What is the Coefficient of Kinetic Friction in a Sled Pulling Problem?

[ScullyX51]

Homework Statement

You are pulling a child up hill on a sled, walking at a constant v. The incline makes the angle theta with the horizontal, and the rope is parallel to the incline. The mass of the sled plus the child is M, and you pull with a constant force T.
Start with making a sketch of the situation, and then draw a free body diagram of the forces acting on mass M (sled +child). Find the coefficient of kinetic friction between the sled and the snow, purely in terms of the given quantities and the acceleration of gravity g.

Homework Equations

F=ma

The Attempt at a Solution

For the free body diagram..I have as follows:
mg pointing straight down in the negative g direction, the normal force pointing diagonally to the left,( positve j and negative i direction), and then Kinetic friction pointing down diagonally to the left (negative j and negative i direction.) Does this sound correct? What I am confused about is since the sled and child are treated as one system...there is no tension force right? What does it mean when they say the tension is constant?

 

[LowlyPion]

Homework Statement

You are pulling a child up hill on a sled, walking at a constant v. The incline makes the angle theta with the horizontal, and the rope is parallel to the incline. The mass of the sled plus the child is M, and you pull with a constant force T.
Start with making a sketch of the situation, and then draw a free body diagram of the forces acting on mass M (sled +child). Find the coefficient of kinetic friction between the sled and the snow, purely in terms of the given quantities and the acceleration of gravity g.

Homework Equations

F=ma

The Attempt at a Solution

For the free body diagram..I have as follows:
mg pointing straight down in the negative g direction, the normal force pointing diagonally to the left,( positve j and negative i direction), and then Kinetic friction pointing down diagonally to the left (negative j and negative i direction.) Does this sound correct? What I am confused about is since the sled and child are treated as one system...there is no tension force right? What does it mean when they say the tension is constant?

Consider the angle of the slope. The tension T will be in balance with the gravity down the slope and the resistance of friction.

Constant velocity just means that there is no acceleration which would imply a net force.

 

[Rake-MC]

Consider the child and the sled as one inseperable mass, I believe it's only there for the wording of the question. There is a force to the right (T), I think you could possibly be interpreting the question as saying that it is the child who is pulling the sled (correct me if I'm wrong). You can take the force T as being from outside of the system.

Just to confirm your forces, they are all correct (provided you are taking the height of the hill to increase as you move to he right), the normal force is perpendicular to the plane, and the force due to kinetic friction is parallel to the plane.
(This is probably what you meant in your description though).

Constant tension simply means that it's not a varying force, if you were given numerical values, you could solve for an exact value of T and it wouldn't change. The question also states velocity is constant. What does that tell you about the sum of all forces?

 

[ScullyX51]

Ok. this is what I have now. I decided to go with a tilted axis, since there is an incline.
so keeping in mind that the axis are tilted as I have:
N pointing up in the positive j direction.
gravity pointing down with components in (positive i and negative j)
t pointing to the right in the positive i direction.
and kinetic friction pointing to the left in the negative i direction.
Then the components I came up with are:
Normal force= Nj
Fgrav= mg(cos theta i- sin theta j)
Kinetic friction= ( - MkN i
tension= Ti

so for all the terms in the i direction I have:
i= mgcos(theta)-MkN+T
j= N-mgsin(theta)

Theh to solve for the coefficient of kinetic friction:
N-mgsin(theta)= 0
N=mgsin(theta)
then I plugged this value of N into the i equation:
Mgcos(theta)-Mk(mgsin(theta))+T=0
Mgcos(theta)+T= Mk(mgsin(theta))
Mk= mgcos(theta)+T/mg(sin(theta)
and I canceled out the mg's and am left with:
Mk= cos(theta)+T/sin(theta)

How does this look?

 

[Rake-MC]

For force of gravity, do you have your sines and cosines backwards?

 

[ScullyX51]

yeah. I think I was confused with my picture. Now I have come up with the following:
Normal force= Nj
Fgrav= mg(-sin(theta)i-cos(theta)j)
T=Ti
Kinetic Friction= -MkN

i= -mgsin(theta)+T-MkN
j= N-mgcos(theta)

-mgsin(theta)+T=Mk(mgcos(theta)
-mgsin(theta)+T/mgcos(theta)=Mk
the mg's cancel..so the answer is:
-tan(theta)+T=Mk

Does this look like it's fixed now? The last part of the question wants to know how you would measure the static friction between the sled and the snow, and we can allow the experimenter some additional equipment...I don't even know what this is asking!
Thanks for all you help~!

 

[Rake-MC]

Your answer looks good now.

Hmmm that's a pretty open-ended question, perhaps use google to find different methods of measuring friction between surfaces (provided I've interpreted the question correctly).

 

[ScullyX51]

ok. thanks for all your help! :)