If a curve with radius of 60 m is properly banked for a car travelling 60 km/h, what must be the coefficient of static friction for a car not to skid when travelling at 90
km/h?
Here is a similar problem. Work thru this one to understand. https://www.youtube.com/watch?v=A4om4K0uyII
Well, well, well, if it isn't the need for speed! Let's crunch some numbers and put on our thinking caps, shall we?
We have a curve with a radius of 60 m and a car zooming along at 60 km/h. To keep this car from skidding, we need to find the coefficient of static friction when the car is traveling at a whopping 90 km/h.
Now, to keep it simple, let's assume that the car perfectly follows the curve and doesn't tip over like a sad clown on a tightrope. In this scenario, the static friction provides the necessary centripetal force that keeps the car on the road.
We know that the centripetal force is given by the formula:
F = mv² / r,
where F is the centripetal force, m is the mass of the car, v is the velocity, and r is the radius of the curve.
Since we're dealing with a curve and not a straight line, we need to consider the force of gravity as well. The normal force (N) and the force of gravity (mg) work together to provide the necessary friction to prevent skidding.
Now, let's compare the two scenarios:
Scenario 1: Car traveling at 60 km/h:
- F₁ = mv₁² / r
- F₁ = (m * (60 km/h)²) / 60 m
Scenario 2: Car traveling at 90 km/h:
- F₂ = mv₂² / r
- F₂ = (m * (90 km/h)²) / 60 m
To prevent skidding, we need the maximum static friction force (Ff) to be equal to the centripetal force (F) in both scenarios. Let's put on our clown noses, do some math, and find out what the coefficient of static friction (μ) needs to be:
Ff₁ = μN₁ = F₁
Ff₂ = μN₂ = F₂
Since the normal force (N) is the same in both scenarios, we can cancel it out:
μ = F₁ / N₁ = (m * (60 km/h)²) / (60 m * g)
μ = F₂ / N₂ = (m * (90 km/h)²) / (60 m * g)
So, if you plug in the values of the masses of the car and the acceleration due to gravity (g), you'll be able to find the coefficient of static friction for the car not to skid!
But remember, my dear friend, this is just a theoretical calculation. In the real world, variables like road conditions, tire quality, and the driver's skills can affect the outcome. So, stay safe, keep your clown car grounded, and remember: skidding is for amateurs!
To determine the coefficient of static friction, we can use the equation for the banking angle of a curved road:
\( \theta = \tan^{-1} \left( \frac{{v^2}}{{g \cdot r}} \right) \)
Where:
- \( \theta \) is the angle of banking
- \( v \) is the velocity of the car
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\))
- \( r \) is the radius of the curve
First, let's convert the given velocities from km/h to m/s:
- \( v_1 = 60 \, \text{km/h} \times \frac{{1000 \, \text{m}}}{{1 \, \text{km}}} \times \frac{{1 \, \text{h}}}{{3600 \, \text{s}}} = 16.67 \, \text{m/s} \)
- \( v_2 = 90 \, \text{km/h} \times \frac{{1000 \, \text{m}}}{{1 \, \text{km}}} \times \frac{{1 \, \text{h}}}{{3600 \, \text{s}}} = 25 \, \text{m/s} \)
Now we can plug the values into the equation to find the banking angle for a velocity of 16.67 m/s:
\( \theta = \tan^{-1} \left( \frac{{(16.67 \, \text{m/s})^2}}{{(9.8 \, \text{m/s}^2) \cdot (60 \, \text{m})}} \right) \)
Using a calculator, we get:
\( \theta \approx 16.6^\circ \)
Now we can calculate the coefficient of static friction for a velocity of 25 m/s:
\(\mu_s = \tan( \theta ) \)
Let's substitute the value of \(\theta\) into the equation:
\(\mu_s = \tan( 16.6^\circ ) \)
Using a calculator, we get:
\(\mu_s \approx 0.293\)
Therefore, the coefficient of static friction must be approximately 0.293 for the car not to skid when traveling at 90 km/h.
To determine the coefficient of static friction required for a car not to skid when travelling at a certain speed on a properly banked curve, we can use the concept of centripetal force.
First, let's understand the key principles involved in this scenario:
1. Centripetal Force (Fc): This force is responsible for keeping an object moving in a curved path. In the case of a car on a banked curve, the centripetal force is provided by the horizontal component of the net force acting on the car.
2. Frictional Force (Ff): The frictional force acts between the tires of the car and the surface of the road. It functions to provide the necessary centripetal force to keep the car on the curved path.
Now, let's derive the equation to solve the problem:
1. We know that the centripetal force is given by the formula:
Fc = (m * v^2) / r
Where:
- Fc is the centripetal force,
- m is the mass of the car,
- v is the velocity of the car, and
- r is the radius of the curve.
2. The frictional force (Ff) is equal to the product of the coefficient of static friction (µs) and the normal force (Fn):
Ff = µs * Fn
3. The gravitational force (Fg) acting on the car can be resolved into two components, perpendicular and parallel to the road surface. The perpendicular component is equal to the normal force.
Fg (perpendicular) = Fn
4. For proper banking, the normal force can be calculated as:
Fn = m * g * cos(θ)
Where:
- g is the acceleration due to gravity, and
- θ is the angle of banking.
Now, let's calculate the coefficient of static friction for the given scenario:
Given:
- Radius of the curve (r) = 60 m
- Velocity of the car (v) = 90 km/h = 25 m/s (converted to meters per second)
Step 1: Calculate the normal force (Fn)
Assuming the angle of banking (θ) is not mentioned, we'll consider it as 0°, which means the curve is unbanked.
Fn = m * g * cos(0°) = m * g
Step 2: Calculate the centripetal force (Fc)
Fc = (m * v^2) / r
Step 3: Equate the centripetal force to the frictional force
Fc = Ff
(m * v^2) / r = µs * Fn
Step 4: Substitute the expressions for Fn and Fc
(m * v^2) / r = µs * (m * g)
Step 5: Simplify and solve for µs
µs = (m * v^2) / (r * g)
Since we don't have the mass (m) of the car, we can't determine the coefficient of static friction (µs) without this information. The mass of the car is a crucial parameter required to solve this problem.