Explanation Take the point of contact of the sign with the wall as the rotation point. The two torques acting on the sign about this rotation point are the torque due to the weight of the sign and the torque due to the tension in the rope. The sign is in equilibrium so that the net torque must be zero. This implies the two torques acting on the sign about the rotation point must be equal and opposite. Write the equation for the net torque.00000 τ → = τ → g + τ → T Here, τ → is the net torque, τ → g is the torque due to the weight of the sign and τ → T is the torque due to the tension in the rope. Equate the above equation to zero. τ → g + τ → T = 0 | τ → g | = | τ → T | (I) The diagram is shown in figure 1. Write the equation for torque. τ → = r → × F → (II) Here, τ is the torque, r → is the vector representing the perpendicular distance from the axis of rotation to the point of action of the force and F → is the force. Use equation (II) to write the expression for the torque due to the weight. τ → g = r → g × F → g (III) Here,, r → g is the perpendicular distance between the axis of rotation and the point where the weight acts and F → g is the weight. Write the equation for F → g . F → g = − m g j ^ Here, m is the mass of the sign and g is the acceleration due to gravity. The perpendicular distance between the weight and the axis of rotation is the radius of the circular sign. Refer to figure 1 and write the equation for r → g . r → g = R i ^ Here, R is the radius of the circular sign. Put the above two equations in equation (III). τ → g = ( R i ^ ) × ( − m g j ^ ) = − m g R k ^ (IV) Use equation (I) to write the expression for the torque due to the tension in the rope. τ → T = r → × F → T (V) Here, r → is the vector denoting the perpendicular distance between the axis of rotation and the point where the tension acts and F → T is the tension. Refer to figure 1 and write the equation for F → T . F → T = − F T sin 60 ° i ^ + F T cos 60 ° j ^ = − 3 2 F T i ^ + 1 2 F T j ^ (VI) Here, F T is the magnitude of the tension

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN: 9781133939146

Author: Katz, Debora M.

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Simple harmonic motion

Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards…

Simple Pendulum

A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.

Oscillation

In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion tak…

Videos

Question

Chapter 14, Problem 37PQ

To determine

The tension in the cable.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14, Problem 36PQ

Chapter 14, Problem 38PQ

Students have asked these similar questions

No chatgpt pls will upvote

Chapter 14 Solutions

Physics for Scientists and Engineers: Foundations and Connections

Ch. 14.1 - A rubber duck floats in a bathtub. Imagine moving...Ch. 14.1 - Prob. 14.2CE Ch. 14.2 - CASE STUDY Hanging a Plane from a Single Point In...Ch. 14.2 - Prob. 14.4CE Ch. 14.4 - Imagine two vertical rods initially of equal...Ch. 14 - What Is Static Equilibrium? Problems 13 are...Ch. 14 - Prob. 2PQ Ch. 14 - Two identical balls are attached to a...Ch. 14 - While working on homework together, your friend...Ch. 14 - Consider the sketch of a portion of a...

Ch. 14 - Prob. 6PQ Ch. 14 - Prob. 7PQ Ch. 14 - Prob. 8PQ Ch. 14 - The keystone of an arch is the stone at the top...Ch. 14 - Prob. 10PQ Ch. 14 - Stand straight and comfortably with your feet...Ch. 14 - Prob. 12PQ Ch. 14 - Prob. 13PQ Ch. 14 - Prob. 14PQ Ch. 14 - Prob. 15PQ Ch. 14 - Prob. 16PQ Ch. 14 - Prob. 17PQ Ch. 14 - Prob. 18PQ Ch. 14 - Prob. 19PQ Ch. 14 - Prob. 20PQ Ch. 14 - Prob. 21PQ Ch. 14 - The inner planets of our solar system are...Ch. 14 - Two Boy Scouts, Bobby and Jimmy, are carrying a...Ch. 14 - Prob. 24PQ Ch. 14 - A painter of mass 87.8 kg is 1.45 m from the top...Ch. 14 - Consider the situation in Problem 25. Tests have...Ch. 14 - Children playing pirates have suspended a uniform...Ch. 14 - Prob. 28PQ Ch. 14 - Prob. 29PQ Ch. 14 - A 5.45-N beam of uniform density is 1.60 m long....Ch. 14 - A wooden door 2.1 m high and 0.90 m wide is hung...Ch. 14 - A 215-kg robotic arm at an assembly plant is...Ch. 14 - Problems 33 and 34 are paired. One end of a...Ch. 14 - For the uniform beam in Problem 33, find the...Ch. 14 - Prob. 35PQ Ch. 14 - A square plate with sides of length 4.0 m can...Ch. 14 - Prob. 37PQ Ch. 14 - At a museum, a 1300-kg model aircraft is hung from...Ch. 14 - A uniform wire (Y = 2.0 1011 N/m2) is subjected...Ch. 14 - A brass wire and a steel wire, both of the same...Ch. 14 - In Example 14.3, we found that one of the steel...Ch. 14 - A carbon nanotube is a nanometer-scale cylindrical...Ch. 14 - A nanotube with a Youngs modulus of 1.000 1012 Pa...Ch. 14 - Consider a nanotube with a Youngs modulus of 2.130...Ch. 14 - Prob. 45PQ Ch. 14 - Use the graph in Figure P14.46 to list the three...Ch. 14 - Prob. 47PQ Ch. 14 - A company is testing a new material made of...Ch. 14 - Prob. 49PQ Ch. 14 - Prob. 50PQ Ch. 14 - Prob. 51PQ Ch. 14 - Prob. 52PQ Ch. 14 - Prob. 53PQ Ch. 14 - Prob. 54PQ Ch. 14 - Prob. 55PQ Ch. 14 - Prob. 56PQ Ch. 14 - A copper rod with length 1.4 m and cross-sectional...Ch. 14 - Prob. 58PQ Ch. 14 - Prob. 59PQ Ch. 14 - Bruce Lee was famous for breaking concrete blocks...Ch. 14 - Prob. 61PQ Ch. 14 - Prob. 62PQ Ch. 14 - Prob. 63PQ Ch. 14 - A One end of a metal rod of weight Fg and length L...Ch. 14 - Prob. 65PQ Ch. 14 - A steel cable 2.00 m in length and with...Ch. 14 - Prob. 67PQ Ch. 14 - Prob. 68PQ Ch. 14 - Prob. 69PQ Ch. 14 - Prob. 70PQ Ch. 14 - Prob. 71PQ Ch. 14 - Prob. 72PQ Ch. 14 - Prob. 73PQ Ch. 14 - We know from studying friction forces that static...Ch. 14 - Ruby, with mass 55.0 kg, is trying to reach a box...Ch. 14 - An object is being weighed using an unequal-arm...Ch. 14 - Prob. 77PQ Ch. 14 - A massless, horizontal beam of length L and a...Ch. 14 - A rod of length 4.00 m with negligible mass is...Ch. 14 - A rod of length 4.00 m with negligible mass is...Ch. 14 - A horizontal, rigid bar of negligible weight is...Ch. 14 - Prob. 82PQ Ch. 14 - Prob. 83PQ Ch. 14 - Prob. 84PQ Ch. 14 - Prob. 85PQ

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning