0 attempts
0% avg
UBQ Credits
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."
A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
A massless rigid rod of length [katex]3d[/katex] is pivoted at a fixed point [katex]W[/katex], and two forces each of magnitude [katex]F[/katex] are applied vertically upward as shown above. A third vertical force of magnitude [katex]F[/katex] may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude [katex]F[/katex] is applied at point?
The moment of inertia of a uniform solid sphere (mass \( M \), radius \( R \)) about a diameter is \( \frac{2}{5}MR^2 \). The sphere is placed on an inclined plane (angle \( \theta \)) and released from rest.
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
