If a curve with a radius of 88m is perfectly banked for a car traveling 75km/h, what must be the coefficient of static friction for a car not to skid when traveling at 95km/h?
I'm with Emily. I've been through about 8 explanations on this site for this specific type of problem and none of them make sense to a beginner like us. Sorry for not being as smart as you are or already having a grasp on this stuff. If we did we probably wouldn't be asking for help in the first place. We are seriously putting forth the effort to learn here and you reply with that?! An explanation with numbers wouldn't hurt. In fact, I'm pretty sure it would help out quite a bit with understanding this concept. I've been through explanation after explanation using only variables, and it's not helping. As soon as I think that I've interpreted your explanation of delicate genius and I input an answer, I get shot down with INCORRECT on my assignment. So drop the arrogant nerd attitude, and provide some real help.
Hi there! I am a currently crying ap physics student who is trying their best in this subject but it is just not clicking for me. Seeing responses from people here who are just like 'it's not that hard' is incredibly disheartening. I tutor kids for math, music, and general topics for school, and that is absolutely the worst approach you can have for teaching. For those who act as though this problem is simple for everybody, you need to realize that some of us have not had access to the same teachers you do. We need help, and shame is not the solution.
No idea if anybody will ever need this, but after spending an hour and a half, as well as several thousand brain cells and around 4 gigantic water glasses along the way, I have finally figured it out.
Terms of service do not prohibit giving out complete and final answers here, so I will use numbers.
The first thing you do is you figure out the θ angle that makes up for the "perfect banking". We know that the "perfect banking" is at v1 = 75 km/h = 20.8 m/s. Now, the problem itself makes no sense unless "perfect banking" means that the generation of centripetal force is supplied solely by the X component of the normal force, Fnx.
At the same time, the vertical forces acting on the car then are Fny and the gravitational force, mg.
So, in the X direction we have: Fnx = m*v^2/r.
As for the Y direction we have: Fny - mg = 0 (it must be 0, as the car's not floating).
Applying tricky geometry, we figure that Fnx = Fn*sinθ. Substitute for the X direction, and you get: Fn*sinθ = m*v^2/r.
Using the same logic, Fny = Fn*cosθ. Substitute for the Y direction, and you get: Fn*cosθ = mg. Solve this last equation for Fn, and you get: Fn = mg/cosθ.
Now substitute the acquired Fn value to what we have got for the X direction:
Fn*sinθ = m*v^2/r;
mg*sinθ/cosθ = m*v^2/r; divide both sides by the mass;
g*tanθ = v^2/r;
tanθ = v^2/(gr).
So, θ = tan^-1 (v^2/(gr)).
You substitute numbers and should get the value of θ = 26.4.
After you have found θ, you consider the scenario with v2 = 90 km/h = 26.4 m/s.
Here, because now the banking is not sufficient so that Fnx alone can support centripetal motion, the friction forces kick in.
The main friction force itself is directed downwards the slope, soooo this means that you resolve for its components in order to procede further - these are Ffrx and Ffry (I wish I could attach PNGs here for greater illustration).
This means that the X direction now is: Fnx + Ffrx = m*v^2/r.
All the while the Y direction right here is: Fny - mg - Ffry = o.
Now, look at the Y direction. We know for a fact that Fny = Fn*cosθ. We also know that Ffr = μs*Fn. The Ffry value, using tricky maths again, is: Ffry = Ffr*sinθ.
We can therefore transform the equation for the Y direction into this: Fn*cosθ - Ffr*sinθ = mg;
Fn*cosθ - μs*Fn*sinθ = mg; solve for Fn;
Fn = mg/(cosθ - μs*sinθ).
Now, since Ffrx = Ffr*cosθ = μs*Fn*cosθ, the X direction turns out like this:
Fn*sinθ + μs*Fn*cosθ = m*v^2/r.
All you have to do next is to substitute the Fn value (the last Fn value we've got) in the last equation. After this, you solve it for μs, and it's solved!
I've got my μs = 0.215..., which was suiting enough for me to write it off as 0.22, which is the correct answer.
Honestly, out of all the problems in Giancoli for Chapter 5, this one was the hardest. Tasks 22-24 don't stand ever near this one, especially so doesn't the 23rd.
I hope this will help somebody in the future. At the same time, controversially, I don't care if any of the smart-hats are going to call me out or if you didn't understand something in here, because let's be real, I won't visit this page ever again in my lifetime.
Good luck to anyone struggling with physics and always remember two things: persevere no matter what and choose Halliday and Walker over Giancoli.
I know that, but you like help me figure it out like what to do; please!!
You gotta think logically, people make this way harder than it needs to be. Draw a FBD, label all the components to each force, then use Newton's 2nd Law to break up the problem into 1. sum of all forces in "x" direction , 2. sum of all forces in "y" direction. Solve each for a specific equation, then substitute in values such as Ffs = usFn. That's an important one for this problem. Plug and chug
Credible source? 5 on the AP Physics Exam... So yeah
To determine the coefficient of static friction required for a car not to skid when traveling at 95 km/h on a curved road, we can start by using the concept of centripetal force.
The centripetal force required to keep the car moving in a curved path is provided by the friction between the tires and the road. If the frictional force is not sufficient, the car will skid.
First, let's calculate the required centripetal force for the car.
Given:
Radius of the curve (r) = 88 m
Speed of the car (v) = 95 km/h = 26.4 m/s
Centripetal force (Fc) is given by the formula:
Fc = (m * v²) / r
where:
m is the mass of the car
v is the velocity of the car
r is the radius of the curve
As we don't have the mass of the car, we can express it in terms of gravity and the weight of the car.
Weight (W) = m * g
where:
m is the mass of the car
g is the acceleration due to gravity (approximately 9.8 m/s²)
Let's substitute the value of mass from the weight equation into the centripetal force equation:
Fc = (W * v²) / (m * r)
Now, let's find the centripetal force required by substituting the given values:
Fc = (W * v²) / (m * r)
= (m * g * v²) / (m * r)
= (g * v²) / r
Now, we can calculate the required centripetal force:
Fc = (9.8 * 26.4²) / 88
Fc ≈ 75.1584 N
Now, let's consider the maximum frictional force that can be exerted on the car to prevent skidding. The maximum frictional force is given by the formula:
Ffriction = μ * N
where:
μ is the coefficient of static friction
N is the normal force
The normal force can be calculated using:
N = m * g
Now, we have the maximum frictional force, which is also the centripetal force required:
Ffriction = μ * N
= μ * m * g
Substituting the values, the equation becomes:
Ffriction = μ * m * g
= (g * m)
The maximum frictional force required is equal to the centripetal force:
Ffriction = Fc
= (g * m)
Equating the two expressions:
(g * m) = (g * v²) / r
Now, we can solve for the coefficient of static friction (μ):
μ = ((g * v²) / (g * r))
Let's substitute the given values to find μ:
μ = ((9.8 * 26.4²) / (9.8 * 88))
After performing the calculation, we find:
μ ≈ 0.267
Therefore, the coefficient of static friction for the car not to skid when traveling at 95 km/h on a curve with a radius of 88 m is approximately 0.267.