Friction and toboggans

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  • #1
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Homework Statement



A 15kg toboggan is pulled across the snow at a constant speed by a horizontal force of 22N.

a) What is the force of gravity on the toboggan?
b) What is ##μ_k##?
c) How much more force is needed to pull the toboggan if 105kg is added to the toboggan?

Homework Equations



##m = 15kg##
##a = 9.8 m/s^2##
##F_A = 22N## [Applied force]
##F_N## [Force normal]
##F_K## [Kinetic force/Friction]


The Attempt at a Solution



a) ##F_G = ma = (15 kg)(9.8 m/s^2) = 147N##

b) The toboggan travels at a constant velocity ( a = 0 ) with no ups and downs. This implies that ##F_N = F_G = 147N## and ##F_K = F_A = 22N##

Now ##F_K = μ_KF_N \Rightarrow μ_K = \frac{F_K}{F_N}##

So that ##μ_K = \frac{22N}{147N} = 0.15##.

c) Hm so the total weight of the toboggan will now be ##m = 120kg## and we have the kinetic friction coefficient ##μ_K = 0.15##.

Would it just be :

##F_K = μ_Kma = (0.15)(120 kg)(9.8 m/s^2) = 176.4N##

So we would require 176.4N+22N = 198.4N more to keep the toboggan at a constant velocity. Is this okay?
 

Answers and Replies

  • #2
PeterO
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Homework Statement



A 15kg toboggan is pulled across the snow at a constant speed by a horizontal force of 22N.

a) What is the force of gravity on the toboggan?
b) What is ##μ_k##?
c) How much more force is needed to pull the toboggan if 105kg is added to the toboggan?

Homework Equations



##m = 15kg##
##a = 9.8 m/s^2##
##F_A = 22N## [Applied force]
##F_N## [Force normal]
##F_K## [Kinetic force/Friction]


The Attempt at a Solution



a) ##F_G = ma = (15 kg)(9.8 m/s^2) = 147N##

b) The toboggan travels at a constant velocity ( a = 0 ) with no ups and downs. This implies that ##F_N = F_G = 147N## and ##F_K = F_A = 22N##

Now ##F_K = μ_KF_N \Rightarrow μ_K = \frac{F_K}{F_N}##

So that ##μ_K = \frac{22N}{147N} = 0.15##.

c) Hm so the total weight of the toboggan will now be ##m = 120kg## and we have the kinetic friction coefficient ##μ_K = 0.15##.

Would it just be :

##F_K = μ_Kma = (0.15)(120 kg)(9.8 m/s^2) = 176.4N##

So we would require 176.4N+22N = 198.4N more to keep the toboggan at a constant velocity. Is this okay?

Are you sure you should add 22N at the end.

Have you calculated the force needed for the loaded toboggan, or the the extra force needed for the load?
 
  • #3
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I think PeterO says it all
 
  • #4
STEMucator
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You're right. My first instinct was 176.4N was the answer. Then I second guessed myself for a moment.

So we would need 176.4N to pull to toboggan.
 
  • #5
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Be careful of the question. It asks for how much MORE force is needed, not how much force is needed.
 
  • #6
STEMucator
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Be careful of the question. It asks for how much MORE force is needed, not how much force is needed.

Then I would need 176.4N - 22N = 154.4N MORE force to pull the toboggan.

176.4N Is the force needed to pull the toboggan while 154.4N is how much more force I need to move the toboggan as compared to when the toboggan didn't have as much mass ( It only required 22N ).
 
  • #8
haruspex
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Then I would need 176.4N - 22N = 154.4N MORE force to pull the toboggan.

176.4N Is the force needed to pull the toboggan while 154.4N is how much more force I need to move the toboggan as compared to when the toboggan didn't have as much mass ( It only required 22N ).
You seem to have introduced an error of 0.4N. 105kg is exactly 7 times 15kg, so the extra force should be exactly 7 times 22N.
 

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