Section I: 2D/3D Motion & Force Flashcards
Explain how to find 𝓥ₓ at t = 1.1s when:
An object of mass 1.26 kg is subject to a force that is always directed either toward the East or West and whose magnitude changes sinusoidally with time. With the positive x − axis pointed towards the East, the x−component of the force is given as follows: Fₓ = F₀ cos(ωt), where F₀ = 4N and ω = 1.2 rad/s
NOTE:
t₀ = 0s
X₀ = 0m
𝓥₀ = 0 m/s.
Start by noting that F₀ cos(ωt) = ma so → F₀ cos(ωt) = m(𝒹𝒱ₓ/𝒹t)
manipulating out differential and taking the integral we have:
∫ F₀ cos(ωt)𝒹t = ∫ m𝒹𝒱ₓ → F₀sin(ωt)/ω = m𝒱ₓ
Then solving for 𝒱ₓ:
𝒱ₓ = F₀sin(ωt)/ωm → 4Nsin(1.2 ⋅ 1.1) / (1.2s ⋅ 1.26 kg) = 2.56m/s
Explain how to find X-Coordinate (position) at t = 1.1s when:
An object of mass 1.26 kg is subject to a force that is always directed either toward the East or West and whose magnitude changes sinusoidally with time. With the positive x − axis pointed towards the East, the x−component of the force is given as follows: Fₓ = F₀ cos(ωt), where F₀ = 4N and ω = 1.2 rad/s
NOTE:
t₀ = 0s
X₀ = 0m
𝓥₀ = 0 m/s.
Start by noting that F₀ cos(ωt) = ma , converting a to a differential then integrating both sides and solving for the function of 𝒱ₓ:
F₀ cos(ωt) = m(𝒹𝒱ₓ/𝒹t) → ∫ F₀ cos(ωt)𝒹t = ∫ m𝒹𝒱ₓ → F₀sin(ωt)/ω = m𝒱ₓ
𝒱ₓ = F₀sin(ωt)/ωm
Now convert 𝒱ₓ to the differential (𝒹𝒳/𝒹t), move the 𝒹t to the right side and integrate:
∫ 𝒹𝒳 = ∫ F₀sin(ωt)/ωm 𝒹t → 𝒳 = - F₀cost(ωt)/ω²m | [t = [0 , 1.1]
Then solve from the lower bound t₀ = 0s, to the lower bound t₁ = 1.1s
A toy car rolls down a ramp at a constant velocity. The car’s mass is m = 1.9 kg and the ramp makes an angle of θ = 13 degrees with respect to the horizontal. Assume the rolling resistance is negligible.
(A) What is the Cars Acceleration?
(B) What is the sum of forces in the x-direction?
(C) Assuming Air Resistance (Fᵣ) what is the equation for the sum of Forces in the x-direction ?
(A): The acceleration is 0 since there is a constant velocity
(B): ΣFₓ = 0 since there is no acceleration in the x-direction
(C): ΣFₓ = mgsin(θ) - Fᵣ
If the position of the Crystal is moved to θ after a car accelerates, what is the best function to find the amount the vehicle has accelerated ?
a = gtan(θ)
Since the chains are of equal length and the height (h) of the mass and length of the board (L) is given find:
(A): θ
(B): Tension of Y component of one chain (T₁y)
(C): Tension of 1 Chain Fₜ
(A) θ is found by dividing L (L₂) in half and taking the tan⁻¹(L₂/h) = θ
(B) Since y-component is predicated on the mass, we know T₁y = .5mg
(C) Knowing the Y component and θ, we can solve for the Fₜ using trig identities:
tan(θ) = T₁y/Fₜ → Fₜ = T₁y/cos(θ)
A student throws a water balloon at an initial angle θ above the horizontal with an initial speed 𝓥₀ from a height of h. The target is located on the ground at a horizontal distance d from the student’s feet. Assume that the balloon moves without air resistance.
(A) How to find the 𝓥₀ if given h, d, and θ
Start with the Horizontal Distance Equation to eliminate t₁:
d = 𝓥₀cosθt → t₁ = d/𝓥₀cosθ
Then plug our new value of t₁ into the Vertical DIstance Equation:
0 = h + 𝓥₀sinθt - ½gt² → 0 = h + 𝓥₀sinθ[d/𝓥₀cosθ] - ½g[d/𝓥₀cosθ]² →
[- h - dtanθ][𝓥₀²] = -½gd²/cos²θ
𝓥₀ = √(-½gd²/[cos²θ][- h - dtanθ])
Cranes use a system of two pulleys to provide mechanical advantage. Mechanical advantage is a factor which multiplies the applied force, and hence reduces the force that must be applied. Three possible pulley configurations are shown in the figure, but of course the number of loops may be increased. A crane is attempting to lift a compact car with a mass of m = 905kg against the force of gravity. The crane’s pulley system produces a mechanical advantage of 10.
(A) How many times is it wrapped if the advantage is 10 ?
(B) What is the equation for sum of force in y-direction: ΣFᵧ if the force by the
crane’s lift is F𝒸
(C) What is minimum force needed to lift the car ?
(D) What is equation to solve for acceleration ?
(A): 5 times; since it is wrapped around 2 pulleys
(B): ΣFᵧ = 10F𝒸 - mg
(C) Since ΣFᵧ = 10F𝒸 - mg, we know that F𝒸 must be a value high enough that the it times 10 exceeds mg, thus: F𝒸 = mg/10
(D) Since: F = ma, we know: 10F𝒸 - mg = ma so we can solve for a using
a = [10F𝒸 - mg]/m
Given m₁, m₂, & μₖ find:
(A) Expression for sum of forces in y-direction (ΣFᵧ)
(B) Expression for sum of forces in x-direction (ΣFₓ)
(C) How would Acceleration be found (a)
(D) How would Tension be found (T)
(A): ΣFᵧ = T - m₂
(B): ΣFₓ = T - m₁gμₖ
(C): a = ΣFₑₓ/Σm
(D): T = m₂(g - a)
Explain solution to:
A gymnast with a mass of 47-kg hops on a trampoline. What force does a trampoline (Ft) have to apply to accelerate her straight up at 7.1 m/s² in Newtons?
Since there is a gravitational acceleration of -9.81m/s² the trampoline must have a force that negates the gravitational force
Setting “a” to a = 7.1m/s²:
Ft = F + mg → Ft = 47kg(7.1 + 9.81)m/s²
Ft = 794.77N
Find the (A) Tension (T) upon breakage and (B) The height the rock travels if:
m = 0.87kg
𝓥 = 2.7m/s
L(string) = 0.61m
(A) Since our equation to centripetal force (Fc) = m𝓥² / r, we can plug and chug for the Tension in the string:
Fc = T = (0.87kg ⋅ 2.7²m²/s²) / 0.61m = 10.40m/s²
(B) Assuming it breaks as the 𝓥₀ tangent to our force at break is 2.7m/s, we can use our kinematic equations to solve for the ∆y at 𝓥₁ = 0.
0 = 𝓥₀² + 2a∆y → 0 = (2.7m/s)² - 2(9.81m/s²)∆y
∆y = 0.37m
Circular turns of radius r in a race track are often banked at an angle θ to allow the cars to achieve higher speeds around the turns. Assume friction is not present.
(A) What is the Y-Component of the normal force in terms of N ?
(B) What is the X-Component ?
(C) What is the value of tanθ in terms of Velocity (𝓥), Radius (r), and Gravity (g)?
(A): Nᵧ = Ncosθ, since N follows the angle of the angle, we can find its magnitude aligned with the axis using basic trig identities.
(B): Nₓ = Nsinθ, similair to (A)
(C): tanθ will have to be the normal force in the y-direction divided by the normal force in the x-direction.
tanθ = Nsinθ/Ncosθ, where: Nsinθ = m𝓥²/r, and Ncosθ = F𝑔 = m𝑔 →
tanθ = (m𝓥²/r) / m𝑔 = 𝓥²/r𝑔
A student jumping on a trampoline reaches a maximum height of h = 1.11 m. The maximum height is measured from the surface of trampoline in its unstretched position. The student has a mass of m = 69 kg.
(A) Explain how to solve for the Velocity 𝓥 near height h
(B) Explain how to find the Spring Constant (K) if the distance she falls below the trampoline line is d = 0.75m
(A): We can find velocity by taking our two Kinetic Energy Equations then solving for 𝓥
½m𝓥² = F𝑔h → 𝓥 = √(2𝑔h)
𝓥 = √([2][9.81m/s²][1.11m]) = 4.67m/s
(B): We can find the Spring Constant by using our Spring Kinetic equation with our Gravitational Kinetic Equation:
m𝑔(h + d) = ½Kd² → K = [2m𝑔(h + d)]/d²]
K = (2[69kg][9.81m/s²][1.86m])/([0.75m]²) =
4476N/m
It takes a student a time Δt to push a block of mass m at a constant velocity over a distance d. The coefficient of friction between the block and the ground is μ.
(A) What is the value of the student’s force: F𝘴 in terms of the given variables ?
(B) What is the Energy produced in the system ?
(C) What is the Power produced in the system ?
(A) Since the Block moves at a constant velocity we must assume ΣF = 0 and thus F𝘴 = F𝘧. We can solve for F𝘴 as:
F𝘴 = m𝑔μ
(B) We know KE is just F∗Δd so KE = m𝑔μΔd
(C) Since we know Power is ΔKE/Δt :
Ｐ = (m𝑔μd) / Δt
A car m = 1500 kg is traveling at a constant speed of 𝓥 = 21 m/s. The car experiences a drag force (air resistance) with magnitude Fｄ = 480 N.
(A) Solve for power needed to maintain constant Velocity.
(B) Solve for the power needed to maintain a constant Velocity at a 9º upward angle.
(A) Start by manipulating our Power Function Ｐ = ΔKE/Δt, to get a function relative to our current variables: ΔKE/Δt → F∗Δd/Δt
Since Δd/Δt = 𝓥, our power function can be rewritten as Ｐᵢ = F𝓥, We can from here plug in our values and solve for:
Ｐᵢ = 480N ∗ 21m/s = 10080W
(B) This can easily be found by adding the gravitational force caused by the new angle mgsinθ
Ｐ = [Fｄ + m𝑔sinθ]𝓥 →
Ｐ = (480N + [9.81m/s²][1500kg][sin(9º)]) ∗ 21m/s
Ｐ = 58421W
A top fuel dragster of mass m = 1000.0kg can accelerate and cover 305m in 3.95s from a dead stop. How much Power is the engine producing at the wheels at the end of the race, assuming the acceleration is constant ?
Start by solving for the Constant Acceleration and Velocity at t using the Kinematics equation:
Δd = ½at² → a = 2[305m]/[3.95s²] = 39.1m/s²
𝓥 = at → 𝓥 = [39.1m/s²][3.95s] = 154.44m/s
Then solve for the Power using the Power Equation:Ｐ = F𝓥 = ma𝓥
Ｐ = [1000kg][39.1m/s²][154.44m/s] = 6038604W
A toy car is rolling down the ramp as shown in the figure. The car’s mass is 1.2kg and the ramp makes an angle of θ = 11 degrees with respect to the horizontal plane. Assume the car rolls without friction.
(A) What is Velocity in Cartesian Form ?
(A) a = {gsinθcosθi + gsinθsinθj}
(B) |a| = gsinθ
(A) If Acceleration is on the Y axis and Mass is on the X-axis, which graph describes our relationship ?
(B) If Acceleration is on the Y-axis and Force is on X-axis, which graph describes our relationship ?
(A) Graph 2, and F = x₁y₁
(B) Graph 1, and m = Δx/Δy
The blocks have masses M₁ and M₂ Block 1 is a distance r from the center of the frictionless surface.
What does 𝓥₁ (Velocity of Block 1) have to be for Block 2 to remain stationary ?
The force which keeps Block 1 going about a Uniform Circular Motion (F₁) must be equal to the force which pulls Block 2 downward (F₂):
F₁ = (m₁ 𝓥₁²)/r
F₂ = m₂𝑔
When F₁ = F₂ → (m₁ 𝓥₁²)/r = m₂𝑔 →
𝓥₁ = √( [m₂𝑔r]/m₁ )
If there is also a Frictional Force keeping the object moving at a constant upward speed:
(A) What is the ΣFᵧ
(B) What is the ΣFₓ
(C) What is Magnitude of Fₐ in terms of μ, m, 𝑔 and θ
(A) ΣFᵧ = Fₐsinθ - F𝑔 - ƒ
(B) ΣFₓ = **Fₐcosθ - N
(C) Since we want our force in terms of μ, m, 𝑔 and θ, we will use our answer for (A) to solve for Fₐ
and since ƒ = μN, & N = Fₐcosθ:
Fₐsinθ - F𝑔 - ƒ = 0 → Fₐsinθ - F𝑔 - μN = 0 →
Fₐsinθ - F𝑔 - μFₐcosθ = 0
We can now solve for Fₐ using algebra:
Fₐsinθ - μFₐcosθ = F𝑔 → Fₐ(sinθ - μcosθ) = F𝑔
Fₐ = F𝑔/(sinθ - μcosθ)
A baseball of mass m is spun vertically on a massless string of length L. The string can only support a tension of Tmax before it will break.
(A) Explain how Maximum Velocity at the top of the circle (𝓥ₜmax) is found
(B) What is the Maximum Velocity at the bottom of the circle (𝓥𝔟max)
(A) Analyzing the Centripetal Motion, we can draw two forces at the top that make up or Centripetal Force: Tension: T(↓) and Gravity: F𝑔(↓) so:
F𝔠 = T + F𝑔 → m𝓥²/r = T + mg → 𝓥 = √( r[T + mg]/m )
(B) At bottom, the forces making up the Centripetal Force are Tension: T(↑) and Gravity: F𝑔(↓) so:
F𝔠 = T - F𝑔 → m𝓥²/r = T - mg →
𝓥 = √( r[T - mg]/m )
Two objects, A and B, hang from a system of ropes and ideal pulleys, as shown. Their masses are mᴀ and mʙ, and the angle in the diagram is ϕ.
How can we find the value of mᴀ in terms of ϕ and mʙ ?
We can start by identifying the values of Tension (T) assuming that the system is static:
mᴀ𝑔 = 2Tcosϕ → T = mᴀ𝑔/2cosϕ
Now we know that mʙ = T so mʙ𝑔 = mᴀ𝑔/2cosϕ →
mʙ = mᴀ/2cosϕ
A jet dives vertically at a speed v before pulling out of the dive after following a portion of a circular arc of radius R as illustrated in the image. The pilot can safely withstand an acceleration of a = 8.9g, where g is the acceleration due to gravity. His mass is m.
(A) How do you find minimum R for safe turn ?
(B) If the jet goes around, what is the force from the seat at the top of the turn ?
(A) To find the minimum R, we just set mg = F𝔠 with the knowledge that
1g = 9.81m/s²
m[a][g] = m𝓥²/R → R = 𝓥²/[a][g]
(B) To find the force on the seat, we must recognize this force as the Normal Force (N). Then we must realize at the top of the route F𝔠 = m𝑔 + N solving this:
F𝔠 - m𝑔 = N → m[𝓥²/R - 𝑔] = N
