Dynamics - Normal and Tangential Motion

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Discussion Overview

The discussion revolves around a homework problem related to dynamics, specifically focusing on the maximum acceleration of a car on a curved road given certain parameters such as frictional force, speed, and road curvature. Participants are exploring the calculations for normal and tangential acceleration.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents a detailed calculation for the maximum acceleration of a car, including both normal and tangential components.
  • Another participant agrees with the initial calculations but suggests that the question specifically asks for the maximum tangential acceleration, which they state is 10.04 ft/s².
  • A third participant initially agrees with the previous response but later corrects themselves, stating that the tangential acceleration is actually 14.85 ft/s².
  • A fourth participant reiterates the corrected tangential acceleration of 14.85 ft/s² and acknowledges a typographical error in their earlier post.

Areas of Agreement / Disagreement

There is some confusion regarding the correct value of the maximum tangential acceleration, with participants initially agreeing on one value before correcting to another. The discussion reflects uncertainty about the interpretation of the problem and the calculations involved.

Contextual Notes

Participants have not resolved the discrepancies in their calculations, and there are indications of missing assumptions or potential errors in the initial interpretations of the problem.

aaronfue
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Homework Statement



The tires on a car are capable of exerting a maximum frictional force of 1753 lb. If the car is traveling at 75 ft.s and the curvature of the road is ρ=560 ft, what is the maximum acceleration that the car can have without sliding?

Homework Equations



ƩFn = man

The Attempt at a Solution



Ff = 1753 lb
v = 75 ft/s
ρ=560 ft
wcar = 3150 lb

an = \frac{v^2}{ρ} = \frac{75^2}{560} = 10.04 ft/s2

I believe that the acceleration would be the magnitude of the tangential and normal acceleration.

ƩFn = man = \frac{3150}{32.2}*10.04 = 982.2 lb

1753 = √Ft2 + 982.22

Solving for Ft = 1452 lb;

Now solving for at → 1452 = \frac{3150}{32.2}*at
at = 14.85 ft/s2

a = √at2 + an2 = √14.852 + 10.042 = 17.90 ft/s2

I'd appreciate it if someone could verify my work.
 
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I get the same answer as you. I think the question is asking for the maximum tangential acceleration which is 10.04 ft/sec^2.
 
LawrenceC said:
I get the same answer as you. I think the question is asking for the maximum tangential acceleration which is 10.04 ft/sec^2.

Tangential was actually 14.85 ft/s^2.

And it makes sense too.

Thanks.
 
Last edited:
aaronfue said:
Tangential was actually 14.85 ft/s^2.

And it makes sense too.

Thanks.

I typed the wrong number...should have typed 14.85 ft/s^2.
 

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