# Question on calculated friction force between a ramp and a cart

1. Nov 16, 2013

### JackSac67

I have a question regarding a friction experiment, so here's what I have so far:

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

Calculate the approximate force of friction between the wheels of a cart and a ramp it is rolling on by only using the cart's mass and different values of acceleration up and down the ramp. Here was the general procedure used to collect the acceleration data:

1. Hold cart at rest directly before the sensor on the ramp that will calculate acceleration, then give it a little push so it rolls up the ramp past the sensor. Do this five times and average the value of deceleration.

2. Hold cart at rest directly above the sensor on the ramp that will calculate acceleration, then let it go so it begins rolling down the ramp and past the sensor. Do this five times and average the value of acceleration.

Cart's mass = .265kg
Cart's acceleration on the way down the ramp = .369m/s^2
Cart's acceleration on the way up the ramp = -.435m/s^2

2. Relevant equations

By choosing the acceleration of the cart down the ramp as the positive direction:

ƩFx as the cart rolls down the ramp = .265(9.8)sinθ - Ff = .265(.365)
ƩFx as the cart rolls up the ramp = .265(9.8)sinθ + Ff = .265(-.435)

3. The attempt at a solution

Subtracting the bottom equation from the top equation yields:

-2Ff = .265(.804)
Ff = -.107N (Rounding to significant figures)

Now my question deals with that resultant friction force. I was confused as to why the force came out negative, but unless I made an error somewhere, I think I have an idea as to why. Because the wheels of the cart and the ramp never changed, the coefficient of friction between them stayed constant. And because the mass of the cart never changed, the normal force exerted on the cart stayed the same, and thus there is a constant normal force and coefficient of friction during both runs of the experiment, so the friction force in both runs of the experiment was the same. Because of this fact, must I consider Ff between the ramp and cart as |Ff|, i.e. the sign doesn't matter in this context? This would make sense to me, because I would expect an very small force of friction to act on small, smooth wheels of a cart. If anyone can confirm or help me understand this better, that would be greatly appreciated, thanks.

2. Nov 16, 2013

### haruspex

You have some sign errors. For gravity, you have consistently put down as positive, which is fine. But your two acceleration measures are taken in opposite directions. Clearly, the actual acceleration was positive down ramp in each case, but you are showing the up ramp acceleration is negative, so it must have been measured in the direction of movement. Correct the sign of that and you will find the frictional force is positive.

Btw, the question is misleading. Even if the two accelerations were the same there would be a frictional force on the wheels from the ramp, but it would be directed up the ramp and be of equal magnitude in both cases. (With no friction, the wheels would rotate at constant speed.) The reason they are not the same is torsional friction at the wheels' axles. If the torsional friction is τ and the wheels' radius is r, this leads to an extra frictional force at the ramp of τ/r, opposing direction of movement of the cart.
So the frictional force between ramp and wheels is:
- without torsional friction: F, always up the ramp
- with torsional friction, going up: F-τ/r up the ramp
- with torsional friction, going down: F+τ/r up the ramp
What you are calculating is τ/r (multiplied by number of wheels).
The magnitude of F depends on the moment of inertia of the wheels.

3. Nov 16, 2013

### JackSac67

Ah, I see now. I forgot I had to keep the direction of acceleration consistent for both equations. In that case I found Ff to be approximately .009N, which certainly makes more sense as a friction force between a smooth ramp and small smooth wheels. I understand what you're saying in regards to the torsional friction, but my class hasn't yet talked about these forces.