Problem 76 - Precalculus with Limits

Question

BRAKING LOAD A sport utility vehicle with a gross weight of 5400 pounds is parked on a slope of \(10^{\circ}\). Assume that the only force to overcome is the force of gravity. Find the force required to keep the vehicle from rolling down the hill. Find the force perpendicular to the hill.

Step-by-Step Solution

Answer

The force needed to keep the vehicle from rolling down the slope (F_{parallel}) is calculated to be 5400 * sin(10^{\circ}) pounds, and the force acting perpendicular to the slope (F_{perpendicular}) is 5400 * cos(10^{\circ}) pounds.

Step 1: Calculate the force of gravity

The force of gravity acting on the car can be calculated using the equation: F_{gravity} = m * g, where m is mass and g is gravity. The given weight of the car is actually its mass times gravity (F_{gravity} = m*g), so we can use that as the gravitational force directly. So, F_{gravity} = 5400 pounds.

Step 2: Calculate the force to keep the vehicle from rolling down

The force needed to prevent the car from rolling down the slope equals the component of the gravitational force acting along the slope. It can be calculated using the formula: F_{parallel} = F_{gravity} * sin(\theta), where \theta is the inclination angle of the slope, here \(10^{\circ}\). Hence, F_{parallel} = 5400 * sin(10^{\circ}).

Step 3: Calculate the force perpendicular to the hill

The force acting perpendicular to the slope, i.e., the normal force is the component of the gravitational force acting perpendicular to the slope. It can be calculated with the formula: F_{perpendicular} = F_{gravity} * cos(\theta), where \theta is the inclination angle. Hence, F_{perpendicular} = 5400 * cos(10^{\circ}).

Key Concepts

Force of Gravity

Everything that has mass is attracted to the earth by the force of gravity. This is a fundamental concept in physics. The force of gravity is essentially the pull that the Earth exerts on objects. It is determined by the equation \( F_{gravity} = m \times g \). Here, \(m\) is the mass of the object, and \(g\) is the acceleration due to gravity. On Earth, \(g\) is approximately \(32.2\) feet per second squared.
However, when calculating the gravitational force, sometimes it is more practical to use the weight of the object directly as it already incorporates both mass and gravitational pull. In our exercise, the car's weight is given as \(5400\) pounds, which represents the gravitational force acting on the car.
When dealing with inclined planes, it’s crucial to understand how gravity is split into components parallel and perpendicular to the plane. We'll explore more on this within the context of an inclined plane.

Inclined Plane

An inclined plane is simply a flat surface tilted at an angle to the horizontal. It helps us to explore how forces act on objects on sloped surfaces. When an object is placed on an inclined plane, the force of gravity pulls it down, bending its path due to the angle of the slope.
The gravitational force acting on the object can be resolved into two components:

• The component parallel to the slope, which tries to pull the object down the slope. (This is often why cars roll downhill.)
• The component perpendicular to the slope, which presses the object against the slope but doesn't move it along the slope.

Understanding these components helps in determining the forces needed to keep an object stationary on a slope, or what additional support might be required to prevent it from sliding.
Inclined planes reduce the effort needed to raise objects by extending the distance over which the force is applied.

Trigonometric Components

Trigonometry provides us with tools to analyze forces on an inclined plane, breaking them into measurable parts. Every force on an incline can be split into components that are parallel and perpendicular to the plane, thanks to trigonometry.
The trigonometric functions \(\sin\) and \(\cos\) help in this process. Consider an angle \(\theta\) in the inclined plane. The gravitational force is split into:

• \( F_{parallel} = F_{gravity} \times \sin(\theta) \), which is the component along the slope (the slope's steepness determines how much gravitational force pulls the object downward).
• \( F_{perpendicular} = F_{gravity} \times \cos(\theta) \), which is the component normal to the slope (this supports the object's weight against the inclined surface).

This breaking down of force into components allows for easier calculations and helps identify the exact amount of force required to counteract gravity's pull in each direction.
Understanding trigonometric components is essential in solving physics problems related to inclined planes, specifically when evaluating the stability of objects on slopes.