/*! 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 99 CALC Minimizing the Tension. A h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 99

CALC 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 slightly beyond the girder's midpoint to minimize tension.

Step by step solution

01

Define the Problem

We have a girder of length \( L \) with multiple objects hanging from it. The girder is supported by a pivot at the left end and a cable attached at a distance \( h \) above its center. We need to find the position along the girder where the cable should be attached to minimize tension.
02

Consider the Forces Involved

The forces at play include the weight of the girder, the weights of the objects, the tension in the cable, and the reaction from the pivot. The tension must counteract the torque produced by the other forces about the pivot for equilibrium.
03

Set Up Torque Equation

Choose the pivot as the axis of rotation and set the torque due to tension equal to the sum of torques due to girder's weight and any hanging objects.\[ T \sin(\theta) \cdot x = \text{Sum of other torques} \]where \( T \) is the tension and \( \theta \) is the angle between the cable and the girder, \( x \) is the attachment point distance from the pivot.
04

Express the Geometry

Using trigonometry, \( \tan(\theta) = \frac{h}{x} \), so \( \sin(\theta) = \frac{h}{\sqrt{h^2+x^2}} \). Substitute this into the torque equation.
05

Minimize the Tension

Express tension \( T \) as \( T = \frac{\text{Sum of other torques}}{\frac{h}{\sqrt{h^2+x^2}} \cdot x} \). Differentiate \( T \) with respect to \( x \) and set the derivative to zero to find the critical points. Use constraints \( 0 \leq x \leq L \).
06

Solve the Derivative

Solving \( \frac{dT}{dx} = 0 \) gives us the position \( x \) where \( T \) is minimized. Calculations may involve evaluating derivatives and simplifying the resulting expressions.
07

Evaluate and Conclude

Check the boundary conditions and the evaluated critical point(s) to ensure the minimum tension is identified under the constraint \( 0 \leq x \leq L \). Compare tension values calculated for critical points and endpoints.

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 91影视!

Key Concepts

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

Equilibrium
Equilibrium is a crucial concept when analyzing forces in physics, especially when dealing with structures like girders. A system is in equilibrium when all the forces and torques acting on it are balanced. This means:
  • There is no net force acting on the system.
  • The sum of all torques around any point is zero.
In our exercise, equilibrium involves balancing the forces on the girder using the pivot and the cable. The tension in the cable and the reaction force at the pivot must counteract the weights of the objects and the girder itself.
Ensuring equilibrium is essential to prevent rotational motion and to keep the girder stable. By analyzing the torques about the pivot point, we can determine how the tension in the cable plays a role in maintaining this equilibrium.
Tension
Tension is the force exerted by a cable, string, or similar object when it is pulled tight by forces acting from opposite ends. It is a common force in problems dealing with suspensions or support structures.
In this exercise, tension is experienced by the cable supporting the girder. Our goal is to minimize this tension, which requires an understanding of how tension changes with the angle and placement of the cable.
We express tension in its relationship to torque by using the equation:\[ T = \frac{\text{Sum of other torques}}{\frac{h}{\sqrt{h^2+x^2}} \cdot x} \]By modifying the attachment point of the cable, you can find the optimal position where this tension is the smallest possible, given all constraints. This approach helps in understanding how suspensions can be optimized for efficiency and safety.
Trigonometry
Trigonometry often plays a pivotal role in physics problems that involve angles and distances. In our exercise, it helps us express the relationship between the angle of the cable and its attachment to the girder.
The equation \( \tan(\theta) = \frac{h}{x} \) is used to describe the angle \( \theta \) relative to the girder. From here, we can find:
  • \( \sin(\theta) = \frac{h}{\sqrt{h^2+x^2}} \), which is critical for calculating the torque due to tension.
These trigonometric expressions allow us to convert geometric information into a form that can be analyzed in terms of the physical forces and equilibrium conditions. Understanding this relationship is key to solving the exercise effectively, as it directly affects our ability to minimize the tension by selecting the proper attachment point.

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

A uniform rod is 2.00 \(\mathrm{m}\) long and has 1.80 \(\mathrm{kg} . \mathrm{A}\) \(2.40-\) -kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 \(\mathrm{m}\) from the left-hand end of the rod?

\(\mathrm{A} 1.05-\mathrm{m}-\operatorname{lon} \mathrm{g}\) rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. P11.91). The cross-sectional area of \(A\) is 2.00 \(\mathrm{mm}^{2}\) and that of \(B\) is 4.00 \(\mathrm{mm}^{2} .\) Young's modulus for wire \(A\) is \(1.80 \times 10^{11} \mathrm{Pa}\) ; that for \(B\) is \(1.20 \times 10^{11} \mathrm{Pa}\) . At what poing the rod should a weight \(w\) be suspended to produce (a) equal stresses in \(A\) and \(B\) and (b) equal strains in \(A\) and \(B ?\)

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 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?

Stress on a Mountaineer's Rope. A nylon rope used by mountaineers elongates 1.10 \(\mathrm{m}\) under the weight of a 65.0 -kg climber. If the rope is 45.0 \(\mathrm{m}\) in length and 7.0 \(\mathrm{mm}\) in diameter, what is Young's modulus for nylon?

A uniform 300 -N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Famous Physicists

Read Explanation

Classical Mechanics

Read Explanation

Fields in Physics

Read Explanation
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.