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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$F_{\parallel} = F\cos(\theta)$$ | The horizontal force \( F \) is resolved into a component along the inclined plane. The component along the plane is given by \( F\cos(\theta) \). |
| 2 | $$F_{g\,\parallel} = mg\sin(\theta)$$ | The gravitational force has a component down the incline given by \( mg\sin(\theta) \). |
| 3 | $$N = mg\cos(\theta)+ F\sin(\theta)$$ | The normal force \( N \) is found by resolving forces perpendicular to the incline. Gravity gives \( mg\cos(\theta) \) and the horizontal force contributes \( F\sin(\theta) \) pushing the block into the plane. |
| 4 | $$F_{f} = \mu N = \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)$$ | The frictional force is given by the coefficient of friction \( \mu \) times the normal force. |
| 5 | $$m\,a = F\cos(\theta)- mg\sin(\theta)- \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)$$ | Applying Newton’s second law along the incline, the net force is the sum of the component of \( F \) along the plane minus both the gravitational and frictional forces. |
| 6 | $$a = \frac{F\cos(\theta)- mg\sin(\theta)- \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)}{m}$$ | This is the final expression for the block’s acceleration \( a \) up the incline in terms of \( m,\,\theta,\,\mu,\,F, \) and \( g \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$0 = F\cos(\theta)- mg\sin(\theta)- \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)$$ | For the block to slide up the plane with constant velocity, the acceleration must be zero. Hence the net force along the incline is zero. |
| 2 | $$F\cos(\theta)- \mu F\sin(\theta) = mg\sin(\theta)+ \mu mg\cos(\theta)$$ | Rearrange the equation by grouping the terms involving \( F \) on the left and the gravitational terms on the right. |
| 3 | $$F \left(\cos(\theta)- \mu \sin(\theta)\right) = mg\left(\sin(\theta)+ \mu\cos(\theta)\right)$$ | Factor out \( F \) on the left-hand side and \( mg \) on the right-hand side for clarity. |
| 4 | $$F = \frac{mg\left(\sin(\theta)+ \mu\cos(\theta)\right)}{\cos(\theta)- \mu\sin(\theta)}$$ | Solve the equation for \( F \) by dividing both sides by \(\cos(\theta)- \mu \sin(\theta)\). |
| 5 | $$\cos(\theta)- \mu\sin(\theta)> 0 \quad \Longrightarrow \quad \tan(\theta) < \frac{1}{\mu}$$ | For \( F \) to be physically meaningful (i.e., a positive real number), the denominator must be positive. Rearranging the inequality yields the condition \( \tan(\theta) < \frac{1}{\mu} \). |
| 6 | $$\boxed{F = \frac{mg\left(\sin(\theta)+ \mu\cos(\theta)\right)}{\cos(\theta)- \mu\sin(\theta)}}$$ | This is the final expression for the magnitude of the applied horizontal force required to make the block slide with a constant velocity, including the physical condition on \( \theta \) and \( \mu \). |
Just ask: "Help me solve this problem."
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are μs=0.6 and μk=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?
Why do raindrops fall with constant speed during the later stages of their descent?
When a horizontal force of \( 4.5 \, \text{N} \) acts on a block on a resistance-free surface, it produces an acceleration of \( 2.5 \, \text{m/s}^2 \). Suppose a second \( 4.0 \, \text{kg} \) block is dropped onto the first. What is the magnitude of the acceleration of the combination if the same force continues to act? Assume that the second block does not slide on the first block.
A spacecraft somewhere in between the earth and the moon experiences 0 net force acting on it. This is because the earth and the moon pull the spacecraft in equal but opposite directions. Find the distance D away from Earth, such that the spacecraft experiences zero net force. The distance between the Moon and Earth is ~3.844 x 108 m.
NOTE: You may need the mass of the earth and moon. You can find this in the formula table.
A pulley system consists of two blocks of mass 5 kg and 10 kg, connected by a rope of negligible mass that passes over a pulley of radius 0.1 meters and mass 2 kg. The pulley is free to rotate about its axis. The system is released from rest, and the block of mass 10 kg starts to move downwards. Assuming that the coefficient of kinetic friction between the pulley and the rope is 0.2, and neglecting air resistance, determine
a) \( a = \frac{F (\cos \theta – \mu \sin \theta) – mg (\mu \cos \theta + \sin \theta)}{m} \)
b) \( F = \frac{\mu m g \cos \theta + \mu F \sin \theta + m g \sin \theta}{\cos \theta} \)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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