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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

A train enters a curved horizontal section of track at a speed of \(100 \mathrm{km} / \mathrm{h}\) and slows down with constant deceleration to \(50 \mathrm{km} / \mathrm{h}\) in 12 seconds. An accelerometer mounted inside the train records a horizontal acceleration of \(2 \mathrm{m} / \mathrm{s}^{2}\) when the train is 6 seconds into the curve. Calculate the radius of curvature \(\rho\) of the track for this instant.

A toy helicopter is flying in a straight line at a constant speed of \(4.5 \mathrm{m} / \mathrm{s}\). If a projectile is launched vertically with an initial speed of \(v_{0}=28 \mathrm{m} / \mathrm{s}\), what horizontal distance \(d\) should the helicopter be from the launch site \(S\) if the projectile is to be traveling downward when it strikes the helicopter? Assume that the projectile travels only in the vertical direction.

A small airplane flying horizontally with a speed of \(180 \mathrm{mi} / \mathrm{hr}\) at an altitude of \(400 \mathrm{ft}\) above a remote valley drops an emergency medical package at \(A\) The package has a parachute which deploys at \(B\) and allows the package to descend vertically at the constant rate of 6 ft/sec. If the drop is designed so that the package is to reach the ground 37 seconds after release at \(A,\) determine the horizontal lead \(L\) so that the package hits the target. Neglect atmospheric resistance from \(A\) to \(B\)

Ball 1 is launched with an initial vertical velocity \(v_{1}=160 \mathrm{ft} / \mathrm{sec} .\) Three seconds later, ball 2 is launched with an initial vertical velocity \(v_{2}\). Determine \(v_{2}\) if the balls are to collide at an altitude of \(300 \mathrm{ft} .\) At the instant of collision, is ball 1 ascending or descending?

At time \(t=0,\) the position vector of a particle moving in the \(x\) -y plane is \(\mathbf{r}=5 \mathbf{i} \mathrm{m} .\) By time \(t=0.02 \mathrm{s}\) its position vector has become \(5.1 \mathrm{i}+0.4 \mathrm{jm}\) Determine the magnitude \(v_{\mathrm{av}}\) of its average velocity during this interval and the angle \(\theta\) made by the average velocity with the positive \(x\) -axis.
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