Calculating Force on a Pin with Ropes and Tensions?

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Homework Help Overview

The problem involves calculating the total force exerted on a pin supporting a precast concrete wall held in place by ropes with known tensions. The subject area includes concepts of static equilibrium and vector addition in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss whether the tensions from additional ropes affect the force on the pin and how to properly add the force vectors. There is uncertainty about the relevance of angles in vector addition.

Discussion Status

The discussion is ongoing, with participants providing insights about the nature of forces acting on the pin and the method of vector addition. Some guidance has been offered regarding the calculation of force components without needing to find angles explicitly.

Contextual Notes

Participants are navigating assumptions about the forces acting on the pin and the implications of the given tension values. There is a focus on understanding how to approach the problem without additional information on angles.

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Homework Statement



A precast concrete wall is temporarily kept in its vertical position by ropes. Find the total force exerted on the pin at position A. The tensions in AB and AC are 420 lbs and 650 lbs.

The Attempt at a Solution



I'm not sure where to start, because I don't know whether or not the tension from DB to DC also counts towards the force on the pin at A. I am also not sure on what values to add. Like if I add vectors AB and AC, is the result at the same angle?
 

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warfreak131 said:
… I don't know whether or not the tension from DB to DC also counts towards the force on the pin at A.

Hi warfreak131! :smile:

There's no force-at-a-distance … the only forces at A are the forces at A. :wink:
I am also not sure on what values to add. Like if I add vectors AB and AC, is the result at the same angle?

You add the force vectors, not the position vectors.

(All the angles are either equal or opposite, so yes you could take a short-cut for each component separately … but I wouldn't recommend it.)
 
tiny-tim said:
Hi warfreak131! :smile:

There's no force-at-a-distance … the only forces at A are the forces at A. :wink:


You add the force vectors, not the position vectors.

(All the angles are either equal or opposite, so yes you could take a short-cut for each component separately … but I wouldn't recommend it.)

ok, but DC, and DB are pulling on the wall too, so wouldn't that put extra tension in the cables AB and AC and cause the pin to experience a greater force?
 
warfreak131 said:
ok, but DC, and DB are pulling on the wall too, so wouldn't that put extra tension in the cables AB and AC and cause the pin to experience a greater force?

ah, but that's taken account of in the question
warfreak131 said:
The tensions in AB and AC are 420 lbs and 650 lbs.

… it may well put extra tension in AB and AC, but you've been given figures that include that! :wink:
 
tiny-tim said:
ah, but that's taken account of in the question


… it may well put extra tension in AB and AC, but you've been given figures that include that! :wink:

very true sir... very true,

so should i just find the angles that the cables make with the fore and background edges, and then add the force vectors?
 
Yes, except that you don't actually need to find the angles, you can get the cosines just by using the coordinates, and dividing by the hypotenuse. :wink:
 
tiny-tim said:
Yes, except that you don't actually need to find the angles, you can get the cosines just by using the coordinates, and dividing by the hypotenuse. :wink:

ok, thanks a bunch
 

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