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Chapter 2: Problem 221

Determine the vertical rise \(h\) of the load \(W\) during 10 seconds if the hoisting drum draws in cable at the constant rate of \(180 \mathrm{mm} / \mathrm{s}\)

Short Answer

Expert verified
The vertical rise of the load is 1.8 meters.

Step by step solution

01

Understand the Problem

We are given the rate at which a cable is being drawn in, which is 180 mm/s. We need to find out how much the load, denoted as \( W \), will move vertically upwards in 10 seconds. This vertical movement is termed as the 'rise' \( h \).
02

Convert Rate to Consistent Units

The rate of cable being drawn in is given as 180 mm/s. Since we may want the rise in meters for more standard units, convert this rate from mm to meters:\( 180 \text{ mm/s} = 0.18 \text{ m/s} \).
03

Multiply by Time to Find Rise

Use the constant rate of 0.18 m/s over a span of 10 seconds to determine the rise \( h \). Multiply the rate by the time to calculate the vertical rise: \[h = (0.18 \text{ m/s}) \times (10 \text{ s}) = 1.8 \text{ m}\].
04

Conclude with Your Result

After performing the calculations, conclude that the vertical rise of the load \( W \) over 10 seconds is 1.8 meters.

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Key Concepts

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

Vertical Motion
Vertical motion in engineering mechanics often deals with objects moving straight up or down. In this exercise, the focus is on a load \(W\) moving upwards when the cable is hoisted. Understanding vertical motion is critical for solving problems about lifts, cranes, or any mechanical setup where direction is solely upward or downward. In such scenarios, gravity acts as a one-directional force, either aiding or opposing the motion. Here, we neglect air resistance and assume ideal conditions to simplify the calculations. The direction of the load movement can describe how far or fast the object moves in the vertical axis, enhancing our mechanical designs or predictions.
Constant Rate
A constant rate implies that an action, such as drawing in cable in this exercise, occurs uniformly over time. For this problem, we see the cable is drawn in at 180 mm/s. This means every second, the cable retracts by precisely 180 mm without variation. Constant rates simplify calculations because they eliminate the need for additional factors such as acceleration.
  • Simplicity: Each second has an identical change, simplifying predictions and adjustments.
  • Predictability: Knowing the rate of movement helps in planning to avoid possible mishaps in systems involving cranes or lifts.

Identifying when this constant change occurs in real-life applications can assist in maintenance, timing, and ensuring the safety of operations.
Unit Conversion
Unit conversion is a fundamental skill in engineering, especially when working with varied measurement systems. In the exercise, we converted the rate of cable movement from millimeters per second (mm/s) to meters per second (m/s). This is essential because calculations are typically more manageable when using standard units like meters.
  • To convert mm to m, remember that \(1 ext{ m} = 1000 ext{ mm}\).
  • Divide the millimeter measure by 1000 to find the equivalent in meters.

For instance, \(180 ext{ mm/s} = 0.18 ext{ m/s}\). Correct conversion ensures uniformity and accuracy, reducing errors in computation and interpretation.
Step-by-Step Solution
Step-by-step solutions break down complex problems into digestible parts. For the given exercise, we start by identifying what's known and what's needed, simplifying a potentially daunting problem. This method allows students to tackle one portion at a time, fostering understanding.
  • Step 1: Identify the given information and what needs to be calculated.
  • Step 2: Perform necessary unit conversions, ensuring calculations align with standard units.
  • Step 3: Apply the constant rate and compute the vertical rise by multiplying by the time factor.
  • Step 4: Analyze the result and conclude with the correct answer.

By dissecting the problem this way, we not only solve for the current scenario but build problem-solving skills applicable in various engineering situations.

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Most popular questions from this chapter

An amusement ride called the "corkscrew" takes the passengers through the upside-down curve of a horizontal cylindrical helix. The velocity of the cars as they pass position \(A\) is \(15 \mathrm{m} / \mathrm{s}\), and the component of their acceleration measured along the tangent to the path is \(g \cos \gamma\) at this point. The effective radius of the cylindrical helix is \(5 \mathrm{m},\) and the helix angle is \(\gamma=40^{\circ} .\) Compute the magnitude of the acceleration of the passengers as they pass position \(A\)

The acceleration of a particle is given by \(a=-k t^{2}\) where \(a\) is in meters per second squared and the time \(t\) is in seconds. If the initial velocity of the particle at \(t=0\) is \(v_{0}=12 \mathrm{m} / \mathrm{s}\) and the particle takes 6 seconds to reverse direction, determine the magnitude and units of the constant \(k .\) What is the net displacement of the particle over the same 6 -second interval of motion?

The rotating element in a mixing chamber is given a periodic axial movement \(z=z_{0} \sin 2 \pi n t\) while it is rotating at the constant angular velocity \(\dot{\theta}=\omega\) Determine the expression for the maximum magnitude of the acceleration of a point \(A\) on the rim of radius \(r .\) The frequency \(n\) of vertical oscillation is constant.

The acceleration of a particle is given by \(a=-k s^{2}\) where \(a\) is in meters per second squared, \(k\) is a constant, and \(s\) is in meters. Determine the velocity of the particle as a function of its position \(s\). Evaluate your expression for \(s=5 \mathrm{m}\) if \(k=0.1 \mathrm{m}^{-1} \mathrm{s}^{-2}\) and the initial conditions at time \(t=0\) are \(s_{0}=3 \mathrm{m}\) and \\[ v_{0}=10 \mathrm{m} / \mathrm{s} \\]

At time \(t=0\) a small ball is projected from point \(A\) with a velocity of \(200 \mathrm{ft} / \mathrm{sec}\) at the \(60^{\circ}\) angle. Neglect atmospheric resistance and determine the two times \(t_{1}\) and \(t_{2}\) when the velocity of the ball makes an angle of \(45^{\circ}\) with the horizontal \(x\) -axis.
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