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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."
The moment of inertia of a solid cylinder about its axis is given by [katex]I = \frac{1}{2}mR^2[/katex]. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A hoop with a mass [katex]m[/katex] and unknown radius is rolling without slipping on a flat surface with an angular speed [katex]\omega[/katex]. The hoop encounters a hill and continues to roll without slipping until it reaches a maximum height [katex]h[/katex].
An object is experiencing a nonzero net force. Which of the following statements is most accurate?
Consider a solid uniform sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?
Angular momentum cannot be conserved if
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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