Calculating the Coefficient of Friction in a Sled-Pulling System at Equilibrium

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Homework Help Overview

The problem involves calculating the coefficient of friction for a sled being pulled at a constant speed on ice, with a rope inclined at an angle. The sled's mass and the tension in the rope are provided, and the context is set within the framework of equilibrium conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equilibrium condition and the forces acting on the sled, including the tension in the rope and the normal force. There is a focus on resolving forces correctly and considering the impact of the rope's angle on the normal force calculation.

Discussion Status

Some participants have provided feedback on the original poster's calculations, suggesting corrections to the normal force and the coefficient of friction. There is an ongoing evaluation of the approach taken, with some participants expressing uncertainty about the accuracy of the calculations and encouraging further verification.

Contextual Notes

There is a mention of the need to resolve both components of the applied force and the implications of the rope's angle on the normal force. The discussion reflects a learning process where assumptions about forces and their contributions to the system's equilibrium are being critically examined.

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Homework Statement


A sled of mass 480kg is pulled along the ice at a constant speed by means of a rope inclined upward at 25° to the horizontal. If the force of tension in the rope is 350N, what is the coefficient of friction between the sled and the ice?


Homework Equations





The Attempt at a Solution



Since the system is at equilibrium,

Ffr = Fa * cos25°
= 350N *cos25°
= 317.2N

FN = 480kg * 9.8m/s2
= 4704N

μK= Ffr/Fn
= 317.2N/4704N
=0.067

Anyone mind taking a look at this to see if I approached this wrong?
 
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Net force is zero... Because of constant velocity constraint.
0.067 looks good

Edit: Lol oops, pesky only other dimension...
 
Last edited:
Jimbo57 said:

Homework Statement


A sled of mass 480kg is pulled along the ice at a constant speed by means of a rope inclined upward at 25° to the horizontal. If the force of tension in the rope is 350N, what is the coefficient of friction between the sled and the ice?


Homework Equations





The Attempt at a Solution



Since the system is at equilibrium,

Ffr = Fa * cos25°
= 350N *cos25°
= 317.2N

FN = 480kg * 9.8m/s2
= 4704N

μK= Ffr/Fn
= 317.2N/4704N
=0.067

Anyone mind taking a look at this to see if I approached this wrong?

The normal reaction force will be less than mg, since the rope angled up is providing some of the upward force needed to "balance" the weight of the sled.!
 
For an evaluation of the answer, draw a fbd.
You have only resolved the x component of the force applied.
You should also resolve the other component too.
When resolving forces, remember to include both.
 
PeterO said:
The normal reaction force will be less than mg, since the rope angled up is providing some of the upward force needed to "balance" the weight of the sled.!

Ah yes, I messed up a basic concept!

So the Fn = mg - 350N*sin25°
=4556.08N

Therefore,

μK= 317.2N/4556.08N = 0.07

Does this look more accurate PeterO or anyone else?
 
Jimbo57 said:
Ah yes, I messed up a basic concept!

So the Fn = mg - 350N*sin25°
=4556.08N

Therefore,

μK= 317.2N/4556.08N = 0.07

Does this look more accurate PeterO or anyone else?

You have allowed for everything [I don't have a calculator handy] so provided you pressed all the right buttons on your calculator that should be correct - and if the answer is exactly 0.07 that would attract me as well.
 
PeterO said:
You have allowed for everything [I don't have a calculator handy] so provided you pressed all the right buttons on your calculator that should be correct - and if the answer is exactly 0.07 that would attract me as well.

Pretty close to 0.07 anyways. It's 0.0696 rounded to 3 sig figs = 0.07. Not so attractive :)

Thanks for the help!
 

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