Static Equilibrium Problem Dealing with Torque

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Homework Help Overview

The problem involves a static equilibrium scenario with a crane model supported by two strings. It includes a uniform horizontal bar and an object being lifted, requiring analysis of forces and torques to find the tension in one of the strings.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply torque equations to find the tension in string A but expresses confusion about the inclusion of "1/2" in their equation. Some participants question the concept of resultant load and its application to the uniform beam's weight distribution.

Discussion Status

The discussion is ongoing, with participants exploring the implications of weight distribution and the definition of resultant load. Clarifications are being sought regarding the original poster's understanding of the problem setup.

Contextual Notes

There is a mention of using hints provided in the problem statement, and the participants are navigating through the definitions and assumptions related to the uniform beam and its resultant forces.

PeachBanana
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Homework Statement



The figure shows a model of a crane that may be mounted on a truck. A rigid uniform horizontal bar of mass = 90.00 kg and length = 5.200 m is supported by two vertical massless strings. String A is attached at a distance = 1.600 m from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass = 3000 kg is supported by the crane at a distance = 5.000 kg from the left end of the bar.

Throughout this problem, positive torque is counterclockwise and use 9.807 m/s^2 for the magnitude of the acceleration due to gravity.


Find the tension in string A.

Homework Equations



This problem let's you use hints and it suggested I use the following equation with having the Tension in String "B" as the origin.

The Attempt at a Solution



Tension A(1.6000m) - (90.00 kg)(9.807 m/s^2) 1/2(5.200 m) - (3000 kg)(9.807 m/s^2)(5.000 m) = 0

I don't understand the "1/2" part in front of the length mass one. Is that because of where it is placed or its moment of inertia?
 

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If the weight of the 90 kg beam is uniformly distributed over its length, where does its resultant load act?
 
I'm sorry, I'm not sure what you mean by resultant load.
 
PeachBanana said:
I'm sorry, I'm not sure what you mean by resultant load.
If a uniform beam that is 2 m long has a mass of 0.5 kg/m, its total mass is 1 kg and it's resultant or total weight is 9.8 N. It can be represented as a single force of 9.8 N acting at its center of gravity.
 

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