# Homework Help: Highschool Dynamics Lab. Consistency problem?

1. Sep 29, 2013

### Fifty

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

There is a box pulled up a ramp at an angle, θ via a pulley system with a mass m1 at the end, accelerating downward under the influence of gravity and tension in the rope connected to the box. The goal is to find the coefficient of kinetic friction between the leather bottom of the box and the laquered oak of the ramp.

The forces acting on the box on the ramp are as follows:

Tension force, forward; Normal force, perpendicular to ramp which is elevated 6.9* from horizontal; friction force, backward; and force of gravity, straight down but 6.9* deviation from normal force.

Sorry for the lack of a diagram! The description was the best excuse for FBDs I can come up with at the moment!

The numbers are:

μ = 0.52
m2 = 0.293 kg
m1 = 0.400 kg
ameasured = 3.1 m/s2
μcalculated = 0.48
θ = 6.9 [ramp above the horizontal]

2. Relevant equations

ƩF(Box) (horizontal component (colinear to ramp surface)) = Friction + Gravity (horizontal component) + Tension

let box use subscript 2, let mass on the end of the pulley system use subscript 1

Mathematically:
ƩFNET 2 = FKF2 + FG2x + FT2 = m2a2
For the mass on the end of the pulley (only moving in vertical component):

ƩFNET 1 = FT1 + FG1 = m1a1

where FT1 = - FT2 and |a1| = |a2|

3. The attempt at a solution

I rearrange both statements to solve for a, and since the magnitude of both are equal, I set the equations equal to each other and sub:

(FT1 + FG1)/m1 = (FKF2 + FG2x - FT1)/m2

Then solve for FT1 (I didn't add FT2 on the right, because I subbed it in with - FT1

This is what I end up with:

FT1 = m1(FKF +FG2x - FG1)/(m1 + m2

I calculate FT1 to be 1.3N, which gives me an acceleration of 6.1 m/s2, which is almost double what the measured acceleration was. I don't understand why the theoretical acceleration is higher if the theoretical coefficient of friction is higher. I mean I do understand why -- there is friction in the pulley, air resistance on the box, air resistance on the falling mass -- but I think I am making an error in my calculation because it is almost twice as high as the measured acceleration! (The highest measurement for acceleration was 3.51, average was 3.1). I have checked and re-checked my calculations. Sometimes I get different answers, like 8 N for the tension, which sounds ridiculous because it would be moving up and it clearly was not.

Furthermore, my teacher is not consistent with his ƩF statements in respect to integer signs. For example, for a vertical F net statement he might state: ƩF = FN - Fg instead of FN + Fg so in his equations, the negatives are already accounted for in the algebraic statements, so I get confused as to whether or not to account for the vector direction when inputting the variables. I prefer to add the variables in the statement, then account for the direction of every single vector when subbing in the variables.

The point of the above paragraph is that the equations he gives me are next to useless when checking my algebra. I don't even use it to verify my answers because I don't pay attention to his methods in class (I'm usually too busy doing a sample problem myself to follow his method, then comparing answers). That is why I do not know when he accounts for vector direction while making the F net statements or after, when plugging in variables.

What is the force of tension in the rope connecting the box on the ramp and the hanging mass?

Thanks for the help guys/ladies!!

I proofread this post, but it's pretty early in the morning here and I'm running on fumes, sorry for lack of grammar/poor sentence structure!

Last edited: Sep 29, 2013
2. Sep 29, 2013

### Simon Bridge

You should have cancelled out the tensions instead of the accelerations.
The acceleration is what you want to have left over.

You don't need all those subscripts, and you can pick your directions on your diagram - just make the accelerations consistent.

... so for the mass on the slope:

$T- \mu N - F = ma$ ... see? T is tension, N is normal force, F for the weight-component.
$N = mg\cos\theta$
$F=mg\sin\theta$

(I also use: $f=\mu N$ for friction. Notice it's a f with a tail - for when you do that freehand.)

So your equation is:
$T-\mu mg\cos\theta - mg\sin\theta = ma$ ...(1)

for the hanging mass:
$Mg-T=Ma$ ...(2)

... notice the directions have been chosen consistent with above, and the equations are numbered.
Its also a lot easier to read - so you make fewer mistakes in your algebra.

Cancel the Tension in the equations and solve for acceleration.

If it is still twice what you get in the experiment, then you should have another look at the theoretical coefficient of friction and the assumptions in your experiment setup.

Last edited: Sep 30, 2013