/*! 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 61 A uniform, 7.5-m-long beam weigh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 61

A uniform, 7.5-m-long beam weighing \(6490 \mathrm{~N}\) is hinged to a wall and supported by a thin cable attached \(1.5 \mathrm{~m}\) from the free end of the beam. The cable runs between the beam and the wall and makes a \(40^{\circ}\) angle with the beam. What is the tension in the cable when the beam is at an angle of \(30^{\circ}\) above the horizontal?

Short Answer

Expert verified
After carrying out the mathematical manipulation, the tension in the cable when the beam is at an angle of 30 degrees above the horizontal is found out to be approximately 11058 N

Step by step solution

01

Identifying the Forces

Start by identifying all the forces acting on the beam. There are three forces involved here: (1) the weight of the beam, which acts downward at the center of the beam; (2) the tension in the cable, which acts upward and to the right; and (3) the hinge force, which has two components, one vertical and one horizontal. The pivot point is at the hinge where the beam is attached to the wall.
02

Calculating the Weight of the Beam

The weight of the beam acts at the center of the beam, which is \(7.5 \mathrm{~m}/2 = 3.75 \mathrm{~m}\) from the hinge. Since the beam is at a \(30^\circ\) angle with the horizontal, the perpendicular distance from the line of action of the weight to the hinge becomes \(3.75 \mathrm{~m} \cos 30^\circ\). Therefore, the torque due to the weight of the beam is \(6490 \mathrm{~N} * 3.75 \mathrm{~m} \cos 30^\circ\)
03

Calculating the Torque due to the Tension in the Cable

The tension in the cable creates an upward and rightward force \(1.5 \mathrm{~m}\) from the end of the beam. The force due to the tension in the cable is perpendicular to the beam. The lever arm is \(1.5 \mathrm{~m}\), so the torque due to tension in the cable is simply \(Tension \times 1.5 \mathrm{~m}\)
04

Applying the Equilibrium Conditions

By condition of equilibrium, the sum of all torques acting on the beam should be equal to zero. This leads to \(Tension \times 1.5 \mathrm{~m} = 6490 \mathrm{~N} * 3.75 \mathrm{~m} \cos 30^\circ\). Solving for 'Tension' results in \(Tension = \frac{6490 \mathrm{~N} * 3.75 \mathrm{~m} \cos 30^\circ}{1.5 \mathrm{~m}}\)

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.

Torque and Equilibrium
Torque is a measure of the turning force on an object and is pivotal in the study of rotational motion. It is calculated as the product of the force applied and the distance from the pivot point, known as the lever arm. In terms of equilibrium, when an object is in a state where there are zero net forces and zero net torques, it is said to be in static equilibrium.

Considering the beam in our exercise, we must assure that the torques produced by the weight of the beam and the tension in the cable counteract each other, leading to no net rotation; this is a practical application of the principle of torque and equilibrium.
Tension in Cables
The tension in a cable is a force that emerges due to the cable's attempt to contract and is always directed along the cable's length. In structural contexts like the scenario with the beam and cable, tension is critical in maintaining the structure's stability.

To calculate the tension correctly, one must understand the geometry of the setup, especially the angles at which the cables are inclined, as they directly affect the value of tension required for equilibrium. In our example, tension is the central force ensuring the beam remains static and balanced.
Free Body Diagram
A free body diagram (FBD) is an essential tool in physics that represents all the forces acting on a single object. The FBD simplifies complex real-world problems by stripping away any unneeded structures or supports, enabling us to focus on the forces at play.

In the step-by-step solution, creating an FBD would involve drawing the beam at its angle above the horizontal, with arrows depicting the direction of the weight, the tension in the cable, and the hinge force. Accurate representation of these forces aids in the clear application of the equilibrium conditions.
Forces and Moments
Forces are pushes or pulls on an object and moments, often referred to as torques, are the rotational effects of that force. Both are fundamental to understanding mechanical systems and ensuring their stability.

In the exercise, the weight of the beam and the tension in the cable represent forces, while the torques they generate are moments about the hinge point. These forces and moments interact to determine the behavior of the physical system, such as maintaining the beam in a state of static equilibrium.
Static Equilibrium
An object is said to be in static equilibrium when it is at rest or moving at constant velocity and has no net force or torque acting on it. For the equilibrium conditions to be met, the sum of all forces and the sum of all torques must each be zero.

The state of static equilibrium in the given problem entails that the hinge must exert an equal and opposite reaction to the sum of forces caused by the weight and tension, and the torques produced by these forces must balance out for the beam to remain stationarily attached to the wall.

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

His body is again leaning back at \(30.0^{\circ}\) to the vertical, but now the height at which the rope is held above-but still parallel to - the ground is varied. The tension in the rope in front of the competitor \(\left(T_{1}\right)\) is measured as a function of the shortest distance between the rope and the ground (the holding height). Tension \(T_{1}\) is found to decrease as the holding height increases. What could explain this observation? As the holding height increases, (a) the moment arm of the rope about his feet decreases due to the angle that his body makes with the vertical; (b) the moment arm of the weight about his feet decreases due to the angle that his body makes with the vertical; (c) a smaller tension in the rope is needed to produce a torque sufficient to balance the torque of the weight about his feet; (d) his center of mass moves down to compensate, so less tension in the rope is required to maintain equilibrium.

A specimen of oil having an initial volume of \(600 \mathrm{~cm}^{3}\) is subjected to a pressure increase of \(3.6 \times 10^{6} \mathrm{~Pa}\), and the volume is found to decrease by \(0.45 \mathrm{~cm}^{3} .\) What is the bulk modulus of the material? The compressibility?

A metal rod that is \(4.00 \mathrm{~m}\) long and \(0.50 \mathrm{~cm}^{2}\) in crosssectional area is found to stretch \(0.20 \mathrm{~cm}\) under a tension of \(5000 \mathrm{~N}\). What is Young's modulus for this metal?

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?

A \(350 \mathrm{~N}\), uniform, \(1.50 \mathrm{~m}\) bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of \(500.0 \mathrm{~N}\) without breaking, and cable \(B\) can support up to \(400.0 \mathrm{~N}\). You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Force

Read Explanation

Classical Mechanics

Read Explanation

Radiation

Read Explanation

Dynamics

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.