Question

Frictional Force \(\quad\) A 5500 -pound vehicle is driven at a speed of 30 miles per hour on a circular interchange of radius 100 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires?

Short answer

The frictional force that the road surface must exert on the tires of the vehicle to keep it from skidding off course is 33541.42 pounds.

Step-by-step solution

Convert speed to an equivalent measure

The speed is given in miles per hour, but we need it in feet per second for this particular problem because the radius is in feet. The conversion factors we'll use are 5280 feet = 1 mile and 3600 seconds = 1 hour. So, we have \(30 \, miles/hour = 30 \, miles/hour \times \frac{5280 \, feet}{1 \, mile} \times \frac{1 \, hour}{3600 \, seconds} = 44 \, feet/second.

Calculate the centripetal force

The net force (centripetal force) exerted on the car while making the turn is given by the formula \(F = \frac{m \cdot v^2}{r}\), where m is the mass of the car (in slugs), v is its speed (in feet per second), and r is the radius of the turn (in feet). However, the weight of the car is given instead of its mass. We can convert weight to mass using the conversion factor that 1 slug = 32.2 lbs. So, the mass of the vehicle is \( \frac{5500 \, lbs}{32.2 \, lb/slug} = 170.81 \, slugs. Then using the centripetal force formula, we have \(F_{c} = \frac{170.81 \, slugs \times (44 \, ft/sec)^2}{100 \, ft} = 33541.42 \, lb.\)

Determine the frictional force

In this case, the frictional force must be equal to the centripetal force required to keep the vehicle in circular motion, so \(F_{f} = F_{c} = 33541.42 \, lb.\)

Key concepts

Centripetal Force

Centripetal force is a crucial concept in physics, particularly when discussing the motion of objects in a circular path. It's not a new force in itself but rather refers to the net force required to keep an object moving in a circle.

This force always points towards the center of the circular path and is perpendicular to the object's velocity. For an object to maintain circular motion, a real force, like tension, gravity, or, as in our textbook exercise, friction, must provide the centripetal force. In the case of a vehicle turning on a circular path, friction between the tires and the road surface must supply the necessary centripetal force to prevent skidding.

The formula to calculate centripetal force is \( F = \frac{m \cdot v^2}{r} \), where \( m \) represents the object's mass, \( v \) its velocity, and \( r \) is the radius of the circular path. Extending this concept to real-world applications, such as ensuring a vehicle safely navigates a turn without losing grip, helps students grasp the tangible impact of physics in daily life.

Force Conversion

Force conversion is an operation crucial for solving physics problems, especially when different measurement systems are involved. In physics, we often use the slug as the unit of mass in the imperial system, and its weight counterpart is the pound (lb). The key to converting between them is understanding that mass and weight are related through gravitational acceleration.

Gravitational acceleration on Earth is approximately \( 32.2 \, ft/s^2 \), which helps us convert a weight in pounds to mass in slugs using the relationship \( 1 \, slug = 32.2 \, lb \). So, to find an object's mass based on its weight, divide the weight by \( 32.2 \, lb/slug \).

For example, a vehicle's weight, given in pounds, is converted to mass in slugs to calculate the centripetal force necessary for circular motion. This conversion is vital in the step-by-step solution when the vehicle's mass is required to compute the centripetal force, and it provides a hands-on application for an otherwise abstract concept.

Unit Conversion

Unit conversion is a fundamental skill in science and engineering which allows us to communicate measurements in different units effectively and is key to solving many physics problems. In the example provided, the speed of the vehicle is initially given in miles per hour (mph), but we need to convert it to feet per second (ft/s) to match the units of the radius given in feet.

The procedure involves multiplying the original speed by conversion factors that cancel the unwanted units and introduce the desired units. Here, we use the facts that there are \( 5280 \, feet \) in a mile and \( 3600 \) seconds in an hour. By using the appropriate conversion factors, \( 30 \, mph \) becomes \( 44 \, ft/s \).

Understanding unit conversion is critical for students, as it not only aids in solving mathematical problems but also ensures they can interpret and apply measurements correctly across various contexts.