Calculate the Coefficient of Friction for a Frozen Turkey

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Discussion Overview

The discussion revolves around calculating the coefficient of friction for a frozen turkey being dragged through snow. Participants explore the application of physics concepts, particularly Newton's laws and frictional forces, in this scenario.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant presents the scenario of dragging a 7.2 kg turkey at a constant velocity of 2 m/s with a horizontal force of 5.3 N, questioning how to find the coefficient of friction.
  • Another participant reiterates the need to use the formula for friction, suggesting that the force of friction (FF) equals the coefficient of friction (u) multiplied by the normal force (fn).
  • A participant seeks clarification on what the force of friction is in this specific example, indicating uncertainty about the values needed for calculation.
  • Another participant references Newton's first law, noting that since the turkey moves at constant velocity, the net force must be zero, implying that the friction force must equal the applied force of 5.3 N.
  • This participant also points out that the normal force is the weight of the turkey, not just its mass, prompting further inquiry into the weight calculation.

Areas of Agreement / Disagreement

Participants generally agree on the application of Newton's laws and the need to calculate the coefficient of friction, but there is no consensus on the specific values for the force of friction or the normal force, leaving the discussion unresolved.

Contextual Notes

Participants have not provided specific calculations for the normal force or the force of friction, leading to uncertainty in the discussion. The problem remains open-ended without definitive numerical answers.

mrs. drake
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I was dragging my frozen 7.2kg turkey due north through the snow at a constant velocity of 2m/s while exerting a horizontal force of only 5.3N. What was the coefficient of friction between the frozen turkey and the snow?

don't you use the formula " FF=ufn" u=coefficient
then mk=ff/fn ?
 
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mrs. drake said:
I was dragging my frozen 7.2kg turkey due north through the snow at a constant velocity of 2m/s while exerting a horizontal force of only 5.3N. What was the coefficient of friction between the frozen turkey and the snow?

don't you use the formula " FF=ufn" u=coefficient
then mk=ff/fn ?
Yes, m'am, that you do, but what, pray tell, is the force of friction in this example? (It is not often that I know the gender of posters or responders, so I am assuming from your name that you are of the female variety. If not, my apologies for respectfully referring to you as 'madam' :biggrin:
 
PhanthomJay said:
Yes, m'am, that you do, but what, pray tell, is the force of friction in this example? (It is not often that I know the gender of posters or responders, so I am assuming from your name that you are of the female variety. If not, my apologies for respectfully referring to you as 'madam' :biggrin:

so

fn= 7.2kg
ff= ?

and yes i am a girl:)
 
OK, Mrs. Drake , Newton's first law of motion tells us that if a frozen turkey is moving at constant velocity, there must be no net unbalanced force acting on that bird. So if you're dragging the darn thing forward with a horizontal force of 5.3 N on a level surface, the friction force must be acting backwards with a friction force of _ :confused:?___ N ? But, wait, there's more: The mass of the turkey is 7.2 kg...It's normal force is it's weight, not its mass, where its weight is equal to ___:confused:?____ N.?

(I am assuming this is a homework problem, so I don't want to supply the answers).:wink:
 

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