Coefficient of friction (between road and wheels of accelerating truck)

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Homework Help Overview

The discussion revolves around determining the coefficient of friction between the road and the wheels of a truck that is accelerating. The truck has a mass of 1400 kg and is experiencing an acceleration of 2 m/s², with an engine force of 4000 N applied to the wheels.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to resolve the problem due to having two unknowns, the frictional force (Ff) and the coefficient of friction (mu). Some participants suggest drawing a free body diagram and applying Newton's second law to find the resultant force in the horizontal direction.

Discussion Status

Participants are exploring different approaches to understand the problem better. While one participant claims to have found an answer from an external source, the discussion remains focused on the reasoning process rather than a definitive resolution.

Contextual Notes

The original poster indicates a desire for assistance without receiving a complete answer, highlighting the educational context of the inquiry. There is an emphasis on understanding the underlying concepts rather than simply obtaining a solution.

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Homework Statement


What is the coefficient of friction between the road and the wheels of a 1400kg truck accelerating at 2 m/s/s if the force applied by the engine to the wheels is 4000N.


Homework Equations


Fn = m * g
Ff = mu * Fn


The Attempt at a Solution


I do not know how to resolve this as I don't know Ff or mu (i.e. two variables)
Any asisstance (not answer) would be appreciated.
 
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First draw a free body diagram.

Then find the resultant force in the horizontal (x) direction.

Newton's 2nd Law states that ∑Fx= ma
 
thanks rock.freak667,
I have found the answer it was on a website called http://www.physics247.com/physics-homework-help/friction-and-weight.php" the answer was 0.09 as is shown here:

mu = Ff/Fn
Ff = (ma) - Fengine
mu = ((ma) - Fengine)/-(mg)
mu = ((1400 * 2) - 4000)/-(1400 * 9.81)
mu = (2800 - 4000)/-13734
mu = -1200/-13734
mu = 0.085...
mu = 0.09
 
Last edited by a moderator:
Once you understand the thinking process, as that will allow you do similar problems or slightly more complex ones.
 

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