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 | h = L \sin(\theta) | Calculate the height h of the ladder against the wall using the sine function where \theta is the angle with the ground. |
| 2 | h = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 \, \text{m} | The angle \theta is given as 60^\circ . The \sin(60^\circ) = \frac{\sqrt{3}}{2} . |
| 3 | w_{\text{lad}} = mg w_{\text{lad}} = 20 \times 9.8 = 196 \, \text{N} |
Calculate the weight of the ladder using its mass m and gravitational acceleration g . |
| 4 | w_{\text{person}} = m_{\text{person}}g w_{\text{person}} = 80 \times 9.8 = 784 \, \text{N} |
Calculate the weight of the person using the person’s mass m_{\text{person}} and gravitational acceleration g . |
| 5 | \text{Moment at the bottom} = \text{Moment at the top} | 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 | F_{\text{wall}} \times h = w_{\text{lad}} \times \frac{L}{2} \cos(\theta) + w_{\text{person}} \times d \cos(\theta) | The moment (or torque) at the top due to the force from the wall F_{\text{wall}} must counterbalance the moments generated by the weight of the ladder and person. L is the ladder length, d is the distance where the person stands from the bottom. |
| 7 | F_{\text{wall}} \times 4.33 = 196 \times \frac{5}{2} \times \frac{1}{2} + 784 \times 4 \times \frac{1}{2} | Substitute values for L = 5 \, \text{m}, d = 4 \, \text{m}, \cos(60^\circ) = \frac{1}{2}, h \approx 4.33 \, \text{m} . |
| 8 | F_{\text{wall}} \times 4.33 = 98 \times 2.5 + 784 \times 2 | Simplification of the equation to compute the force exerted by the wall. |
| 9 | F_{\text{wall}} \times 4.33 = 245 + 1568 | Total moments at the top due to the weight of both the ladder and person. |
| 10 | F_{\text{wall}} \times 4.33 = 1813 | Add the moments for the final calculation. |
| 11 | F_{\text{wall}} = \frac{1813}{4.33} \approx 418.71 \, \text{N} | Calculate the force exerted by the wall by dividing the total moment by the height h . |
| 12 | F_{\text{wall}} \approx 419 \, \text{N} | Finding the final value and rounding off to the nearest whole number, providing the force in Newtons. |
Phy can also check your working. Just snap a picture!
A point on the edge of a disk rotates around the center of the disk with an initial angular velocity of 3 rad/s clockwise. The graph shows the point’s angular acceleration as a function of time. The positive direction is considered to be counterclockwise. All frictional forces are considered to be negligible.
What condition(s) are necessary for static equilibrium?
A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?
An object’s angular momentum changes by 10\,\text{kg-m}^2\text{/s} in 2.0 s. What magnitude average torque acted on this object?
A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above. The outside edge of the disk is moving at a linear speed v, and the clay is moving at speed \frac{v}{2}. The clay sticks to the outside edge of the disk. How does the angular momentum of the system after the clay sticks compare to the angular momentum of the system before the clay sticks, and what is an explanation for the comparison?
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_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
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} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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