The maximum speed the truck can go without sliding

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Homework Help Overview

The problem involves determining the maximum speed of a truck rounding a circular bend without causing a crate of eggs to slide due to static friction. The context includes concepts from circular motion and friction, specifically focusing on the relationship between speed, radius, and frictional forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the maximum speed using the relationship between static friction and centripetal force, while some participants provide feedback on the terminology used in the explanation.

Discussion Status

The discussion is active, with participants reviewing the original poster's reasoning and suggesting minor adjustments to terminology. There is no explicit consensus on the correctness of the final answer, but the feedback indicates engagement with the reasoning process.

Contextual Notes

Participants are focused on ensuring clarity in the application of terms related to the truck and the crate, indicating a potential area of confusion in the problem setup.

goli12
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Homework Statement


A crate of eggs is located on the back of a truck. The truck rounds a circular bend in the road with radius of 35 meters. If the coefficient of static friction between the crate and the truck is 0.6, what is the maximum speed the truck can go without the crate sliding?

Homework Equations


F=mv2/r
Fs=μsn

The Attempt at a Solution


Hey guys, can you please check my answer for this question, I'm not entirely sure it is correct...

Firstly, I took the normal force of the truck to get n=mg, as it is not moving vertically. This results in the force of static friction (Fs) to be: Fs=μsmg.

Secondly, this static friction is providing the centripetal force of the truck to go around in a circular motion and so resulting in the following equation: Fs=F=mv2/r = μsmg.

So with algebra I made velocity the subject. The masses cancel out leaving: v2/r = μsg and so resulting in v=√(μsgr)

Now I enter the variables v=√(0.6*9.8*35), giving the velocity of 14.35ms-1.
 
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Looks fine to me :)
 
Solution looks ok, however, substitute the word 'truck' with 'crate' in your attempt.
 
PhanthomJay said:
Solution looks ok, however, substitute the word 'truck' with 'crate' in your attempt.

Ahh yes, my mistake many thanks :-p
 

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