Help for physics static and kinetic friction

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Homework Help Overview

The discussion revolves around a physics problem involving static and kinetic friction. The scenario includes a penguin on a sled, with specific weights and coefficients of friction provided, and seeks to determine the maximum horizontal force that can be applied without the penguin sliding off the sled.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of friction forces and the application of Newton's second law. Questions are raised about the maximum static friction force and the associated acceleration. There is also a request for clarification on the original poster's calculations and assumptions.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, suggesting a need to consider the maximum acceleration of the penguin and the forces acting on it. Multiple interpretations of the problem and calculations are being explored, but no consensus has been reached.

Contextual Notes

There is a noted discrepancy in the weight of the penguin, with one participant pointing out the difference between 70 N and 70.1 N. Additionally, the original poster has requested a prompt response, indicating urgency in resolving the problem.

sundeepsingh
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help for physics static and kinetic friction.
Question:
A 70 Newtons Penguin is on a 50 N. sled. the static friction between the penguin and sled is 0.671 and the kinetic friction between the sled and the SNOW is 0.119. FInd the maximum horizontal force that can be applied without the penguin sliding off the sled?
please post the answer A.S.A.p
 
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Please post your work to get help.

Hint: What is the maximum value of the static friction that the sled can exert on the penguin? What acceleration is associated with that force? Apply Newton's 2nd law to find the applied force on the sled.
 
this is what i did.
the friction force between sled and snow is= FN X u=(70.1+50)*0.119=14.2919 N.
the friction force between the penguin and sled = (70.1)*0.671=47.0371, therefore, the max horizontal force that can be applied to the sled without causing the penguin to slide off the sled=14.2919+47.0371=61.329 N. But this is wrong, where m i wrong please tell me.
 
Did someone say penguin :D
 
sundeepsingh said:
the friction force between sled and snow is= FN X u=(70.1+50)*0.119=14.2919 N.
OK. (But the penguin weighs 70N, not 70.1N.)
the friction force between the penguin and sled = (70.1)*0.671=47.0371,
OK. (Same comment.)
therefore, the max horizontal force that can be applied to the sled without causing the penguin to slide off the sled=14.2919+47.0371=61.329 N.
Nope. First find the maximum acceleration of the penguin. (What force acts on the penguin?) Then figure out what force must be applied to the sled so that the sled+penguin has that acceleration. (Don't forget to consider the friction force between sled and snow that must be overcome.)
 
THANKS a lot DOC AL, i LUV U lol :P
 

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