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Chapter 12: Problem 82

A \(95,000-\mathrm{kg}\) train locomotive starts across a 280 -m-long bridge at time \(t=0 .\) The bridge is a uniform beam of mass \(23,000 \mathrm{kg}\) and the travels at a constant 80.0 \(\mathrm{km} / \mathrm{h}\) . What are the magnitudes of the vertical forces, \(F_{\mathrm{A}}(t)\) and \(F_{\mathrm{B}}(t),\) on the two end supports, written as a function of time during the train's passage?

Short Answer

Expert verified
The force on support A decreases and on B increases as the train moves across. Expressed as functions of time: \[ F_A(t) = 112,815 - \frac{3.328 t}{280} \times 931,950 \] \[ F_B(t) = 112,815 + \frac{3.328 t}{280} \times 931,950 \].

Step by step solution

01

Convert speed to m/s

The train's speed is given as 80 km/h. First, we need to convert this to meters per second (m/s) by using the conversion factor: 1 km/h = 0.27778 m/s. 80 km/h = 80 * 0.27778 m/s = 22.22 m/s.
02

Calculate time to cross the bridge

Determine the time it takes for the train to cross the entire length of the bridge. This can be calculated using the formula for time: \[ t = \frac{d}{v} \]Where \( d \) is the distance (280 m) and \( v \) is the velocity (22.22 m/s).\[ t = \frac{280 \, \text{m}}{22.22 \, \text{m/s}} = 12.6 \, \text{s} \]
03

Understand the forces acting at the supports

There are vertical forces at supports A and B, denoted as \( F_A(t) \) and \( F_B(t) \). As the train moves, different portions of its weight contribute to these forces. The bridge's weight also acts evenly on both supports.
04

Calculate the force equilibrium

The sum of forces in the vertical direction must be zero due to equilibrium. The bridge's weight (\( W_b = 23,000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 225,630 \, \text{N} \)) is evenly distributed, giving each support an initial force of:\[ \frac{225,630 \, \text{N}}{2} = 112,815 \, \text{N} \]The train's weight (\( W_t = 95,000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 931,950 \, \text{N} \)) adds to these forces as it moves across the bridge.
05

Consider train's position effect

Consider the position of the train as it moves across the bridge. At any point \( t \), the center of mass of the train is at some point on the bridge. If \( x(t) \) is the distance of the train's center of mass from support A at time \( t \), then\[ x(t) = 22.22t \]
06

Calculate moments about support

Apply the principle of moments about one of the supports, such as A. For equilibrium:\[ F_B(t) \times 280 = 225,630 \times 140 + 931,950 \times \left(\frac{x(t)}{2}\right) \]Rearrange to solve for \( F_B(t) \), considering geometry and balance. Then, find \( F_A(t) \) using equilibrium of forces. Use a system of simultaneous equations based on equilibrium conditions to find both forces.
07

Express forces as functions of time

Using calculations and substitutions derived from the balance of forces and moments, express the forces as functions of time:\[ F_A(t) = 112,815 - \frac{3.328 t}{280} \times 931,950 \]\[ F_B(t) = 112,815 + \frac{3.328 t}{280} \times 931,950 \]Here, the coefficients account for the appropriate lever arms and force distributions.

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Key Concepts

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

Force Equilibrium
In mechanics, force equilibrium is a fundamental concept. It refers to the state where the sum of all forces acting on a system is zero, resulting in no net motion. For the train crossing a bridge, we consider vertical forces to understand the equilibrium.

At equilibrium, the forces acting upwards on the supports of the bridge must balance the combined weight of the train and the bridge itself. Let's break this down:
  • The weight of the bridge acts evenly across its length. This means each support shares the bridge's weight.
  • The weight of the train affects these forces as it moves. It changes the distribution of forces due to its motion across the bridge.
When these conditions are met, the system is in equilibrium, and we can calculate the supporting forces needed to balance the system using the principle of moments.
Center of Mass
The center of mass is a key concept when analyzing mechanics problems. It is the point at which the total mass of a body or system is considered to be concentrated for the purposes of movement analysis.

In the case of a train on a bridge, the center of mass affects how weight is distributed at different positions as the train travels:
  • As the train moves from one end of the bridge to the other, the center of mass of the train shifts, altering the forces on the supports.
  • We define the position of the center of mass relative to the support points to find out how these forces change with time.
This understanding is crucial in determining how the load applied by the train changes dynamically as it passes over the bridge.
Moment of Force
The moment of force, also known as torque, is an important concept when discussing how forces affect rotation or pivoting around a point. It is defined as the perpendicular distance from a point to where the force is applied, multiplied by the force itself.

For the train on the bridge, we consider moments about one of the supports to maintain equilibrium. Calculating the moment allows understanding of leverage and how forces are distributed. For instance:
  • The moment caused by the bridge's weight about a support point is constant because the bridge doesn't move.
  • The train, however, causes varying moments as it changes position. We calculate these moments to find out how support forces vary with the movement of the train.
Using balance and geometry, we can find expressions for the forces at both supports during the train's crossing.
Uniform Beam
A uniform beam means a structure that has a consistent mass distribution along its length. In our scenario, the bridge acts as a uniform beam. This affects how forces are calculated.

Some implications of the bridge being a uniform beam include:
  • The bridge's weight is evenly distributed, leading to equal support reactions initially.
  • Knowing it is uniform simplifies calculation since every section is equally heavy, making calculations straightforward when breaking down forces.
Uniformity allows simpler use of mechanics principles, like force equilibrium and torque, to evaluate situations involving distributed loads, such as this one.
Train Dynamics
Train dynamics involves understanding how trains move and the forces they exert on structures like bridges. This involves both static and dynamic considerations.

Key aspects to consider include:
  • The train travels at a constant speed across the bridge, affecting how and when forces change on the supports.
  • The weight of the train plays a significant role in calculating the dynamic response of the bridge.
  • As the train moves, forces on the supports change, affecting the equilibrium state of the bridge.
Understanding these dynamics is essential for determining not just the instantaneous forces, but how they vary over time as the train progresses along the path.

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Most popular questions from this chapter

A tightly stretched "high wire" is \(36 \mathrm{~m}\) long. It sags \(2.1 \mathrm{~m}\) when a 60.0 -kg tightrope walker stands at its center. What is the tension in the wire? Is it possible to increase the tension in the wire so that there is no sag?

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