Section #2: Force & Motion Flashcards
What is the equation to Gravity ?
F₉ = m𝑔
Where gravity (𝑔) = 9.81m/s²
What is the equation to Force ?
F = ma
What is the equation for Weight
W = m𝑔
What does normal force describe ?
Normal force is the complete opposite to an active force. It describes the force which is opposed to another:
What is the equation to Normal Force ?
Static System:
N = m𝑔
Dynamic System:
N = m(𝑔 + a) ; When acceleration is given in a system, we can add it to the equation to know how much force a body has on another.
What does Tension describe ?
The force of a rope on an object: (example: The force the rope makes on Tarzan as he swings; this is typically parallel to the actual rope.)
What is the formula for Tension when acceleration is Downward ?
What is the formula for Tension when acceleration is Upward ?
Downward: T = 𝓂(𝑔 - a)
Upward: T = 𝓂(𝑔 + a)
How is Tension found in this Problem when M and m are given?
Start by finding setting Right/Downward motion as our positive direction.
Then apply our equation: aₛ= ΣFₑₓₜ / Σmass
Then once we obtain a we may plug it into T = Ma
How to find the Tension in of the rope in this system
Since the Normal force is perpendicular to the slope we can create a right triangle with the block as m𝑔 as the magnitude.
Tension can then be solved using
T = m𝑔sinθ
N = m𝑔cosθ
How would the Tension be found in The 3 Cords
Cord 3: We can use T₃ = M(a)
For Cord 1 and 2 we can use 2 equations and 2 unknowns (noting that the system is stationary):
∑T𝓍 = -T₂cosθ₂ + T₁cosθ₁ = 0
∑T𝓎 = T₂sinθ₂ + T₁sinθ₁ - T₃ = 0
Then use system of equations to solve for T₁ & T₂
How would the Net Force acting on a swinging object be found ?
- We start by considering the F𝑔 of the object as the Normal force (N) created by that gravity.
- We can than use the Force parallel to the rope as our tension as our magnitude.
We can now solve for the Tension using trig since:
T = N / sinθ
What is the equation of friction ?
ƒ = μN
Where μ represents the coefficient of static friction and N is the Normal Force
If A & B accelerate forward in this system, how can we find the minimum mass of C needed to make this system static given a friction ƒₛ between A and the surface ?
Note that when a system is static: Tension in Block A is equal to ƒₛ: (T = ƒₛ)
It is also true that Tension in Block B is equal to the weight of Block B: (T = m₆𝑔)
Lastly the Normal force N in this system will be N = mₐ𝑔 + m꜀𝑔
Thus we can use the equations to form
ƒₛ = m₆𝑔 → μₛ(mₐ𝑔 + m꜀𝑔) = m₆𝑔
How do you find, with a given μₛ, the maximum volume a cylinder can be filled to before substance starts to leak ? Knowing that V = Ah/3 ?
Start by Identifying the volume of a Cylinder as πr²h/3
Since a cylinder can be cut into a right triangle, putting θ at the bottom corner, we know that r is the adj and h is the opp, this h = rtanθ.
Now now we can establish that the normal force on the edge will be perpendicular to the hypotenuse, and the Weight of the cylinder being directed downward. This will create a right triangle with:
m𝑔cosθ = N
m𝑔sinθ = ƒₛ
Now we can solve for h starting with ƒ = μN → m𝑔sinθ = (μ)m𝑔cosθ → μ = tanθ
We can plug this back into the equation after establishing h = rμ
πr³μ/3
What does Drag Force describe ?
Drag Force is the force from air as a body travels through it. It typically slows down the object until it reaches a terminal velocity.
How are the Centripetal Force, Normal Force, and Frictional Force oriented in a situation where a car goes about a circular curve at constant velocity ?
Where:
F = m(V²/R)
N = F𝑔
ƒₛ = μN
How is Tension found in both strings along with the Centripetal Force of the Ball ?
Note: Mass of the ball (m) along with the Tension of the top rope (T₁) is given.
Start by noting that as the ball spins, the X and Y directions does not change. So:
∑F𝓍 = 0, ∑F𝓎 = 0. We can then line these equations up with all the given components:
∑F𝓎: T₁𝓎 - T₂𝓎 - m𝑔 = 0 → T₂sinθ = m𝑔 - T₁sinθ
∑F𝓍: - T₁𝓍 - T₂𝓍 + Fc = 0 → T₁cosθ + T₂cosθ = Fc
How can the Work done on the body in this system be found given the magnitude of each force and the displacement (d) of the body.
Start by finding the sum of the forces in their respective directions:
∑F𝓍 = -F₁ - F₂sinθ + F₃cosθ
∑F𝓎 = -F₂cosθ + F₃sinθ
Then find the magnitude F, F = √(∑F𝓍² + ∑F𝓎²).
We can now use our eqxn for work: 𝓦 = Fd
If given L (the length of the cord in its relaxed state), 𝓶 (the mass of the jumper), and the height of the bridge from the ground. How could one find distance (𝒳) the jumper will be from the ground at the lowest point of the jump ?
Start by noting that the lowest point of the jump will be when the velocity is 0. Since we know that energy is predicated on the length a certain force is conducted, we can use our Work equations:
ΔKE = ΔU
Since U is both Elastic and predicated on Gravity, we know ΔU = Uₑ + U₉. This makes our equation:
0 = Uₑ + U₉.
Noting that U₉.(downward) = - mg(L + 𝒳) and Uₑ = ½K𝒳₁² we can write our equation as:
0 = ½K𝒳₁² - mg(L + 𝒳) and solve for 𝒳 using quadratic solutions.
Given the diagram below, At what speed will he arrive at the top of the slope of the lower peak (ignore friction).
How can we approximate the Coefficient of Kinetic Friction between the Snow and Skies in order to make the skier top just at the top of the lower peak?
Since we are starting at the top of the Higher Peak, and the energy doesn’t change due to friction, we can set our equation:
U𝑔₀ = U𝑔₂ + KE₂ and solve by plugging in what we know:
m(9.81m/s²)(850m) = m(9.81m/s²)(750m) + ½m𝓥₂² → 𝓥₂ = √(1962m²/s²)
𝓥₂ = 44.3m/s
To find the Coefficient of Kinetic Friction, we know ƒₖ𝒳 = Eₜₕ and that Eₜₕ must = ΔEₘₑ in order for our skies to stop: (note 𝒳 = 3200)
ΔEₘₑ = Eₜₕ → U𝑔₀ - U𝑔₂ = ƒₖ𝒳 →
m𝑔(850m) - m𝑔(750m) = μₖm𝑔cos(30)(3200m)
μₖ = 0.036
