Banked Curve: Finding the normal force with friction

Click For Summary

Homework Help Overview

This discussion revolves around a level 1 physics problem focused on centripetal motion on banked curves and the forces acting on a car navigating a curve. The original poster has completed two parts of a five-part question, calculating the net centripetal force and the centripetal force provided by the banking of the road.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the normal force and friction force while expressing confusion over the equations involving these forces. Some participants suggest using a free body diagram to better organize the equations and resolve forces along vertical and horizontal axes.

Discussion Status

The discussion is ongoing, with participants providing guidance on organizing equations and suggesting methods to approach the problem. There is no explicit consensus yet, as the original poster continues to seek clarity on the calculations and relationships between the forces involved.

Contextual Notes

The original poster has provided specific values such as the mass of the car, radius of the curve, and angle of incline, but expresses uncertainty about the correctness of their calculations and the relationships between the forces involved.

WitteVrouw
Messages
1
Reaction score
0
This is a five part question, of which I have completed 2 parts, which I will explain below. This is a level 1 physics problem concerning centripetal motion on banked curves and the forces that apply to objects in these scenarios.


PROBLEM:

A car travels at a speed of 27m/s around a curve with a radius of 43m (keep in mind that acceleration due to gravity is 9.8m/s2

What is the net centripetal force needed to keep the car from skidding sideways?
Answer in units of N.


38,993.02326N
My solution to part one ----> Correct


Were there no friction between the car's tires and the road, what centripetal force could be provided just by the banking of the road?
Answer in units of N.

15,782.67791N My solution to part two ------> Correct


Now suppose the friction force is sufficient to keep the car from skidding. Calculate the magnitude of the normal force (Fn) exerted on the car by the road's surface. Hint: Check the correctness of your answer to the first part before proceeding with this and the following questions.

Solution to Part three -------> Unknown

Calculate the magnitude of the friction force.
Answer in units of N.

Solution to Part four ---------> Unknown



Calculate the lowest possible value of the static friction coefficient mus that would prevent the car from skidding.

Solution to Part five ---------> Unknown


NOW, here is where I am stuck. I am fairly sure that parts 3-5 deal with the components of the Normal Force and the Force of Static Friction but I keep getting lost with my equations (somehow I end up with far too many sin-cos-tan arrangments with Fn, etc, divided by one another).

SO,

Here is my data so far:

Fc (centripetal force) needed to keep car from skidding sideways: 38,993.02326N
Fc provided by just the banking of road (without friction): 15,782.67791N
Magnitude of Normal Force: Unknown
Magnitude of Ff (friction force): Unknown
Static friction coefficient mus: Uknown
Mass of car: 2300kg
Radius: 43m
Ac (centripetal acceleration): 16.95348837m/s2
Velocity: 27m/s
Angle of Incline: 35 degrees

Help would be so greatly appreciated! Sorry to inconvenience anyone, I know the problem is quite extensive! Thank you very much!
 
Last edited:
Physics news on Phys.org
You can organize your equations better with a free body diagram. Did you draw one for the banked turn? I suggest that you resolve the forces along vertical and horizontal axes in the banked turn case. Note that the vertical component of normal force must be equal to the weight whilst the horizontal component provides the centripetal acceleration required to negotiate the turn.
 
  • Like
Likes   Reactions: Delta2
The presence of friction complicates things. I think the system of two equations by resolving forces in horizontal and vertical components are
$$F_N\cos 35-mg-T\sin 35=0$$
$$F_N\sin 35+T\cos 35=F_C$$ where ##F_C## the answer in first question.
 
WitteVrouw said:
but I keep getting lost with my equations
We can help with that, but only if you show us your working so far.
 
  • Haha
Likes   Reactions: jack action

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
17
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
16
Views
3K
Replies
9
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 3 ·
Replies
3
Views
10K