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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. |
Just ask: "Help me solve this problem."
An object is experiencing a nonzero net force. Which of the following statements is most accurate?
A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?
A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 1.00 N·m? Assume that the moment of inertia of a rectangle is [katex] I = \frac{1}{12}M(a^2+b^2) [/katex]
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to ends of a rod of negligible mass of length \( 3x \). The rod is free to turn about a vertical axis through point \( P \). The first mass, \( m \), is located \( x \) away from point \( P \), and therefore the other mass, of \( 2m \), is \( 2x \) from point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_i \) about \( P \). The system is gradually brought to rest by friction.
Derive expressions for the following quantities in terms of \( \mu \), \( m \), \( x \), \( g \), and \( \omega_i \).
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
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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
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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