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Objective: Determine the minimum pressing force required to hold the picture against the wall.
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F_{\text{friction}} = \mu_s F_{\text{normal}}[/katex] | The maximum static frictional force is the product of the coefficient of static friction and the normal force. |
| 2 | [katex]F_{\text{gravity}} = mg[/katex] | The force of gravity acting on the picture. |
| 3 | [katex]F_{\text{friction}} = F_{\text{gravity}}[/katex] | To prevent the picture from sliding, the frictional force must equal the gravitational force. |
| 4 | [katex]\mu_s F_{\text{normal}} = mg[/katex] | Combine steps 1 and 3. |
| 5 | [katex]F_{\text{normal}} = \frac{mg}{\mu_s}[/katex] | Solve for the normal force, which is the pressing force required. |
| 6 | [katex]F_{\text{normal}} = \frac{(1.42\text{ kg})(9.8\text{ m/s}^2)}{0.62}[/katex] | Substitute the values for [katex]m[/katex], [katex]g[/katex], and [katex]\mu_s[/katex]. |
| 7 | [katex]F_{\text{normal}} = \frac{13.916\text{ N}}{0.62}[/katex] | Calculate the gravitational force. |
| 8 | [katex]F_{\text{normal}} = 22.445\text{ N}[/katex] | Calculate the minimum pressing force. |
Final answer for the minimum pressing force: [katex]\boxed{F_{\text{normal}} = 22.45 \text{ N}}[/katex]
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On a harsh winter day, a 1500 kg vehicle takes a circular banked exit ramp (radius R = 150 m; banking angle of 10 degrees) at a speed of 30 mph, since the speed limit is 35 mph. However, the exit ramp is completely iced up (= frictionless). To make matters worse, a wind is blowing parallel to the ramp in a downward direction. The wind exerts a force of 3000 N. Under these conditions, can the driver continue to follow a safe horizontal circle on the exit ramp and stay below the speed limit? To convert mph into m/s use 1 mi = 1607 m and 1 hr is 3600 s.
An object has a mass of 10 kg. For each case below answer the questions and provide an example.
A 1.5 kg block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is 0.300 and the coefficient of static friction is 0.400. Calculate the amount of time that passes from when the force is applied to when the block stops.
The block is moving horizontally at a constant velocity. There are two applied forces on the object as shown in the image. In which direction is the friction force acting on the object?
When the speed of a rear-wheel-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
22.45 N
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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