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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 \( \mu \). |
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You pull a box with a constant force across a frictionless table using an attached rope held horizontally. If you now pull the rope with the same force at an angle to the horizontal (with the box remaining flat on the table). Does the acceleration of the box increase, decrease, or remain the same if the rope is pulled at an angle? Explain.
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 net force of \( 8.0 \) \( \text{N} \) accelerates a \( 4.0 \) \( \text{kg} \) body from rest to a speed of \( 5.0 \) \( \text{m s}^{-1} \). Which of the following is equal to the work done by the force?
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.
The Earth’s radius is \(6.37 \times 10^{6} \, \text{m}\). What is the radius of a planet that has the same mass as Earth but on which the free-fall acceleration is \(5.50 \, \text{m/s}^2\)?
Which of the following must be true for an object at translational equilibrium?
A car is going through a dip in the road whose curvature approximates a circle of radius \( 200 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 20\% \) more than their normal weight \( (1.2\,W) \)?
The gravitational force that the Moon exerts on Earth is often cited as the source of the tides we witness. However, the gravitational force the Sun exerts on Earth is over \(100\) times greater than the force the Moon exerts on Earth.
Why is the force from the Moon credited for the tides, and not the force from the Sun?
In the diagram shown, a \(20 \, \text{N}\) force is applied to block \(B\) (\(7 \, \text{kg}\)). Block \(A\) has a mass of \(3 \, \text{kg}\). Assume frictionless conditions.
A westward–moving car is changing its speed. The net force on the car ____.
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] |
Metric Prefixes
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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