# Banked Curve-Determine smallest static coefficient of friction

• Alexanddros81
In summary: Good job!In summary, the flatbed railway car travels at a constant speed of 80 km/h around a curve with a radius of 55m and a bank angle of 15 degrees. The smallest static coefficient of friction between the crate and the car that would prevent the crate of mass M from sliding is 0.52. To determine the critical speed at which the friction force is zero, the equation v^2cosθ - gRsinθ = 0 can be used, resulting in a critical speed of 12.02 m/s.
Alexanddros81

## Homework Statement

13.49 The flatbed railway car travels at the constant speed of 80 km/h around a curve
of radius 55m and the bank angle 15deg. Determine the smallest static coefficient of friction between the crate and the car that would prevent the crate of mass M from sliding.

## The Attempt at a Solution

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Am I at the right track? If yes how do I proceed?

It looks like you got confused with the algebra. You cannot have μs and NA together in your final expression. It might be cleaner to use a coordinate system where the x-axis is aligned with the incline. Then the normal force and the force of static friction are in separate equations. Find an expression for NA from the y-equation and put it in the y-equation in μsNA. Don't forget that the acceleration has both x and y components in this system of axes.

TSny
kuruman's suggestion is good. If you decide to stay with your orientation of axes, you will need to correct the following:

The horizontal component of friction does not equal μs times the horizontal component of the normal force.

One more thing you need to consider is the direction of the force of friction. For a given coefficient of friction, radius of curvature and banking angle there is a range of speeds that will keep the car on track. How do you know that it is towards the center of the circle? If the car goes "too fast", yes the force of friction is towards the center, but if the car goes "too slow", friction needs to point away from the center. In short, how do you determine whether the speed is too fast or too slow?

TSny said:
kuruman's suggestion is good. If you decide to stay with your orientation of axes, you will need to correct the following:
View attachment 211960

The horizontal component of friction does not equal μs times the horizontal component of the normal force.

Ah, yes I got it wrong! It should be ##Fs_x = F_s cosθ = μ_s N_A cosθ##

Also in response to kuruman: I don't know how to determine whether the speed is too fast or too slow

My thought is if the railway car travels at the constant speed of 80Km/h around a the curve then the crate will have the tendency to move away from the center of the circle causing a friction force pointing to the center of the circle.

Last edited:
Alexanddros81 said:
My thought is if the railway car travels at the constant speed of 80Km/h around a the curve then the crate will have the tendency to move away from the center of the circle causing a friction force pointing to the center of the circle.
Suppose the angle of the incline were 90o and the car has enough speed as it goes around to stay at fixed height (daredevils on motorcycles do this all the time.) Then surely fs must point up the incline to cancel the weight. Assuming no sliding at any time, as you decrease the angle keeping the speed constant, fs will become smaller, go through zero and then change direction. The critical angle is matched to the particular speed. Now suppose you fix the angle and change the speed. There is a critical speed at that angle at which the friction is zero. Can you find the critical speed for this problem? That's what determines whether the given speed is too fast or too slow and hence the direction of friction.

kuruman said:
Suppose the angle of the incline were 90o and the car has enough speed as it goes around to stay at fixed height (daredevils on motorcycles do this all the time.) Then surely fs must point up the incline to cancel the weight. Assuming no sliding at any time, as you decrease the angle keeping the speed constant, fs will become smaller, go through zero and then change direction. The critical angle is matched to the particular speed. Now suppose you fix the angle and change the speed. There is a critical speed at that angle at which the friction is zero. Can you find the critical speed for this problem? That's what determines whether the given speed is too fast or too slow and hence the direction of friction.

Doing some research on the web regarding the problem in question I found two good sources of info:
1)https://physics.stackexchange.com/q...-find-direction-of-friction-on-a-banked-curve

According to the above the critical speed can be found when the frictional force equals zero so that the component of
W parallel to the slope is equalthe centripetal force.
so ## Wsinθ=ma_c## => ## mgsinθ=m\frac {v^2} {r}## => ##v = \sqrt {gsinθr} = 11.82 m/s ##

Since the speed of the car is bigger than the critical speed the friction force should be parallel to the slope
to the left (towards the center of curve).

Your expression for μs is correct. Your expression for the critical speed is not. If μs = 0 (frictionless case), then we v = vcrit.. Your expression then gives ##v^2_{crit.}\cos \theta-g R \sin \theta=0##. If you solve for ##v_{crit.}##, what do you get?

kuruman said:
Your expression for μs is correct. Your expression for the critical speed is not. If μs = 0 (frictionless case), then we v = vcrit.. Your expression then gives ##v^2_{crit.}\cos \theta-g R \sin \theta=0##. If you solve for ##v_{crit.}##, what do you get?
Ah, yes.
It should be ##v^2_{crit} cosθ - gRsinθ = 0## => ##v^2_{crit}=\frac {gRsinθ} {cosθ}## => ##v_{crit} = \sqrt {gRtanθ}## => ##v_{crit} = 12.02 \frac {m} {s}##

So my result 0.52 for ##μ_s## is correct?

Alexanddros81 said:
Ah, yes.
It should be ##v^2_{crit} cosθ - gRsinθ = 0## => ##v^2_{crit}=\frac {gRsinθ} {cosθ}## => ##v_{crit} = \sqrt {gRtanθ}## => ##v_{crit} = 12.02 \frac {m} {s}##

So my result 0.52 for ##μ_s## is correct?
That is what I get.

## 1. What is a banked curve?

A banked curve is a curved section of road or track that is angled higher on one side than the other. This angle, known as the bank angle, helps vehicles navigate the curve more safely and efficiently.

## 2. Why is it important to determine the smallest static coefficient of friction on a banked curve?

The smallest static coefficient of friction on a banked curve is important because it indicates the minimum level of friction needed between the vehicle's tires and the road surface to prevent slipping and maintain control while navigating the curve. Knowing this value can help engineers design safer and more efficient roads and vehicles.

## 3. How is the smallest static coefficient of friction calculated on a banked curve?

The smallest static coefficient of friction on a banked curve is calculated using the formula: μ = tanθ, where μ is the coefficient of friction and θ is the bank angle in degrees. This formula assumes a level road surface and does not take into account factors such as tire tread and road conditions.

## 4. What factors can affect the smallest static coefficient of friction on a banked curve?

The smallest static coefficient of friction on a banked curve can be affected by various factors, including the bank angle, the weight and speed of the vehicle, the type and condition of the tires, and the condition of the road surface. Other factors such as weather conditions and the presence of oil or debris on the road can also impact the coefficient of friction.

## 5. How does the smallest static coefficient of friction impact driving on a banked curve?

The smallest static coefficient of friction plays a crucial role in driving on a banked curve. If the coefficient of friction is too low, the vehicle may slip and lose control while navigating the curve. On the other hand, if the coefficient of friction is too high, the vehicle may experience excessive friction, which can cause wear and tear on the tires and decrease fuel efficiency. Therefore, it is important to determine the smallest static coefficient of friction to ensure safe and efficient driving on banked curves.

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