| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | F_{\text{gravity}} = mg | Calculate the gravitational force acting on the crate. This uses the mass of the crate m = 25.0 \, \text{kg} and the acceleration due to gravity g = 9.8 \, \text{m/s}^2. |
| 2 | F_{\text{gravity}} = 25.0 \times 9.8 = 245 \, \text{N} | Performing the calculation gives the gravitational force acting on the crate. |
| 3 | F_{\text{gravity, parallel}} = F_{\text{gravity}} \sin(\theta) | Determine the component of the gravitational force that acts parallel to the slope. Here, \theta = 38.0^\circ. |
| 4 | F_{\text{gravity, parallel}} = 245 \sin(38.0^\circ) = 150.8 \, \text{N} | Calculation of the parallel component of gravitational force using the angle’s sine value. |
| 4.5 | F_{\text{gravity, parallel}} = F_{\text{friction needed}} | Answer – \boxed{150.8 \, \text{N}} . This is the total force frictional force needed to hold the box in place. To verify the static friction can provide enough force we can continue calculate the MAX static friction force. This number should be greater than 150.8 \, \text{N} . |
| 5 | f_{max} = \mu_s F_{\text{normal}} | Maximum force of static friction that can be applied. |
| 6 | F_{\text{normal}} = F_{\text{gravity}} \cos(\theta) + F_{\text{applied}} | The normal force is affected by both the gravitational perpendicular force and the additional applied force of 59 N, which acts perpendicular to the slope, reducing the normal force. |
| 6 | F_{\text{normal}} = 245 \cos(38.0^\circ) + 59 = 252.1 \, \text{N} | Calculation of the normal force considering the cosine of the incline angle and the applied force. |
| 7 | f_s = \mu_s F_{\text{normal}} | Calculation of the max static friction force, where \mu_s is the coefficient of static friction between the crate and the pavement. |
| 8 | f_s = 0.599 \times 252.1 = 151 \, \text{N} | The magnitude of the maximum frictional force is 151 \, \text{N} which is greater than the 150.8 \, \text{N} . to keep it in place. Thus the scenario is possible. |
Phy can also check your working. Just snap a picture!
You are standing on a bathroom scale in an elevator. The elevator starts from rest on the first floor and accelerates up to the third floor, 12 meters above, in a time of 6 seconds. The scale reads 800N. What is the mass of the person?
In the diagram shown a 20 N force is applied to a block B (7 kg). Block A has a mass of 3 kg. Assume frictionless conditions.
A 1000 kg car is traveling east at 20m/s when it collides perfectly inelastically with a northbound 2000 kg car traveling at 15m/s. If the coefficient of kinetic friction is 0.9, how far, and at what angle do the two cars skid before coming to a stop?
A force F is used to hold a block of mass m on an incline as shown in the diagram above. The plane makes an angle of \theta with the horizontal and F is perpendicular to the plane. The coefficient of friction between the plane and the block is µ. What is the minimum force, F, necessary to keep the block at rest?
Which of the following statements about the acceleration due to gravity is TRUE?
150.8 \, \text{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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