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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

A wrecking ball (weight \(=4800 \mathrm{~N}\) ) is supported by a boom, which may be assumed to be uniform and has a weight of \(3600 \mathrm{~N}\). As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is \(32^{\circ}\), and the angle between the boom and the horizontal is \(48^{\circ} .\) Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by the hinge at point \(P\).

Concept Questions The drawing shows two identical systems of objects; each consists of three small balls (masses \(m_{1}, m_{2}\), and \(m_{3}\) ) connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at a different place, as shown. (a) Do the systems necessarily have the same moments of inertia? If not, why not? (b) The same force of magnitude \(F\) is applied to the same ball in each system (see the drawing). Is the magnitude of the torque created by the applied force greater for system A or for system B? Or is the magnitude the same in the two cases? Explain. (c) The two systems start from rest. Will system A or system B have the greater angular speed at the same later time? Or will they have the same angular speeds? Justify your answer. Problem The masses of the balls are \(m_{1}=9.00 \mathrm{~kg}, m_{2}=6.00 \mathrm{~kg}\), and \(m_{3}=7.00 \mathrm{~kg}\). The magnitude of the force is \(F=424 \mathrm{~N}\). (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b) Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force moves with the system and always points along the \(4.00-\mathrm{m}\) rod. What is the angular velocity of each system

A pair of forces with equal magnitudes, opposite directions, and different lines of action is called a "couple." When a couple acts on a rigid object, the couple produces a torque that does not depend on the location of the axis. The drawing shows a couple acting on a tire wrench, each force being perpendicular to the wrench. Determine an expression for the torque produced by the couple when the axis is perpendicular to the tire and passes through (a) point \(\mathrm{A},\) (b) point \(\mathrm{B}\), and (c) point C. Express your answers in terms of the magnitude \(F\) of the force and the length \(L\) of the wrench.

A thin uniform rod is rotating at an angular velocity of \(7.0 \mathrm{rad} / \mathrm{s}\) about an axis that is perpendicular to the rod at its center. As the drawing indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

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?
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