A car manufacturer claims that you can drive its new vehicle across a hill with a 47 degree slope before the vehicle starts to tip. If the vehicle is 2.0 meter wide, how high is its center of gravity?
To determine the height of the center of gravity, we can use trigonometry. The angle of 47 degrees represents the angle between the ground and the slope.
Let's assume that the height of the center of gravity is represented by "h" (in meters).
To find the height, we can use the tangent function:
tan(47 degrees) = h / (2.0 meters)
Rearranging the equation to solve for h:
h = tan(47 degrees) * 2.0 meters
Using a scientific calculator, the tangent of 47 degrees is approximately 1.0723.
Substituting the values:
h ≈ 1.0723 * 2.0 meters
h ≈ 2.1446 meters
Therefore, the center of gravity of the vehicle is approximately 2.1446 meters high.
To determine the height of the center of gravity of the vehicle, we can use trigonometry.
First, let's visualize the problem. Imagine a right triangle, where the width of the vehicle forms the base of the triangle, the height we want to find is the opposite side, and the slope of the hill is the angle of the triangle, which is 47 degrees.
Now, we can use the tangent function to relate the angle and the height of the triangle:
tan(angle) = opposite/adjacent
In this case, the angle is 47 degrees, and the adjacent side is half the width of the vehicle (since the center of gravity will be in the middle):
tan(47) = height/(2/2)
tan(47) = height/1
Now we can solve for the height:
height = tan(47)
Using a calculator, we find that the value of tan(47) is approximately 1.0724.
Therefore, the height of the center of gravity of the new vehicle is approximately 1.0724 meters.
The whole deal is drawing the picture.
the wheel is 1 meter down slope from the center line of the vehicle
g is x meters from the road normal to road surface along that center line
so
tan 47 = x/1
x = tan 47 = 1.07 meters
but do not put passengers on the roof. (I am a ship designer :)