**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 L and weight W = 50 \, \text{N} .

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

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

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

– W is the weight of the ladder acting downwards from its center of mass.

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

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

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

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

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

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

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

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

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

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

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

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

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

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- Statistics

Intermediate

Mathematical

MCQ

A wheel of radius *R* and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses m, M, and 2M, respectively, are mounted on the rim of the wheel, as shown above. If the system is in static equilibrium, what is the value of m in terms of M ?

- Rotational Inertia, Rotational Motion, Torque

Beginner

Mathematical

MCQ

A meter stick of mass .2 kg is pivoted at one end and supported horizontally. A force of 3 N downwards is applied to the free end, perpendicular to the length of the meter stick. What is the net torque about the pivot point?

- Rotational Motion, Torque

Advanced

Proportional Analysis

MCQ

A ice skater that is spinning in circles has an initial rotational inertia I_{i}. You can approximate her shape to be a cylinder. She is spinning with velocity ω_{i}. As she extends her arms she her rotational inertia changes by a factor of x and her angular velocity changes by a factor of y. Which one of the following options best describe x and y.

- Angular Momentum, Rotational Motion

Intermediate

Conceptual

MCQ

A student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is I = MR^2. Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers.

- Rotational Motion, Torque

Beginner

Mathematical

MCQ

An object weighing 120 N is set on a rigid beam of negligible mass at a distance of 3 m from a pivot, as shown above. A vertical force is to be applied to the other end of the beam a distance of 4 m from the pivot to keep the beam at rest and horizontal. What is the magnitude *F* of the force required?

- Torque

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

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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_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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