Finding the Coefficient of friction

  • Thread starter Nanart
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  • #1
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Homework Statement



A 135 kg sled is being pulled along a horizontal surface at a uniform speed by means of a rope making an angle of 19.3 degrees with the horiontal. If the tension in the rope is 119 N Find the Coefficient of friction.

Homework Equations



u=Ff/Fn

The Attempt at a Solution


I have never done a question like this before but i fugured i probably had to find the force from the horizontal..im not sure if i did it right but i used trig to find that force and i got 112.3 N i then put that in the formula U=112.3/135 and i got 0.8319 but I really have no idea if i did it right.
 

Answers and Replies

  • #2
djeitnstine
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You are correct. [tex]\mu F=F_{f}[/tex] so the ratio of the 2 must be the coefficient of friction. I also checked and our numbers match.
 
  • #3
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You are correct. [tex]\mu F=F_{f}[/tex] so the ratio of the 2 must be the coefficient of friction. I also checked and our numbers match.

Awsome...Thanks
 
  • #4
PhanthomJay
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But don't forget that the weight of the block is not 135 N; its mass is 135 Kg. Also, the normal force fron the weight is offset by the upward component of the pulling force.
 
  • #5
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HI i dont really get how u get u=112.3/135
do you use the component method?
here are my calculations...please tell me if they are right
y components
Fnet=Tensiony +Fnormal - Forcegravity
0 =119*sin19.3 + Fnormal -135*9.8
Fnormal= 135*9.8 -119*sin19.3
=1362.33N
Ff=1362.33*u
x components
Fnet= Fappx-Ff
135*a=119cos19.3 - 1362.33
0=119Cos19.3 - 1362.33 (because constant velocity, so acceleration= 0)
U=119Cos19.3/1362.33
=0.0824

I am in grade 12 and I understand tht grade 12 and grade 11 methods are different, but please tell me what i did wrong and why i am not getting the right answer. THank you
 
  • #6
13
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HI i dont really get how u get u=112.3/135
do you use the component method?
here are my calculations...please tell me if they are right
y components
Fnet=Tensiony +Fnormal - Forcegravity
0 =119*sin19.3 + Fnormal -135*9.8
Fnormal= 135*9.8 -119*sin19.3
=1362.33N
Ff=1362.33*u
x components
Fnet= Fappx-Ff
135*a=119cos19.3 - 1362.33
0=119Cos19.3 - 1362.33 (because constant velocity, so acceleration= 0)
U=119Cos19.3/1362.33
=0.0824

I am in grade 12 and I understand tht grade 12 and grade 11 methods are different, but please tell me what i did wrong and why i am not getting the right answer. THank you

You are on the right track i had my teacher explain it to me and the answer is actually 0.0874
 
  • #7
PhanthomJay
Science Advisor
Homework Helper
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HI i dont really get how u get u=112.3/135
do you use the component method?
here are my calculations...please tell me if they are right
y components
Fnet=Tensiony +Fnormal - Forcegravity
0 =119*sin19.3 + Fnormal -135*9.8
Fnormal= 135*9.8 -119*sin19.3
=1362.33Nyou made a simple math error here that carries through the rest of the problem
Ff=1362.33*u
x components
Fnet= Fappx-Ff
135*a=119cos19.3 - 1362.33
0=119Cos19.3 - 1362.33 (because constant velocity, so acceleration= 0)
U=119Cos19.3/1362.33
=0.0824

I am in grade 12 and I understand tht grade 12 and grade 11 methods are different, but please tell me what i did wrong and why i am not getting the right answer. THank you
see note above on math error; otherwise your method is very good.
 
  • #8
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thanks guys...appreciate your help
 

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