UFPM Unit Summary

UFPM 

Unbalanced Force Particle Model

Let's talk about the Unbalanced Force Particle Model! UFPM!
This model will help us to solve for unbalanced forces, even when we don't know the acceleration.

First, let's take a look at Newton's 2nd Law. It tells us that acceleration is directly proportionate to force and inversely proportionate to mass. Using this info, we can find the acceleration, or "a", of any moving object by using the formula a=fnet/m.

Right about now, you might be wondering what an fnet is. Well, fnet stands for net force. You can find the net force by adding together all the forces acting upon your object.

For example: If an 80 kg skier is standing still, waiting for the ski lift, there will be 800 N (Convert kg to N by multiplying the kg amount by 10) of negative gravitational force pulling down on the skier and 800 N of positive normal force pulling up on the skier. So, the net force will be 0.

-800 N + 800 N = 0 N

Now that we know the net force of the skier, we can calculate their acceleration!
Using the formula a = fnet/m, just plug in the amount of force in the fnet and the amount, in kg, of the skier.

a = 0 N/80 kg
a = 0 m/s^2

In this scenario, the force is vertical. However, if the skier were to be observed while they were skiing down a hill, the net force would change.

If the ski slope our skier is racing down a 20 m slope slanted at a 65 degree angle, we can use trigonometry to find the net force and then find the acceleration. In this scenario, the fgx is the fnet because it is the only force acting upon the skier as they go down the slope.

COS (25) = 80 kg/fgx
fgx = 80 (COS (25) )
fgx = 72.50 kg (10) = 725 N

Now that we know the the fnet, we can find the acceleration using our previous formula.

a = fnet/m
a = 725 N/ 80 kg
a = 9.06 m/s^2

Since we know the basics now, we can move on to solving for many other variables. Using kinematics formulas and Newton's 2nd Law, we can solve for acceleration, mass, net force, initial velocity, final velocity, change in time, and displacement.
Here are the equations:

vf = a(change in t) + vi
change in position = 1/2a(change in t^2) + vi(change in t)
vf^2 = vi^2 + 2a(change in position)
ff = coefficient of kinetic friction(fn)

If the skier is traveling at 4 m/s when they begin to descend the slope, how fast will they be going when the reach the bottom of the slope 20 m later?
We know:
vi = 4 m/s
change in position = 20 m
a = 9.06 m/s^2
m = 80 kg

If we use the kinematics formula vf^2 = vi^2 + 2a(change in position)  we can find the value of vf.

vf^2 = 4^2 + (2(9.06))(20)
vf^2 = 16 + 362.4
vf^2 = 378.4
Take the square root of 378.4 to get a vf of 19.5 m/s

We can also find the amount of time it took the skier to descend the mountain using another kinematics formula.

vf = a(change in t) + vi
19.5 = 9.06(change in t) + 4
15.5 = 9.06(change in t)
1.71 = t... approximately

That's a fast skier!
Unless, of course, the snow they are skiing on has friction.

If the coefficient of friction is 0.08, we can find the amount of friction acting upon the skier!
But, first, we have to return to trigonometry to find the fgy which is equal to the fn.

COS (65) = 80 kg/fgy
fgy = 80 (COS (65) )
fgy = 33.81 kg (10) = 338.1 N = fn

ff = coefficient of kinetic friction(fn)
ff = 0.08 (338.1)
ff = 27.05

*BONUS
Atwood Machines!

Atwood Machine Pro-Tips:

Weight = Pull (dangle-y weight)

Mass = Whole System (dangle-y weight + weight on the table)

a = fnet/m

fnet = weight of the hanger (dangle-y weight)

m = mass of system (dangle-y weight + weight on the table)

Everything else you can solve just like a normal object! Just be sure to make free-body diagrams for both weights to help visualize the forces acting upon the weights. 

LA FIN~