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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."
A forward horizontal force of 12 N is used to pull a 240 N crate at constant velocity across a horizontal floor. The coefficient of friction is
An object is moving at constant velocity. Which of the following could be the free-body diagram representing the forces acting on the object?
The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of 38.0° with respect to the horizontal. Suppose a wooden crate with a mass of 25.0 kg is placed on Baldwin Street. An additional force of 59 N must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is 0.599, what is the magnitude of the frictional force?
A pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) 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 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
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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