Calculate the force on the other support

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Homework Help Overview

The problem involves a uniform thin rod supported horizontally by two bricks, with a focus on calculating the force on one support after the other is removed. The subject area includes concepts of static equilibrium, torque, and forces.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for length in torque calculations and question whether it will affect the outcome. There is an exploration of dimensional analysis to predict the relationship between variables. Some participants express confusion about constants and how to derive them from the problem setup.

Discussion Status

The discussion is active, with participants seeking clarification on the role of length and constants in their calculations. Guidance has been offered to work through standard equations, but no consensus has been reached on the specific approach to take.

Contextual Notes

Participants note the absence of certain information, such as the length of the rod, which may be relevant to the problem. There is also a mention of imposed homework rules that may limit the information shared.

H_Ab
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Homework Statement


A uniform thin rod of weight 100 N is supported horizontally by two bricks at its
ends. At t=0 one of these bricks is kicked out quickly. Calculate the force on the
other support immediately thereafter.

Homework Equations


I'm not sure if I am on the right track with this equation since I don't have the length and from what I know, I need the length for torque problems: Mg(L/2)sin(theta)

The Attempt at a Solution


I thought since each brick has a force of 50N that if one brick is locked the force on the supporting brick is -50N since it's a downward force. Please help me with the equation and how to go about it. Thanks
 
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H_Ab said:
I'm not sure if I am on the right track with this equation since I don't have the length and from what I know, I need the length for torque problems: Mg(L/2)sin(theta)
Follow that through and see what happens. Maybe the length will not feature in the answer.
(From dimensional analysis, you can predict that it won't. Given an acceleration (g) a mass and a length as input, and a force to be the result of multiplying and dividing these and raising them to powers, the only combination that works is acceleration * mass * constant.)
 
So what would my constant be? My weight would be 50N since it's for one brick, and my acceleration is 9.81m/s. I'm confused about the constant.
 
H_Ab said:
So what would my constant be?
You need to determine that by working through the usual equations.
Create unknowns for mass, length, angular acceleration and write out the standard equations for linear acceleration versus force and angular acceleration versus torque.
 

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