To kick off this unit, we started with a skateboard challenge to witness an unbalanced model firsthand. Right off the bat, Will, Hank, Ally and I discovered that the more mass you add to the skateboard, the harder it is to accelerate. The more force you use to pull the skateboard, the faster it accelerates. When Ally got on the board it was easy for her to get going, but when Hank got on the skateboard it was more difficult for him to accelerate and once he did, it was harder to stop him. Therefore we concluded that acceleration and mass have an inverse relationship and acceleration and force have a directly proportional relationship.
The next study we did in order to help us discover more unbalanced forces in every day life was the elevator challenge. After Hank stepped on the scale and we rode the elevator up and down a few times, we noticed that his weight increased as we accelerated upward and it decreased when we went back down to Floor 1. This is because the forces are unbalanced and we were then introduced to the formula: Fnet= Fg+Fn. Fg & Fn are both on the horizontal axis and when you add the forces on an axis you can find the Fnet.
Practice w/ N2L
You have a 2kg box being pushed with a net force of 4N.
a) calculate the acceleration of the box
a=Fnet/m
a=4/2
a=2m/s^2
b) how would acceleration compare if the mass box was reduced in 1/2?
a=Fnet/m
a=4/1
a=4 m/s^2 It would double
c) how would acceleration compare if the force was doubled?
a=Fnet/m
a=4/1
a=4 m/s^2 It would double
d) how would the acceleration compare if the force was cut in 1/2?
a=Fnet/m
a=2/2
a= 1 m/s^2 It would cut in half
e) how would the acceleration compare if the mass was doubled?
a=4/4
a= 1m/s^2 It would cut in half
f) how would the acceleration compare if the mass was doubled and the force doubled?
a=8/4
a=2 m/s^2 It would stay the same
g) how would the acceleration compare if the force and the mass were both cut in 1/2?
a=2/1
a=2m/s^2 It would stay the same
h) how would the acceleration compare if the force was doubled and the mass was cut in 1/2?
a=8/1
a=8m/s^2 It would quadruple
i) how would the acceleration compare if the force was cut in 1/2 and the mass were doubled?
a=2/4
a=0.5m/s^2 It would be a quarter of the original acceleration
j) How would the acceleration compare if the force was quadrupled?
a=16/2
a=8m/s^2 It would quadruple
k) Would the acceleration compare if the mass was cut down to 1/4th the original?
a=4/.25
a=8m/s^2 It would quadruple
Important Formulas
vf = at + vi
Δx = 1/2 aΔt^2 + viΔt
vf^2 = vi^2 + 2aΔx
a = Δv/Δt
a = Fnet/mass
Practice Problems
A 4600 kg helicopter accelerates upward at 2.0 m/s^2. Determine the lift force exerted on the propellers by the air.
Fnet=m x a
Fnet= 4600 x 2
Fnet= 9200N
*In order to find Fair you must find Fnet first & then add that to Fg. You also know that Fair > Fg because the helicopter is accelerating upward.*
9200+46,000=55,200N (Fair)
2. If an unwilling cat with a mass of 16kg is being dragged across the floor at a constant velocity by a leash with a force of 56 newtons, and the leash is 45 degrees to the horizontal, what would be the normal force? (by: VDC)
m=16kg
Fpull=56N
Fn= ?
cos(45)=adj/56
56*cos(45)=adj/56*56
Fn=39.59N
3. A monster truck with a mass of 925 kg starts from rest and travels 80.0m in 7 seconds. The truck is uniformly accelerating. Find the Fnet.
m=925kg
Vi=0m/s
∆x=80m
t=7s
∆x=1/2a∆t^2+Vi∆t
80=1/2a(7)^2+0*7
80=a(24.5)
a=3.26m/s^2
Fnet=m*a
Fnet=925*3.26
Fnet=3015.5
Conclusion
After this unit, my classmates and I can confidently say that we can successfully solve an entire unbalanced FBD while finding every missing part. Compared to the beginning of the year or even last unit, this is a huge improvement and long stride towards mastering the concept of physics. After experimenting with the skateboards and testing the fluctuation of weight in the elevator, it is evident that physics is apart of our daily lives.
