What is the required horizontal force to deflect a hanging rod by 30 degrees?

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Homework Help Overview

The problem involves a uniform rod hinged at one end and requires determining the horizontal force needed to deflect it by 30 degrees from the vertical. The subject area includes concepts of static equilibrium and moments in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss taking moments about the hinge point and the application of trigonometric functions in calculating the moments due to the applied force and the weight of the rod. There are questions about the correct use of sine and cosine in the context of the problem.

Discussion Status

The discussion is active with participants providing their attempts at solving the problem and questioning each other's reasoning. Some guidance has been offered regarding the correct application of trigonometric functions, but there is no explicit consensus on the approach yet.

Contextual Notes

There is a mention of a second question regarding the meaning of "a gauge of 56.5 ins" in the context of a railway line, which may indicate additional context or constraints related to the problem.

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A uniform rod AB, of mass 5lb, is hinged at upper end A and hangs down freely. What horizontal force applied at B will be required in order to deflect the rod through 30 degrees from the vertical. Thanks.
My attempt: taking moments about A, Fcos30xAB = wxAC = wcos30xAB/2 therefore F = w/2. F= applied horizontal force, w = weight.A 2nd question what does "a gauge of 56.5 ins" mean in rhe context of a railway line?
 
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John O' Meara said:
My attempt: taking moments about A, Fcos30xAB = wxAC = wcos30xAB/2 therefore F = w/2. F= applied horizontal force, w = weight.A
Careful. Note that the weight acts vertically, so its moment should have a sin30, not a cos30.

2nd question what does "a gauge of 56.5 ins" mean in rhe context of a railway line?
The inner distance between the rails.
 
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I would take a second look and the Fcos30xAB. Shouldn't it be sin to get the correct length of the normal distance?
 
civil_dude said:
I would take a second look and the Fcos30xAB. Shouldn't it be sin to get the correct length of the normal distance?
Actually, that part is correct, since the force is applied horizontally. But the moment due to the weight should be Wsin30xAB/2, not Wcos30xAB/2. My bad, for not reading carefully. :redface: I'll revise my comments in the original post.
 
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