# Kinetic friction incline problem

1. Jun 25, 2013

### Nirupt

1. The problem statement, all variables and given/known data
Mildred pulls a sled loaded with beer along a horizontal surface at constant velocity, as Bubba told her he needed it for he and his buddies to watch the football game. If the sled and beer in it has mass = 75 kg, the coefficient of kinetic friction μk between the runners and the snow is 0.10, and she is pulling a rope inclined at an angle of 42 degrees, what is the magnitude of the tension in the rope?

2. Relevant equations

3. The attempt at a solution

X direction:

μκ = .1

Fκ = .1*Fn

ƩFnetχ = Fτ*cos(42) - μκ*Fn = 0

ƩFnety = Fn + Fτ*sin(42) - mg = 0
Fn = -Fτ*sin(42) + mg

so I plug this for Fn

Fτ*cos(42) - μκ*(-Fτ*sin(42) + mg) = 0

Fτ*cos(42) + μκ*Fτ*sin(42) - μκ*mg = 0
Fτ = μκ*mg/cos(42) + μκsin(42)
Fτ = .1*735/(cos(42) + .1*sin(42))
Fτ = 73.5/.0669
Fτ = 90.7504
F ≈91N

Just checking to make sure this is correct. I feel like I messed up my algebra with Fτ.. I mean.. would it be 2Fτ or Just Fτ like I have it. I feel like basic algebra can be kind of overwhelming at time.

2. Jun 25, 2013

### haruspex

Your working looks correct until you evaluated (cos(42) + .1*sin(42)). cos(42deg) > cos(45deg) = sqrt(2)/2 > .7.