/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 Minimizing the Tension. A heavy ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Minimizing the Tension. A heavy horizontal girder of length \(L\) has several objects suspended from it. It is supported by a frictionless pivot at its left end and a cable of negligible weight that is attached to an I-beam at a point a distance \(h\) directly above the girder's center. Where should the other end of the cable be attached to the girder so that the cable's tension is a minimum? (Hint: In evaluating and presenting your answer, don't forget that the maximum distance of the point of attachment from the pivot is the length \(L\) of the beam.)

Short Answer

Expert verified
Attach the cable at the far end of the beam, such that \( x = L \).

Step by step solution

01

Understand the Physical Setup and Forces Involved

Consider the beam as a uniform horizontal rod of length \( L \), pivoted at one end with various masses suspended from it. The tension in the cable provides support to the beam, while the beam is also supported by the pivot at one end. The tension should balance the torques due to the gravitational forces of the suspended weights and the beam's own weight.
02

Define the Torque and Equilibrium Conditions

For the system to be in equilibrium, the sum of torques around the pivot must be zero. If the other end of the cable is attached at point \( x \) from the pivot, the tension \( T \) in the cable creates an upward force at point \( (L/2, h) \) above the beam's center. The torque due to the cable is \( T \cdot x \) and must balance the torques from the suspended objects and the weight of the beam.
03

Express and Derive Torque Equations

Calculate the gravitational torque. Let \( W \) be the total weight acting at the center of mass \( (L/2) \). The gravitational torque about the pivot is \( W \cdot (L/2) \). The cable tension must satisfy the equation: \( T \times x = W \cdot (L/2) \).
04

Solve for Minimum Tension by Adjusting x

Rearrange the torque equilibrium equation to solve for tension as a function of \( x \): \( T = \frac{W \cdot (L/2)}{x} \). To minimize tension \( T \), maximize \( x \). Since the maximum \( x \) is \( L \), the point of attachment should be at the far end, i.e., \( x = L \).
05

Conclude Optimal Attachment Position

To minimize the tension in the cable, attach the other end of the cable at the farthest possible point from the pivot, which is at the end of the beam, \( x = L \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Torque Equilibrium
When we talk about torque equilibrium, we're diving into the world of balancing rotational forces. Imagine a seesaw perfectly balanced on its point of rotation, or "pivot." For an object like the horizontal girder in our scenario, achieving torque equilibrium means making sure that all forces causing it to rotate one way are perfectly countered by forces trying to rotate it the opposite way.

In physics, torque is calculated by multiplying the force applied with the distance from the pivot point where the force is acting. Mathematically, if we look at a force applied at a distance:
  • Torque = Force × Distance from pivot
If the girder is to remain in a stable or "equilibrium" state, the total torque—in all directions—must equal zero.

This means:
  • The torque caused by the tension in the cable, trying to tilt or spin the girder one way, must be exactly balanced by the torque due to the combined weight of the girder and the objects hanging from it.
Mastering torque equilibrium simplifies the problem of balancing any set of rotational forces, ensuring stability.
Mechanical Equilibrium
Mechanical equilibrium is reached when an object is in a state of complete balance. This ensures the object is not accelerating in any direction. For a structure like a beam with objects hanging from it, mechanical equilibrium involves both translational and rotational aspects.

Let's break it down:
  • Translational equilibrium involves the sum of all forces acting on the object being zero. This means there's no net force pushing it in any direction, ensuring the beam doesn’t slide or move side-to-side.
  • Rotational equilibrium, as mentioned with torque, means all rotational forces are balanced.
In our exercise, mechanical equilibrium is achieved by:
  • The pivot providing support and ensuring no linear motion.
  • The cable providing tension sufficient to counteract any rotational tendencies due to gravity acting on the girder and suspended objects.
Every structure we build, from skyscrapers to simple shelves, relies on these principles of mechanical equilibrium to remain stable and unmovable.
Tension Minimization
Finding the minimal tension in a cable supporting a loaded beam involves strategic placement of the cable's attachment point. The essence of tension minimization is to reduce the force the cable must exert to maintain equilibrium.

Here's how:
  • Tension in the cable should counterbalance the gravitational forces, creating equilibrium. Think of it like a tightrope walker balancing their weight with minimal strain on the rope.
  • From the torque equation, the cable's tension is inversely proportional to its distance from the pivot. The further the cable is attached from the pivot, the less tension is required, as it can "leverage" the distance to create a balancing counter-torque.
The exercise shows that to minimize tension, the cable must be attached at the maximum possible distance from the pivot, or the full length of the beam.
  • This maximizes the stabilizing effect the cable has on the girder, using less force to achieve balance and thus achieving tension minimization efficiently.
In practical engineering, minimizing tension increases the safety and longevity of structures, as less stress is placed on the supporting cables.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two people carry a heavy clectric motor by placing it on a light board 2.00 \(\mathrm{m}\) long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light bnt weighs 200 \(\mathrm{N}\) , with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

Shear forces are applied to a rectangular solid. The same forces are applied to another rectangular solid of the same material, but with three times each edge length. In each case the forces are small enough that Hooke's law is obeyed. What is the ratio of the shear strain for the larger object to that of the smaller object?

Flying Buttress. (a) A symmetric building has a roof sloping upward at \(35.0^{\circ}\) above the horizontal on each side. If each side of the uniform roof weighs \(10,000 \mathrm{N}\) , find the horizontal force that this roof exerts at the top of the wall, which tends to push out the walls. Which type of building would be more in danger of collapsing: one with tall walls or one with short walls? Explain. (b) As you saw in part (a), tall walls are in danger of collapsing from the weight of the roof. This problem plagued the ancient builders of large structures. A solution used in the great Gothic cathedrals during the 1200 s was the flying buttress, a stone support running between the walls and the ground that helped to hold in the walls. A Gothic church has a uniform roof weighing a total of \(20,000 \mathrm{N}\) and rising at \(40^{\circ}\) above the horizontal at each wall. The walls are 40 \(\mathrm{m}\) tall, and a flying buttress meets each wall 10 \(\mathrm{m}\) below the base of the roof. What horizontal force must this flying buttress apply to the wall?

A uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N}\) and is 14.0 \(\mathrm{m}\) long. A cable is connected 3.5 \(\mathrm{m}\) from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge.

Stress on the Shin Bone. Compressive strength of our bones is important in everyday life. Young's modulus for bone is about \(1.4 \times 10^{10}\) Pa. Bone can take only about a 1.0\(\%\) change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 \(\mathrm{cm}^{2} ?\) (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point) (b) Estimate the maximum height from which a \(70-k g\) man could jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 \(\mathrm{s}\) , and assume that the stress is distributed equally between his legs.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.