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Chapter 9: Problem 13

Concept Simulation 9.1 at illustrates how the forces can vary in problems of this type. A hiker, who weighs \(985 \mathrm{~N}\), is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs \(3610 \mathrm{~N},\) and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?

Short Answer

Expert verified
(a) Force at near end = 723 N; (b) Force at far end = 3872 N.

Step by step solution

01

Understand the System

The problem involves a hiker standing on a bridge, creating two external forces on the system at both ends of the bridge from the supports. The hiker's weight and the weight of the bridge are acting downwards, while the upward forces from the supports are exerted by the concrete supports.
02

Determine the Forces

The system is in equilibrium, meaning the sum of forces and the sum of torques (moments) must be zero. The total downward forces from the hiker and the bridge are \[ W_h = 985 \, \text{N} \quad \text{(hiker's weight)}, \quad W_b = 3610 \, \text{N} \quad \text{(bridge's weight)} \].The problem asks for the forces from the supports on the bridge at two points: Near end (R1) and Far end (R2).
03

Calculate Total Torque

We select the far end of the bridge as the pivot point for torque calculation to find the force at the near end, R1. The torques are calculated as follows:- Torque due to hiker: \[ \tau_h = W_h \times \left(\frac{1}{5} \times L\right) \]- Torque due to the bridge's weight at its center: \[ \tau_b = W_b \times \left(\frac{1}{2} \times L\right) \].Using the equilibrium condition for torques, we have:\[ R_1 \times L = \tau_h + \tau_b \].Solving for \(R_1\), we substitute \(L\) with the length parameter where needed.
04

Solve for R1 (Near End Support Force)

Assuming the bridge is of length \( L \), the torques about the far end due to the hiker and the bridge are:\[ \tau_h = 985 \, \text{N} \times \frac{L}{5} \]\[ \tau_b = 3610 \, \text{N} \times \frac{L}{2} \]Equating torques (since net torque should be zero):\[ R_1 \times L = \left(985 \times \frac{L}{5}\right) + \left(3610 \times \frac{L}{2}\right) \]\[ R_1 = \frac{985}{5} + \frac{3610}{2} \].Calculating yields \(R_1 = 723 \, \text{N}\).
05

Calculate Total Forces

From the equilibrium condition for forces, \[ R_1 + R_2 = W_h + W_b \].Substituting knowns: \[ R_1 + R_2 = 985 + 3610 \].
06

Solve for R2 (Far End Support Force)

Using the result from Step 4 for \(R_1\) in the total force equation: \[ 723 + R_2 = 4595 \].Solving for \(R_2\) gives:\[ R_2 = 4595 - 723 \].Calculating yields \(R_2 = 3872 \, \text{N}\).

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Key Concepts

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

Equilibrium
In physics, equilibrium refers to a state where all forces and torques acting on a system are balanced. This means there is no net force or torque, resulting in no acceleration or rotation.
For objects like the hiker and the bridge, this balance involves both vertical forces and rotational effects.
In this context, equilibrium ensures that the sum of all upward forces (from the supports) equals the sum of all downward forces (from the hiker and the bridge). Moreover, the sum of all torques about any point must be zero.
Understanding equilibrium is crucial because it helps us predict how structures behave under various loads. It allows engineers to design safe buildings, bridges, and other constructions by ensuring they are stable and can support the expected forces.
  • Equilibrium in forces: No overall movement.
  • Equilibrium in torques: No rotation.
Force Calculation
Calculating the forces on each support involves understanding how loads are distributed in equilibrium.
The key is to use the principle that the total forces and total torques are zero in a balanced system.

Determining Force at Supports

First, consider the total weight: this will be the sum of both the hiker鈥檚 and the bridge's weight. In our exercise, these forces act downwards, totaling 4595 N.
Since the system is in equilibrium, each support (near and far end) exerts upward forces to counteract these weights.
When calculating the forces:
  • Use torques to find forces at specific points. This involves selecting a pivot point, such as one end of the bridge, to simplify the calculation.
  • The calculation for the near end involves balancing all torques around the far end.
The resulting force at the near end is 723 N, and the far end is 3872 N, achieved by setting up equations for zero torque and solving them with known distances.
Bridge Mechanics
Bridge mechanics concerns how forces are transmitted through a bridge structure. Understanding these principles ensures that bridges can bear loads like vehicles or pedestrians safely.
Here, we deal with a small, uniform bridge resting on two supports. The mechanics involve distributing the load evenly across these supports.

Load Distribution:

This exercise involves understanding how the weight of the bridge itself and additional loads (like the hiker) distribute along the length of the bridge. The bridge's weight acts at its center, while the hiker adds a point load.

Support Reactions:

The reaction forces at the supports must counterbalance these loads to maintain equilibrium. These reactions depend on the position of the hiker and the symmetry of the bridge.
  • The nearer the load to a support, the greater the force that support needs to exert.
  • This understanding is crucial for designing bridges that can handle dynamic loads effectively.

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Most popular questions from this chapter

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

A small 0.500 -kg object moves on a frictionless horizontal table in a circular path of radius \(1.00 \mathrm{~m} .\) The angular speed is \(6.28 \mathrm{rad} / \mathrm{s} .\) The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than \(105 \mathrm{~N}\), what is the radius of the smallest possible circle on which the object can move?

A massless, rigid board is placed across two bathroom scales that are separated by a distance of \(2.00 \mathrm{~m}\). A person lies on the board. The scale under his head reads \(425 \mathrm{~N},\) and the scale under his feet reads \(315 \mathrm{~N}\). (a) Find the weight of the person. (b) Locate the center of gravity of the person relative to the scale beneath his head.

A helicopter has two blades (see Figure 8-12), each of which has a mass of \(240 \mathrm{~kg}\) and can be approximated as a thin rod of length \(6.7 \mathrm{~m}\). The blades are rotating at an angular speed of \(44 \mathrm{rad} / \mathrm{s}\). (a) What is the total moment of inertia of the two blades about the axis of rotation? (b) Determine the rotational kinetic energy of the spinning blades.

Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius. One has the shape of a hoop and the other the shape of a solid disk. Each wheel starts from rest and has a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each makes the same number of revolutions in the same time. (a) Which wheel, if either, has the greater angular acceleration? (b) Which, if either, has the greater moment of inertia? (c) To which wheel, if either, is a greater net external torque applied? Explain your answers.
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