Problem 117 - University Physics with Modern Physics

Question

You are driving a classic 1954 Nash Ambassador with a friend who is sitting to your right on the passenger side of the front seat. The Ambassador hat bench seats. You would like to be closer to your friend and decide to use physics to achieve your romantic goal by making a quick turn. (a) Which way (to the left or to the right) should you turn the car to get your friend to slide closer to you? (b) If the coefficient of static friction between your friend and the car seat is \(0.35,\) and you keep driving at a constant speed of \(20 \mathrm{m} / \mathrm{s},\) what is the maximum radius you could make your turn and still have your friend slide your way?

Step-by-Step Solution

Answer

(a) Turn to the right; (b) Maximum radius is approximately 116.61 m.

Step 1: Understanding the Problem

When the car turns, it creates a centripetal force acting towards the center of the turn, which can cause the passenger to slide if friction is not enough to hold them in place. To have your friend slide towards you, turning right is needed. As the car turns right, the lack of friction will let them slide to the left (towards you).

Step 2: Calculate Maximum Frictional Force

The static frictional force is the product of the coefficient of static friction and the normal force. The normal force here is equal to the weight of your friend, which can be denoted as \( F_f = \mu_s \times m \times g \). Here \( \mu_s = 0.35 \) is the coefficient of static friction, and \( g = 9.8 \, m/s^2 \) is the acceleration due to gravity.

Step 3: Relation of Forces for Sliding

While turning, the required centripetal force \( F_c \) to keep your friend moving in a circle is \( F_c = \frac{m \times v^2}{r} \), where \( v = 20 \, m/s \) and \( r \) is the radius of the turn. For sliding to occur, \( F_c \) must exceed \( F_f \).

Step 4: Equating Forces to Find Maximum Radius

To just overcome static friction, set \( F_f = F_c \). Hence, \( \mu_s \times m \times g = \frac{m \times v^2}{r} \). Simplifying and solving for \( r \) gives \( r = \frac{v^2}{\mu_s \times g} = \frac{(20)^2}{0.35 \times 9.8} \).

Step 5: Solving the Equation

Compute the maximum radius \( r \): \( r = \frac{400}{0.35 \times 9.8} = \frac{400}{3.43} \approx 116.61 \mathrm{m} \).

Key Concepts

Static Friction

Static friction plays a crucial role in determining whether an object remains stationary or begins to slide. It acts between surfaces in contact and prevents sliding. In the context of the exercise, static friction is what keeps your friend from sliding across the car seat.
Static friction depends on two main factors:

• The normal force, which is the force perpendicular to the surfaces in contact. In this case, your friend’s weight acts as the normal force because Earth’s gravity pulls them down onto the seat.
• The coefficient of static friction (\( \mu_s \)), a dimensionless value that represents the friction level between two materials. Here, \( \mu_s = 0.35 \). A larger coefficient indicates more friction.

When you turn the car, static friction tries to hold your friend in place. If the necessary centripetal force to keep the circular motion exceeds this frictional force, the static friction will fail, causing your friend to slide. This scenario explains why turning a car could make someone slide along a smooth bench seat.

Centripetal Acceleration

Centripetal acceleration is the acceleration that acts on an object moving in a circular path, directed toward the center of the circle. In a turning car, any object inside experiences this acceleration. This is why your friend feels pushed towards the outside of the curve.
The formula for centripetal acceleration (\( a_c \)) is:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the velocity and \( r \) is the radius of the circular path. The faster you go and the tighter the curve, the higher the centripetal acceleration.
In this exercise, you're trying to calculate the tightest curve (smallest radius) your car can take without the static friction force keeping your friend in place. If the centripetal acceleration becomes too large, your friend will slide in the opposite direction of the turn. This acceleration challenges static friction's limit.

Newton's Second Law

Newton's Second Law is fundamental to understanding motion, especially when dealing with forces like static friction and centripetal forces. This law states that the force on an object is equal to the mass of the object multiplied by its acceleration.
Mathematically, it's expressed as:\[ F = m \times a \]where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. In the exercise, different forces act on your friend as the car turns.
To make your friend slide towards you, the centripetal force required to move them in a circular path must be calculated using Newton’s Second Law. The centripetal force (\( F_c \)) required can be expressed as:
\[ F_c = \frac{m \times v^2}{r} \]
Setting this equal to the maximum static friction (as previously calculated with static force) shows the conditions needed for sliding to initiate. By equating these expressions, you find the radius (\( r \)) representing the strongest turn to slide your friend. Understanding how these principles intersect helps clarify the exercise’s details and your strategies while driving.