.
Apply this condition, and determine the maximum value for α in terms of
b, h and μs.
(Use any variable or symbol stated above along with the following as necessary: g for acceleration due to gravity.)
α <
To determine the speed at which the SUV will skid out of the turn, we need to consider the forces acting on the SUV and set the maximum friction force equal to the product of the coefficient of static friction and the normal force.
(a) The maximum friction force can be expressed as: F_max = μs * N
The normal force can be found by considering the weight of the SUV. Since the center of mass is midway between the wheels, the weight of the SUV is evenly distributed between the two wheels. The weight (force due to gravity) can be calculated as: N = (h/2 + αh) * mg
The maximum friction force is equal to the centripetal force needed to keep the SUV in the turn. Considering that the centripetal force is given by the equation F_c = mv^2/R, where m is the mass of the SUV, v is the speed, and R is the radius of the turn, we can equate the two expressions:
μs * N = mv^2/R
Substituting the expression for N in terms of h, α, and m, we get:
μs * (h/2 + αh) * mg = mv^2/R
Simplifying and solving for v, we find:
v_skid = sqrt(2 * μs * g * (h/2 + αh) * R)
(b) To determine the speed at which tipping occurs, we need to consider the torque acting on the SUV that tends to tip it over. The torque is given by the force applied at the outer wheel multiplied by the distance between the center of mass and the outer wheel. The maximum force that can be applied is equal to the entire normal force (N).
The torque can be expressed as: T = N * b/2
The torque should be equal to the torque provided by the weight of the SUV, which can be calculated as: T_g = (h/2) * mg
Equating the two expressions and solving for v, we get:
N * b/2 = (h/2) * mg
Substituting the expression for N in terms of h, α, and m, we have:
[(h/2 + αh) * mg] * b/2 = (h/2) * mg
Simplifying and solving for v, we find:
v_tip = b * g / (2(1 + α))
(c) To determine the maximum value for α, we need to compare the expressions for v_skid and v_tip, and establish the condition v_skid < v_tip.
Comparing the expressions derived in parts (a) and (b), we have:
sqrt(2 * μs * g * (h/2 + αh) * R) < b * g / (2(1 + α))
Squaring both sides and simplifying, we get:
2 * μs * (h/2 + αh) * R < b^2 / (4 * (1 + α)^2)
Simplifying further, we find:
μs * (h/2 + αh) * R < b^2 / (8 * (1 + α)^2)
Solving for α, we obtain:
α < (b^2 / (8 * R * μs) - h/2) / h
Therefore, the maximum value for α is given by:
α < [(b^2 / (8 * R * μs) - h/2) / h]