Question

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 \(\alpha h\) above the ground, where \(0<\alpha<1\). The SUV enters a turn at a dangerously high speed, \(v\). The radius of the turn is \(R(R \gg b)\), and the road is flat. The coefficient of static friction between the road and the properly inflated tires is \(\mu_{s}\). 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, \(F \rightarrow \mu_{\mathrm{s}} N\). Determine the speed, \(v_{\text {skid }},\) for which this will occur. Assume no tipping occurs. 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, \(v_{\text {tip }}\), at which this will occur. Assume no skidding occurs. c) It is safer if the SUV skids out before it tips. This will occur as long as \(v_{\text {skid }}

Short answer

Answer: The maximum value for α is α < (μs * b) / (2 * h).

Step-by-step solution

Understand the given information

The SUV has a height \(h\) and a wheelbase of length \(b\). The center of mass is at \(\alpha h\) above the ground. The coefficient of friction between the tires and the road is \(\mu_s\). The car enters a turn with a radius \(R\) and a speed \(v\). It is given that the turn radius \(R\) is much larger than the wheelbase \(b\). We have to determine the speeds \(v_{\text{skid}}\) and \(v_{\text{tip}}\) at which the SUV will skid out of the turn or begin to tip.

Determine the maximum friction force, \(F \rightarrow \mu_{\mathrm{s}} N\)

According to the equilibrium conditions, the net force on the SUV must be zero, which implies that the normal force \(N\) must be equal to the weight of the SUV, \(mg\). Thus, the maximum friction force \(F\) can be written as: \[F_{\text{max}} = \mu_{s}N = \mu_{s}mg\]

Calculate the skidding velocity, \(v_{\text{skid}}\)

The skidding of the SUV will happen due to the centrifugal force, which acts opposite to the friction force. Thus, we can express the centrifugal force \(F_{c}\) using the formula \(F_{c} = m \times a_{c}\), where \(a_{c}\) is the centripetal acceleration. The centripetal acceleration can be expressed in terms of the velocity and turn radius as \(a_{c} = \frac{v^2}{R}\). Thus, we can write: \(F_c = m\frac{v_{\text{skid}}^2}{R}\) When the centrifugal force equals the maximum friction force, the SUV will skid out of the turn: \(m\frac{v_{\text{skid}}^2}{R} = \mu_s mg\) Therefore, the skidding velocity can be found by rearranging this equation: \(v_{\text{skid}} = \sqrt{\mu_s gR}\)

Calculate the tipping velocity, \(v_{\text{tip}}\)

We will now tackle the problem of calculating the tipping velocity. For that, we should first analyze the torque being produced. The tipping will happen when the gravitational force acting on the center of mass produces a torque greater than the torque applied by the horizontal force acting on the outside tire. As the torque due to the horizontal force (\(F_H\)) acts upon the outside tire, we can write the torque equilibrium equation as: \(F_H \times b = m \times g \times (\alpha h)\) The horizontal force \(F_H\) is related to the centrifugal force \(F_c\) as: \(F_H = \frac{1}{2}F_c\) And as before, we have \(F_c = m\frac{v_{\text{tip}}^2}{R}\). So, we can rewrite the torque equilibrium equation as: \(\frac{1}{2}m\frac{v_{\text{tip}}^2}{R} \times b = m \times g \times (\alpha h)\) Now, we solve this equation for \(v_{\text{tip}}\): \(v_{\text{tip}} = \sqrt{2Rg\alpha\frac{h}{b}}\)

Determine maximum \(\alpha\) in terms of \(b, h\), and \(\mu_s\)

For the SUV to be safer, it should skid before it tips, which means that \(v_{\text{skid}} < v_{\text{tip}}\). We can rewrite this inequality using the expressions obtained in Steps 3 and 4: \(\sqrt{\mu_s gR} < \sqrt{2Rg\alpha\frac{h}{b}}\) Square both sides and divide by \(gR\) to find an expression for \(\alpha\): \(\alpha < \frac{\mu_sb}{2h}\) Hence, the maximum value for \(\alpha\) is \(\frac{\mu_sb}{2h}\), given the parameters \(b\), \(h\), and \(\mu_s\).

Key concepts

Static Friction

When a vehicle navigates a turn, static friction plays a critical role in keeping it from sliding. Static friction is the force that opposes the initial motion between two surfaces that are in contact and at rest relative to each other. In the context of vehicle dynamics, it's the force between the tires and the road that allows the vehicle to turn without skidding sideways.

It's important to note that static friction is self-adjusting up to its maximum value, which depends on the coefficient of static friction, \(\mu_s\), and the normal force, usually the weight of the vehicle. For the SUV in our exercise, the maximum static frictional force is given as \(\mu_s mg\). This force is a crucial factor in determining the maximum velocity at which the SUV can turn without skidding, denoted \(v_{\text{skid}}\).

Centripetal Acceleration

To turn, a vehicle must experience centripetal acceleration, which is always directed toward the center of the turn's radius. This acceleration is critical because it changes the direction of the vehicle's velocity without altering its speed. The formula for calculating centripetal acceleration, \(a_c\), is \(a_c = \frac{v^2}{R}\), where \(v\) is the vehicle's velocity, and \(R\) is the radius of the turn.

The need for centripetal acceleration introduces a centripetal force, acting towards the center of the turn's radius. In our SUV's case, this force is provided by static friction up to a certain speed. If the required centripetal force to maintain the turn exceeds the maximum static frictional force, the SUV will skid out of the turn.

Torque Equilibrium

Torque equilibrium is a state where the total torque acting on an object is zero, leading to no rotational acceleration. Vehicles, like the SUV in our exercise, are subjected to various forces that can create torque around their center of mass. When turning, the centripetal force creates a torque that can potentially tip the vehicle over if it is not balanced by other forces and torques.

In the scenario where we analyze the SUV for tipping, we focus on the torque around its outer wheels, which acts as a pivot point. The tipping will occur if this torque exceeds the torque provided by the static friction working at the contact point of the tires with the road. The force opposing the tipping torque is the horizontal component of the static friction, which, through torque equilibrium, allows us to calculate the maximum speed for tipping, \(v_{\text{tip}}\).

Maximum Velocity for Skidding

The maximum velocity for skidding, denoted as \(v_{\text{skid}}\), is the highest speed a vehicle can maintain in a turn without losing traction. This velocity is derived from the balance between the centripetal force needed to keep the vehicle in the turn and the maximum static friction force that can be exerted by the tires on the road. Beyond this velocity, the tires can no longer provide the necessary centripetal force, leading to skidding.

In the given SUV problem, we used the formula \(m\frac{v_{\text{skid}}^2}{R} = \mu_s mg\) to calculate \(v_{\text{skid}}\) by setting the centripetal force equal to the maximum static frictional force. This gives us a valuable measure of a vehicle's handling limit under certain conditions.

Maximum Velocity for Tipping

On the other hand, the maximum velocity for tipping, denoted as \(v_{\text{tip}}\), is the critical speed at which the centripetal force causes the vehicle's center of mass to generate enough torque around the outer wheels (which act like a fulcrum) that could lead to tipping over. This speed is dictated by the height of the center of mass and the wheelbase of the vehicle, among other factors.

In our exercise, we determined \(v_{\text{tip}}\) by considering the torque caused by the gravitational force on the SUV's center of mass and equating it to the torque caused by the horizontal component of the centripetal force. This condition gave us the formula \(v_{\text{tip}} = \sqrt{2Rg\alpha\frac{h}{b}}\), allowing us to understand at what speed tipping would occur.

Safety Conditions in Vehicle Turning

The safety conditions in vehicle turning involve understanding and achieving the right balance between the risks of skidding out and tipping over. Ideally, a vehicle should be designed or operated in such a way that it will skid before it reaches a speed high enough to tip. This is generally a safer failure mode as it typically results in less-severe consequences.

In our SUV scenario, the safety condition is expressed as \(v_{\text{skid}} < v_{\text{tip}}\). By following this guideline, we calculated the maximum permissible value of \(\alpha\), representing how high the center of mass can be relative to the wheelbase and height of the vehicle, to ensure that the SUV will skid before it has the chance to tip. This formula, \(\alpha < \frac{\mu_sb}{2h}\), is vital for vehicle designers and can inform changes to the vehicle's design to improve its turning safety.