Overcoming friction

  • Thread starter mrs. drake
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  • #1
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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  • #2
PhanthomJay
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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 ?
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:
 
  • #3
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:)
 
  • #4
PhanthomJay
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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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