Question

A car manufacturer claims that you can drive their new vehicle across a hill with a 47 slope before the vehicle starts to tip. Part A If the vehicle is 2.0 wide, how high is its center of gravity?

Answer 1

To determine the height of the center of gravity of the car on a 47° slope, we can use the concept of torque. The perpendicular component of the weight can be calculated using trigonometry. Solving for the height of the center of gravity allows us to assess the car's stability on the slope.

Step-by-step explanation:

To determine the height of the center of gravity of the car, we can use the concept of torque. Torque is the product of the force applied and the distance from the pivot point. In this case, the force acting on the car is its weight, and the pivot point is the 47° slope of the hill.

Since the car is on a 47° slope, the weight of the car can be divided into two components - one that acts parallel to the slope and one that acts perpendicular to the slope. The perpendicular component of the weight is equal to the centripetal force that keeps the car from tipping over.

Assuming the car has a uniform distribution of weight, we can use trigonometry to determine the height of the center of gravity. The perpendicular component of the weight can be calculated using the formula mg cos(θ), where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the slope. Setting this equal to the centripetal force, which is equal to mg sin(θ), we can solve for the height of the center of gravity.

For example, if the car is 2.0m wide and has a mass of 1000kg, the height of its center of gravity can be calculated as follows:

Perpendicular component of weight = mg cos(θ) = 1000kg * 9.8m/s² * cos(47°)

Centripetal force = mg sin(θ) = 1000kg * 9.8m/s² * sin(47°)

Setting these two equal to each other, we can solve for the height of the center of gravity.

Answer 2

The height of the car's center of gravity is found by creating a trigonometric relation using the slope angle and the width of the car. With a slope of 47 degrees and a width of 2.0 meters, the height of the center of gravity is approximately 1.072 meters.

Step-by-step explanation:

The question involves calculating the height of the center of gravity of a car based on the maximum slope it can climb without tipping over. This can be determined by understanding the relationship between the slope angle, the width of the vehicle, and the height at which the center of gravity is located.

To find the height of the center of gravity, we can use the concept of static stability. A car will start to tip over when the line of action of the gravitational force passes through the edge of the base of support. With a slope angle (θ) of 47 degrees and the car being 2.0 meters wide, we can set up a right triangle where the width of the car is the base and the height of the center of gravity is the opposite side.

The height (h) can be calculated using the trigonometric relation:

tan(θ) = h / (width/2)

h = tan(θ) × (width/2)

h = tan(47 degrees) × (2.0m/2)

h = tan(47 degrees) × 1.0m

Now we just calculate the tan(47 degrees) and multiply it by 1.0m to find the height of the center of gravity. Using a calculator, we find:

tan(47 degrees) ≈ 1.072

h ≈ 1.072 × 1.0m = 1.072 meters

Therefore, the height of the center of gravity for this car is approximately 1.072 meters above the ground.