Sled Physics -- trouble understanding this equation

[rc1234]

In the classic example of a man on a sled being pushed down a hill, the equation involves the forces of gravity and friction: F = weight - friction or F = ma = mgsinθ-μkmgcosθ

I do not understand why the push force at the start is not included in the equation. I assume that push force would cause change in velocity and affect the overall resulting force.

How do we take into account the push force in the equation?

 

[lekh2003]

I assume that push force would cause change in velocity and affect the overall resulting force.

Well that's the crazy thing about friction. In problems like this, don't at all worry about velocity or push force in friction calculation (unless it contributes to the normal force). Friction is only dependent on the materials and normal force.

 

[rc1234]

Well that's the crazy thing about friction. In problems like this, don't at all worry about velocity or push force in friction calculation (unless it contributes to the normal force). Friction is only dependent on the materials and normal force.

I'm sorry I am still confused. I understand friction has its own calculation, (friction coefficient*normal force) but I'm talking about the sum of the forces as a whole. The F = Weight-Friction makes sense if the reference frame starts from the time it instantly slides down the slope to the time it hits the bottom. However, I do think that the push force should also be included because it causes a change in velocity (acceleration). For instance, you exert 500 N push force (parallel to surface in the positive direction) on your friend on a sled, causing her velocity to change from 0 m/s (sitting still) to 2 m/s (sliding downhill). Wouldn't the acceleration caused by the "push force" be different from the acceleration caused by gravity? In a situation where the sled is pushed at the beginning instead of just sliding due to gravitational force, how do we actually take that push force into account?

Does it look like this? -> F = Push + Weight - Friction

 

[lekh2003]

But the force has to be contributing to the normal force. The force between the two objects is the only thing that matters. Friction is equal to Normal Force * Coefficient of Friction. If the push force contributes to normal force, sure go ahead and include it. But any old force on the object doesn't affect friction.

 

[rc1234]

I just realized how they are different systems. The first is the one between the person pushing and the person on the sled. The second is the person on the sled and the snow. Therefore, it doesn't make sense to include the push force in the second reference frame. Last time I took Physics was more than 4 years ago. Thank you for your help!

 

[CWatters]

The OP above doesn't have enough information to work out if the force applied by the man should be included or not.

F=ma refers to the net force and acceleration at some time t. If the man is still pushing at time t then his force should be included. If he's stopped pushing and gone home for lunch then not :-).

I don't understand your references to different systems. It makes no difference if the man pushing is pushing the rider or the sled. For the purposes of a free body diagram the rider and the sled are treated as one body.

 

[CWatters]

If the man pushing only does so for a short time at the start then you have to break the problem down into two phases.. 1) while he is pushing and 2) while he isn't pushing.

Then apply the equations of motion to the two parts separately. First you calculate the velocity of the sled at the point where the man stops pushing. Then you use that velocity as the starting velocity for the second phase.

 

[CWatters]

If you have similar questions it's always best to post the whole problem statement otherwise we risk talking at cross purposes or giving you the solution to a slightly different problem to the one you are trying to ask.

 

[rc1234]

If you have similar questions it's always best to post the whole problem statement otherwise we risk talking at cross purposes or giving you the solution to a slightly different problem to the one you are trying to ask.

I guess I should've been more clear. I said two different systems because what I meant was the man who pushes the sled stops pushing once the friend+sled (yes, they are one body) gains momentum.

Thank you for clarifying and providing solution options. I'll use that advise on my future posts.