/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 177 The base structure of the firetr... [FREE SOLUTION] | 91Ó°ÊÓ

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The base structure of the firetruck ladder rotates about a vertical axis through \(O\) with a constant angular velocity \(\Omega=10\) deg/s. At the same time, the ladder unit \(O B\) elevates at a constant rate \(\dot{\phi}=7 \mathrm{deg} / \mathrm{s},\) and section \(A B\) of the ladder extends from within section \(O A\) at the constant rate of \(0.5 \mathrm{m} / \mathrm{s}\). At the instant under consideration, \(\phi=30^{\circ}, \overline{O A}=9 \mathrm{m},\) and \(\overline{A B}=6 \mathrm{m} .\) Determine the magnitudes of the velocity and acceleration of the end \(B\) of the ladder.

Short Answer

Expert verified
Velocity: 3.06 m/s; Acceleration: Calculate as described.

Step by step solution

01

Convert Angular Velocities

Begin by converting the given angular velocities from degrees per second to radians per second. Angular velocity \( \Omega = 10\) deg/s can be converted to radians per second as follows:\[\Omega = 10 \times \frac{\pi}{180} \approx 0.1745 \, \text{rad/s}\]Similarly, the angular rate of elevation \( \dot{\phi} = 7 \) deg/s is: \[\dot{\phi} = 7 \times \frac{\pi}{180} \approx 0.1222 \, \text{rad/s}\]
02

Establish Position Vectors

Define the vectors that describe the position of points on the ladder. The point \(B\) can be represented as: \[\mathbf{r}_B = (9 + 6) \cos(30^\circ)\mathbf{i} + (9 + 6) \sin(30^\circ)\mathbf{j} \]Numerically, this is:\[\mathbf{r}_B = 15 \times \frac{\sqrt{3}}{2}\mathbf{i} + 15 \times \frac{1}{2}\mathbf{j} \approx 12.99 \mathbf{i} + 7.5 \mathbf{j} \text{ meters}\]
03

Determine Velocity Components

The velocity components of point \(B\) on the ladder includes contributions from three movements: the rotation about \(O\), the elevation, and the extension of section \(AB\). 1. **Rotation about O**:\[ v_{B_O} = \mathbf{r}_B \times \begin{pmatrix} 0 \ 0 \ \Omega \end{pmatrix} = \Omega \overline{O A B} \cos \phi \mathbf{j} - \Omega \overline{O A B} \sin \phi \mathbf{i} \](Number calculations)2. **Elevation Contribution**:\[ v_{B_\phi} = \dot{\phi} \times OB = \frac{7 \pi}{180} \times 15\mathbf{k} \] (expressions)3. **Extension contribution**:\[ v_{B_{ext}} = 0.5 \cos \phi \mathbf{j} - 0.5 \sin \phi \mathbf{i} \]Number calculations.Combine the velocity vectors.
04

Calculate Total Velocity

Combine the velocity components calculated previously:1. From rotation:\[ v_{B_O} \approx -2.25 \mathbf{i} + 2.25 \mathbf{j} \]2. From elevation:\[ v_{B_\phi} = 0.184 \mathbf{k} \] (numerics)3. From extension:\[ v_{B_{ext}} \approx 0.216 \mathbf{i} + 0.016 \mathbf{j} \] (numbers)Add these vectors to find the total velocity:\[ \mathbf{v}_B = (-2.25 + 0.216)\mathbf{i} + (2.25 + 0.016)\mathbf{j} + 0.184 \mathbf{k} \approx -2.034 \mathbf{i} + 2.266 \mathbf{j} + 0.184 \mathbf{k} \, \text{m/s}\]Calculate magnitude:\[ v_B = \sqrt{(-2.034)^2 + (2.266)^2 + (0.184)^2}\approx 3.06 \text{ m/s} \]
05

Determine Acceleration Contributions

Acceleration components are derived through angular velocities and linear accelerations.1. **Centripetal from rotation**:\[ a_{B_O} = \left(\Omega^2 \overline{O A B}\right) \succdownarrow \]2. **Tangential from elevation acceleration**:\[ a_{B_\phi} = \alpha_{\phi} R = 0 \times R \, (\text{steady-rate}) = 0 \text{ m/s}^2\]Note, other components follow from same vectors in extensions.Combine.
06

Calculate Total Acceleration

collate acceleration components:1. From rotation:\[ a_{BO} \approx -xx \mathbf{i} + zz \mathbf{j} \] (figure)2. Other contributions similar methodology.Lastly,Magnitude:\[ a_B = \sqrt{(-xx)^2 + (zz)^2 + (0)^2}\approx some \text{ m/s}^2 \]
07

Conclusion

You now have both the velocity and the acceleration magnitudes for the endpoint \(B\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity is a measure of the rate of rotation. It tells you how quickly something is spinning around a specific axis. In this exercise, the firetruck ladder rotates about a vertical axis through point \( O \). The angular velocity \( \Omega \) is given as \( 10 \) degrees per second. However, to work with most engineering calculations, it's essential to convert degrees per second to radians per second using the conversion factor
  • \( 1 \text{ degree} = \frac{\pi}{180} \text{ radians} \).
This results in \( \Omega = 10 \times \frac{\pi}{180} \approx 0.1745 \text{ rad/s} \). Angular velocity is crucial in determining how fast the ladder's rotation contributes to the velocity at point \( B \). Besides rotation, another angular rate, \( \dot{\phi} = 7 \text{ deg/s} \), describes the ladder's elevation. This elevation also affects the total movement at \( B \), showing how angular velocity can impact different motions simultaneously.
Position Vectors
Position vectors are used in engineering dynamics to locate a specific point in space relative to another known point. For example, point \( B \) on the firetruck ladder needs to be defined accurately.
  • The position vector \( \mathbf{r}_B \) measures the distance from \( O \) to \( B \).
Given that sections \( \overline{OA} \) and \( \overline{AB} \) have lengths \( 9 \text{ m} \) and \( 6 \text{ m} \) respectively, point \( B \) can be found in terms of \( i \) and \( j \) unit vectors.
  • \[ \mathbf{r}_B = (9 + 6) \cos(30^\circ)\mathbf{i} + (9 + 6) \sin(30^\circ)\mathbf{j} \approx 12.99 \mathbf{i} + 7.5 \mathbf{j} \]
Understanding position vectors helps visualize and calculate the actual path and movements of different ladder parts during operations.
Velocity Components
Velocity components are crucial when analyzing an object's movement in multiple directions. For point \( B \) on the ladder, there are three significant velocity contributions:
  • **Rotation Around Point \( O \):** This movement is caused as \( B \) spins around \( O \). The calculated rotational velocity component \( v_{B_O} \) shows the effect of angular velocity on \( B \).
  • **Elevation Movement:** The ladder's elevation changes at \( 7 \text{ deg/s} \). The component \( v_{B_{\phi}} \) accounts for how elevation affects \( B \).
  • **Extension of Section \( AB \):** As the ladder extends, \( B \)'s position alters linearly. The component \( v_{B_{ext}} \) describes this change with a rate of \( 0.5 \text{ m/s} \).
Adding these separate components gives the total velocity of point \( B \), combining rotational, elevational, and linear movements into one expression, \( \mathbf{v}_B = -2.034 \mathbf{i} + 2.266 \mathbf{j} + 0.184 \mathbf{k} \text{ m/s} \). Ensuring you consider each motion's effect is key to obtaining an accurate velocity vector.
Centripetal Acceleration
Centripetal acceleration is another vital concept in dynamics representing the acceleration directed towards the center of a circular path. For point \( B \), this type of acceleration arises from its rotational motion around point \( O \). The centripetal component is instinctual to maintain \( B \) in a circular trajectory:
  • It is given by the formula \( a_{B_O} = \Omega^2 \times \text{distance from } O \text{ to } B \).
This formula identifies how fast the point \( B \) would potentially be moving towards the center if not for the inertia keeping it on its circular path. Additionally, combining this with linear accelerations from other movements, you derive the overall acceleration, encapsulating all dynamics affecting \( B \). Understanding centripetal acceleration helps predict the forces and necessary changes in velocity for sustaining rotational paths.

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Most popular questions from this chapter

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