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Pro Tip – Draw an FBD to visualize the all forces and lever arms acting on the ladder. Note that you can split either the forces or the lever arm into components as long as the two are are perpendicular to each other.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] h = L \sin(\theta) [/katex] | Calculate the height [katex] h [/katex] of the ladder against the wall using the sine function where [katex] \theta [/katex] is the angle with the ground. |
| 2 | [katex] h = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 \, \text{m} [/katex] | The angle [katex] \theta [/katex] is given as [katex] 60^\circ [/katex]. The [katex] \sin(60^\circ) = \frac{\sqrt{3}}{2} [/katex]. |
| 3 | [katex] w_{\text{lad}} = mg [/katex] [katex] w_{\text{lad}} = 20 \times 9.8 = 196 \, \text{N} [/katex] |
Calculate the weight of the ladder using its mass [katex] m [/katex] and gravitational acceleration [katex] g [/katex]. |
| 4 | [katex] w_{\text{person}} = m_{\text{person}}g [/katex] [katex] w_{\text{person}} = 80 \times 9.8 = 784 \, \text{N} [/katex] |
Calculate the weight of the person using the person’s mass [katex] m_{\text{person}} [/katex] and gravitational acceleration [katex] g [/katex]. |
| 5 | [katex] \text{Moment at the bottom} = \text{Moment at the top} [/katex] | The torque or moment due to the person and the ladder about the point where the bottom of the ladder contacts the ground must be balanced by the force exerted by the wall. |
| 6 | [katex] F_{\text{wall}} \times h = w_{\text{lad}} \times \frac{L}{2} \cos(\theta) + w_{\text{person}} \times d \cos(\theta) [/katex] | The moment (or torque) at the top due to the force from the wall [katex] F_{\text{wall}} [/katex] must counterbalance the moments generated by the weight of the ladder and person. [katex] L [/katex] is the ladder length, [katex] d [/katex] is the distance where the person stands from the bottom. |
| 7 | [katex] F_{\text{wall}} \times 4.33 = 196 \times \frac{5}{2} \times \frac{1}{2} + 784 \times 4 \times \frac{1}{2} [/katex] | Substitute values for [katex] L = 5 \, \text{m}, d = 4 \, \text{m}, \cos(60^\circ) = \frac{1}{2}, h \approx 4.33 \, \text{m} [/katex]. |
| 8 | [katex] F_{\text{wall}} \times 4.33 = 98 \times 2.5 + 784 \times 2 [/katex] | Simplification of the equation to compute the force exerted by the wall. |
| 9 | [katex] F_{\text{wall}} \times 4.33 = 245 + 1568 [/katex] | Total moments at the top due to the weight of both the ladder and person. |
| 10 | [katex] F_{\text{wall}} \times 4.33 = 1813 [/katex] | Add the moments for the final calculation. |
| 11 | [katex] F_{\text{wall}} = \frac{1813}{4.33} \approx 418.71 \, \text{N} [/katex] | Calculate the force exerted by the wall by dividing the total moment by the height [katex] h [/katex]. |
| 12 | [katex] F_{\text{wall}} \approx 419 \, \text{N} [/katex] | Finding the final value and rounding off to the nearest whole number, providing the force in Newtons. |
Just ask: "Help me solve this problem."
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
The object shown in the diagram below consists of a cylinder of mass \( 100 \) \( \text{kg} \) and radius \( 25.0 \) \( \text{cm} \) connected by four thin rods, each of mass \( 5.00 \) \( \text{kg} \) and length \( 0.75 \) \( \text{m} \), to a thin-outer ring of mass \( 20.0 \) \( \text{kg} \). A small chunk of metal of mass \( 1.00 \) \( \text{kg} \) is welded to the outer ring. Determine the moment of inertia of the entire assembly about the center of the inner cylinder, treating the metal chunk as a point mass. Hint: The moment of inertia of a disk about it center is \(\tfrac{1}{2} M R^2\), a thin rod about it center is \(\tfrac{1}{12}ML^2\), and a thin hoop about its center is \(I = MR^2\). Use only algebra and do not use “·” in your math.
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
What condition(s) are necessary for static equilibrium?
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
419 N
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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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