A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) θ = 15o and (b) θ = 32o.
Wc = M*g.
Fn = Mg*Cos15.
a. Fn/Wc = Mg*Cos15/Mg = Cos15 = 0.966.
b. Fn/Wc = Mg*Cos32/Mg = Cos32 = 0.848.
Oh, you've come to the right place for some twisted humor! Let's tackle these uphill battles:
(a) When θ = 15°:
Well, it's no "uphill battle" for me. But for the car, it can be quite a "slope" to overcome! The ratio of the magnitude of the normal force to the weight of the car is like a celebrity couple's relation-"ship." We're talking a 15° angle "tilted" toward the horizontal. So, dear ratio, you may want to put on your seatbelt because it's about to get a little bumpy!
(b) When θ = 32°:
Ah, 32°, the angle where things start getting a bit "uphill" for our car. This time, the ratio of the magnitude of the normal force to the weight of the car is like a rollercoaster ride that just won't end. Picture the car clinging to the hill like a terrified passenger holding onto their popcorn for dear life! It's a bumpy, thrilling adventure where gravity pulls the car down, but the normal force pushes back, saying "Not so fast, my friend!"
Remember, these ratios depend on the angle of the hill; don't let them "drive" you crazy!
To determine the ratio of the magnitude of the normal force to the weight of the car, we need to consider the forces acting on the car on an inclined plane.
The weight of the car, W, acts vertically downward and can be calculated using the equation W = m * g, where m is the mass of the car and g is the acceleration due to gravity.
The normal force, N, acts perpendicular to the inclined plane. The angle θ is the angle between the inclined plane and the horizontal ground. The normal force can be calculated using the equation N = W * cos(θ).
Now, let's calculate the ratio for each given angle:
(a) θ = 15°:
The ratio of the magnitude of the normal force to the weight of the car is N/W.
N = W * cos(θ) = W * cos(15°)
So, the ratio is N/W = W * cos(15°) / W = cos(15°).
(b) θ = 32°:
The ratio of the magnitude of the normal force to the weight of the car is N/W.
N = W * cos(θ) = W * cos(32°)
So, the ratio is N/W = W * cos(32°) / W = cos(32°).
Therefore, in each case, the ratio of the magnitude of the normal force to the weight of the car will be equal to cos(θ), where θ is the given angle.
To determine the ratio of the magnitude of the normal force to the weight of the car, we first need to understand the forces acting on the car.
When a car is on an inclined plane, the weight of the car can be split into two components: one parallel to the plane (known as the component of weight along the incline) and one perpendicular to the plane (known as the normal force). The ratio we are looking for is the magnitude of the normal force divided by the weight of the car.
Let's start with part (a) where θ = 15o.
In this case, the weight of the car has a component along the incline and a component perpendicular to the incline. The weight component parallel to the incline is given by:
W_parallel = mg * sin(θ)
Where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The weight component perpendicular to the incline (which is the magnitude of the normal force) can be found using:
W_perpendicular = mg * cos(θ)
To find the ratio of the magnitude of the normal force to the weight of the car, we can divide W_perpendicular by the weight of the car:
Ratio = W_perpendicular / W = (mg * cosθ) / (mg) = cosθ
Therefore, the ratio is simply equal to cos(θ). In this case, when θ = 15o, the ratio is equal to cos(15o).
For part (b) where θ = 32o, we can follow the same procedure. The ratio will be equal to cos(θ), so when θ = 32o, the ratio is equal to cos(32o).
So, to summarize:
(a) Ratio = cos(15o)
(b) Ratio = cos(32o)
Remember to convert the angles to radians if required in your calculations.
I hope this explanation helps you understand how to find the ratio of the magnitude of the normal force to the weight of the car in each case. Let me know if you have any further questions!
A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) θ = 16o and (b) θ = 35o.
(a) Ratio = Number
0.96
Units?
(b) Ratio = Number
0.82
Units ?