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**Part (a): Drawing a Free Body Diagram (FBD)**

For the purposes of this explanation, drawing of the FBD is described:

– The ladder rests against a wall with length [katex] L [/katex] and weight [katex] W = 50 \, \text{N} [/katex].

– [katex] F_N [/katex] represents the normal force exerted by the wall on the ladder, acting horizontally at the top of the ladder.

– [katex] F_f [/katex] is the frictional force at the base of the ladder, which opposes the sliding movement, acting horizontally towards the wall.

– [katex] N [/katex] stands for the normal force exerted by the ground on the ladder, vertically upwards.

– [katex] W [/katex] is the weight of the ladder acting downwards from its center of mass.

**Part (b): Finding the Minimum Angle [katex]\theta_{\text{min}}[/katex] so the Ladder Does Not Slip**

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | [katex] \text{Sum of horizontal forces: } F_N = F_f [/katex] | The normal force [katex] F_N [/katex] exerted by the wall is balanced by the frictional force [katex] F_f [/katex] at the base, since there is no other horizontal motion. |

2 | [katex] \text{Sum of vertical forces: } N = W [/katex] | The normal force [katex] N [/katex] from the ground balances the gravitational force [katex] W [/katex] because there is no vertical motion. |

3 | [katex] F_f = \mu N [/katex] | The frictional force [katex] F_f [/katex] can be expressed as the product of the coefficient of static friction [katex] \mu [/katex] and the normal force [katex] N [/katex]. |

4 | [katex] \text{Torque about point at base of ladder: } \\ W \frac{L}{2} \cos(\theta) = F_N L \sin(\theta) [/katex] | Taking torque about the base, counterclockwise torques due to the wall’s normal force [katex] F_N [/katex] should balance the clockwise torque due to the weight [katex] W [/katex]. The distance from the base to the CM is [katex] \frac{L}{2} [/katex]. |

5 | [katex] \frac{W}{2} = F_N \tan(\theta) [/katex] | Further simplification. |

6 | [katex] \frac{W}{2} = \mu W \tan(\theta) [/katex] | Replace [katex] F_N [/katex] with [katex] \mu N [/katex] as solved for in step 3, since [katex] F_N = F_f [/katex] as described in step 1. |

7 | [katex] \tan(\theta) = \frac{1}{2\mu} [/katex] | Simplify further by canceling out [katex] W [/katex] and isolating [katex]\theta [/katex]. |

8 | [katex] \theta = \tan^{-1}(\frac{1}{2\mu}) [/katex] | Finally, take the inverse tan to find [katex] \theta [/katex]. |

9 | [katex] \theta = \tan^{-1}(\frac{1}{2 \times .4}) [/katex] | Substitute [katex] \mu = 0.4 [/katex] into the equation derived. |

10 | [katex] \theta = \tan^{-1}(0.8) \approx 51.34^\circ [/katex] | This is the minimum angle to ensure the ladder does not slip. Calculate [katex] \theta [/katex] in degrees. |

11 | [katex] \boxed{\theta_{\text{min}} \approx 51.34^\circ} [/katex] |
Final answer. |

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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

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Two masses, my = 32 kg and mg = 38 kg, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius R = 0.311 m and mass 3.1 kg. Initially my is on the ground and mg rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of me just before it strikes the ground. Assume the pulley bearing is frictionless.

- Rotational Energy, Rotational Inertia, Rotational Kinematics, Rotational Motion, Torque

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A boy is sitting at a distance [katex] d_1 [/katex] from the fulcrum, and girl is sitting at a distance [katex] d_2 [/katex] from the fulcrum, with [katex] d_1 > d_2 [/katex]. The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.

- Rotational Motion, Torque

- An FBD with 5 forces must be drawn or described for this question.
- [katex] \theta_{\text{min}} \approx 51.34^\circ [/katex]

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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

Conversion Example

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] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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