# Pulling A Sled At Equilibrium

1. Aug 2, 2012

### Jimbo57

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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?

2. Aug 2, 2012

### Jakeus314

Net force is zero... Because of constant velocity constraint.
0.067 looks good

Edit: Lol oops, pesky only other dimension...

Last edited: Aug 2, 2012
3. Aug 2, 2012

### PeterO

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.!!

4. Aug 2, 2012

### azizlwl

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.

5. Aug 2, 2012

### Jimbo57

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?

6. Aug 2, 2012

### PeterO

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.

7. Aug 2, 2012

### Jimbo57

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

Thanks for the help!