The Force of Friction of Ice
When the lake behind my house froze over me and my friend Matt decided to do some pond skating. First we had to test out the ice to make sure it was frozen thick enough. While I was out on the ice I decided to try doing some jumping jacks while keeping my feet on the ground.
In order for these "sliding jacks" to be possible there needed to be a force of friction between the soles of my shoes and the ice of the lake. I decided to solve for the force of friction present between my shoes and the ice as my feet slid across the ice. In order to do this I had to split the motion up into four separate parts. The first part is the acceleration of my feet moving outwards. The second is the deceleration of my feet moving outwards as they come to a stop. The third piece of motion is the acceleration of my feet moving inwards and the last is the deceleration as my feet come to a stop in the center. The first thing I did was create force diagrams for each of the points of motion with the "sliding jack".
After this I used logger pro to analyze the video. In order to do this I assumed a few things. The first is that the distance between my legs when they are at the furthest point away from each other is 1 m. I also assumed that the motion and force of each foot was equal to the other. I then uploaded the data into a google sheets document and calculated the acceleration, time, change in velocity, change in position, and final velocity for each of the four pieces of motion.
The graph of X Velocity vs. Time:
Using the data above in google sheets I then created graphs for each of the four parts of the motion shown. I calculated the slope for each graph and used that for the acceleration value in my calculations.
Once I calculated all of these values I began creating force equations for each direction. In order to solve for both the force of my foot as well as the force of friction I needed to create a system of equations. I used energy in order to create this system.
Important equations:
∑F = mass•Acceleration
Initial Mechanical Energy + Work of outside forces = Final Mechanical Energy
KE = 1/2•mass•velocity^2
Part 1: Foot acceleration while moving outwards
Mass = 56.25 Kg
Average Acceleration = .0812
Time = .37
ΔV = 1.108
ΔX = .156
V = 1.189
∑Fy = FN - Fg = N - Fg = 0
N = 275.906 N
∑Fx = Ffoot - friction = (m/2)a
= foot - friction = 2.28375 N
Ei + Wf = Ef
0 + (foot+friction)(distance) = .5*(m/2)v^2
.156forces = 19.88045
Outside forces applied = 127.439 N
F-f = 2.28
F+f = 127.439
Force of the foot = 64.8594 N
Force of friction = -62.5794 N
Coefficient of Friction = .2268
Mass = 56.25 Kg
Average Acceleration = -.155
Time = .33
ΔV = -1.865
ΔX = .363
V = 0.182
∑Fy = FN - Fg = N - Fg = 0
N = 275.906 N
∑Fx = Ffoot - friction = (m/2)a
= foot - friction = -4.35938 N
Ei + Wf = Ef
0 + (foot+friction)(distance) = .5*(m/2)v^2
.363forces = .45805
Outside forces applied = N
F-f = -4.35938 N
F+f = .465806
Force of the foot = -6.76726N
Force of friction = -2.4079 N
Coefficient of Friction = .008
Part 3: Foot accelerating while moving inwards
Mass = 56.25 Kg
Average Acceleration = -.153
Time = .2
ΔV = -.9326
ΔX = -.144
V = -1.121
∑Fy = FN - Fg = N - Fg = 0
N = 275.906 N
∑Fx = Ffoot - friction = (m/2)a
= foot - friction = -4.30313 N
Ei + Wf = Ef
0 + (foot+friction)(distance) = .5*(m/2)v^2
.2forces = 17.6715
Outside forces applied = 88.3576N
F-f = -4.30313 N
F+f = 88.3576 N
Force of the foot = -46.33034765625 N
Force of friction = 42.02722265625 N
Coefficient of Friction = .168
Part 4: Foot decelerating while moving inwards
Mass = 56.25 Kg
Average Acceleration = .165
Time = .2
ΔV = 1.057
ΔX = -.133
V = -0.064
∑Fy = FN - Fg = N - Fg = 0
N = 275.906 N
∑Fx = Ffoot - friction = (m/2)a
= foot - friction = 4.64063 N
Ei + Wf = Ef
0 + (foot+friction)(distance) = .5*(m/2)v^2
.2forces = 0.0567
Outside forces applied = .288N
F-f = 4.64063 N
F+f = .288 N
Force of the foot = 2.46431 N
Force of friction = −2.17631 N
Coefficient of Friction = .0079
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