# Newton's Law of a sled

1. Dec 7, 2015

### mace42

1. The problem statement, all variables and given/known data
This is a problem from a post back in 2006. I have a fairly good understanding of how to do it; however, there is one part about the answer that I cannot figure out at all. (See the relevant equations)

Problem: A 5kg penguin sits on a 10kg sled. A horizontal force of 45 N is applied to the sled, but the penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and snow as well as that between the sled and the penguin is .2 b) Determine the tension in the cord and c) the acceleration of the sled if you try to solve for the tension using F=ma, do you could both the weights as the normal force? or do you only count the penguin's weight? Thanks

2. Relevant equations
On the post it said that the net force in the x direction of the sled was as follows:

Sum of all forces sub x = ma sub x
Fg + F sub f + (-Fx) = ma sub x

I do not understand why Fg was included as a force acting in the x direction. Is this wrong or am I missing something.

This is the equation I used:

Sum of all forces sub x = ma
F sub f + (-Fx) = ma

Any insight would be greatly appreciated!!

2. Dec 7, 2015

### Chandra Prayaga

Confusion arises by just trying to do the problem by looking at the equation for the x-motion. Do the problem completely. You don't have to guess, or be told by anyone, whether you should count both weights or only one. It will all come out as part of the solution.

So the forces on the sled are:
1. Pulling force 45 N
2. Friction from ground
3. Friction from penguin
4. Normal force from ground
5. Normal force from penguin
6. Weight of sled

The last three are in the y-direction, but they decide the friction forces, so let them be there. Draw the force diagram

Forces on the penguin are:
7. Tension from the string
8. Friction from sled
9. Normal force from sled
10. Weight of penguin

Draw the force diagram for this separately

Now use Newton's second law on each object, and then take components. Also use Newton's third law, which gives Force # 3 = #8; #5 = #9. Use this information and the answer will roll out automatically.

3. Dec 7, 2015

### mace42

Thank you so much for the help!! Ultimately I got a=(Fa-ug(Mp+Ms))/Ms as an equation for acceleration and 1.6 m/s/s for an answer. If that isn't correct I can list my work.

4. Dec 7, 2015

### mace42

I just discovered the upload button haha, here's my work.

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