Force tension, friction, and coefficient of friction

  • #1

Homework Statement


A child pulls a friend on a sled. The child maintains a constant tension of 29.4 N in the rope. The combined mass of the friend and sled is 30.0 kg, and the coefficient of sliding friction between the sled and snow is 0.1. How far can the child pull her friend in 6 sec?

Homework Equations


F = ma
fk = muk x FN
Ff = mu x FN
Fw/2 = FT

The Attempt at a Solution


so I got the force weight of the child and the sled is 294 N
so the force normal of the child and sled is 294N
fk = 0.1 x 294 = 29.4
and the force tension is 29.4 so 29.4 = Fw/2
= 58.8 N = Fw but i don't know what weight that is
and i need another equation to put this in but i don't know which one
 

Answers and Replies

  • #2
Doc Al
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Can we assume that the rope is being pulled horizontally?
Fw/2 = FT
I don't recognize that equation. Can you explain what it means?

so I got the force weight of the child and the sled is 294 N
so the force normal of the child and sled is 294N
fk = 0.1 x 294 = 29.4
Good. That's the force of friction.

So what's the net force on the sled+friend?

(I think there is key information missing from this problem statement. Is it from a textbook?)
 
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  • #3
haruspex
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(I think there is key information missing from this problem statement. Is it from a textbook?)
Probably can assume it means from rest.
 
  • #4
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What does Fw/2=Ft mean?
I believe you need to work out the resultant force on the sled. After finding the resultant force, use Newton Second law of motion, F=ma to find the acceleration,a
After that, you can apply formula for motion in constant acceleration. since u are assuming the sled is at rest, use the information that u have and substitute it to a suitable formula(there are 4 formulae for motion in constant acceleration) a,u,t and you can find the displacement
 
  • #5
Can we assume that the rope is being pulled horizontally?

I don't recognize that equation. Can you explain what it means?



Good. That's the force of friction.

So what's the net force on the sled+friend?

(I think there is key information missing from this problem statement. Is it from a textbook?)
I'm actually not sure what the equation is, my teach wrote it on the board and i wrote it down, maybe it was just for a specific problem, I'm not sure.
well on the sled and friend theres the 29.4 N of friction pulling it back then the 29.4 N of FT, but wouldn't those cancel each other out? or is there something else?
 
  • #6
Doc Al
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well on the sled and friend theres the 29.4 N of friction pulling it back then the 29.4 N of FT, but wouldn't those cancel each other out?
Yes, they would give a net force of zero and thus no acceleration. So the problem seems to be a bit off. Is this a textbook problem? Or something your instructor made up?
 

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