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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."
The system above is NOT balanced since \(m_2\) is twice the mass of \(m_1\). Which of the following changes would NOT balance the system so that there is 0 net torque? Assume the plank has no mass of its own.
A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at [katex] 470 \, \frac{rad}{s^2}[/katex]. The ball is released after [katex]\frac{3}{4} [/katex] of a revolution. How high does it go above the center of the wheel?
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
Keep in mind its asking about linear VELOCITY (which is speed + direction and that direction is changing). Use simple explanations and do not mention unit vectors.
A solid metal bar is at rest on a horizontal frictionless surface. It is free to rotate about a vertical axis at the left end. The figures below show forces of different magnitudes that are exerted on the bar at different locations. In which case does the bar’s angular speed about the axis increase at the fastest rate?
Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
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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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