Magnitude of The Force Supported at Point A

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Discussion Overview

The discussion revolves around calculating the magnitude of the force supported at point A due to a 2.9-kN load applied to a bracket, while neglecting friction. Participants explore various approaches to solving the problem, including the use of equilibrium equations and trigonometric functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the vertical force at B and derives components of the force at A using equilibrium equations, initially arriving at FA=1.672 kN.
  • Another participant points out a potential error in the calculations and suggests checking the math, leading to a correction in the vertical force at A to FAy=0.9 kN.
  • There is a discussion about the appropriate trigonometric function to use, with suggestions to consider using cosine instead of sine for certain calculations.
  • After adjustments, a participant recalculates and finds FA=2.09 kN, but expresses uncertainty about the correctness of this result.
  • Another participant confirms the y-component of the force at A as 2.09 kN but notes that the x-component still needs to be determined.
  • One participant calculates the x-component of the force at A as 0.2 kN, leading to a total force of 2.099 kN, which is later noted as incorrect by the system.
  • A final participant resolves the issue by rounding the total force to 2.10 kN, indicating a potential misunderstanding of the rounding rules in the context of the problem.

Areas of Agreement / Disagreement

Participants express various calculations and corrections, indicating that there is no consensus on the final answer. Multiple competing views and calculations remain unresolved throughout the discussion.

Contextual Notes

There are indications of typos and potential misapplications of trigonometric functions, which may affect the calculations. The discussion reflects uncertainty regarding the correct approach to the problem and the final numerical results.

samccain93
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Homework Statement



Calculate the magnitude of the force supported by the point at A under the action of the 2.9-kN load applied to the bracket. Neglect friction in the slot.

Homework Equations



ƩMA=0
ƩFy=0
ƩFx=0

The Attempt at a Solution



I started by finding the moment around A so that I could solve for the vertical force at B:

ΔyAC=(135)sin(34)

MA=0=(2.9)(75.459)-FBy(155)
FBy=-1.41

Then I found FBX using trig functions and the angle between the vertical axis and FB:

(-1.41)tan(56)=-2.09

Then I summed all the forces to find the components of the force at point A:

ƩFy=0=-1.41+FAy
FAy=1.41Kn
ƩFx=0=-2.09+2.99+FAx
FAx=0.9Kn

Then I found FA using the Pythagorean theorem

FA=√(0.92+1.412)=1.672Kn

This was wrong, am I missing a force/forces somewhere, or are my calculations incorrect?

Thanks for any help!
 

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samccain93 said:

Homework Statement



Calculate the magnitude of the force supported by the point at A under the action of the 2.9-kN load applied to the bracket. Neglect friction in the slot.

Homework Equations



ƩMA=0
ƩFy=0
ƩFx=0

The Attempt at a Solution



I started by finding the moment around A so that I could solve for the vertical force at B:

ΔyAC=(135)sin(34)

MA=0=(2.9)(75.459)-FBy(155)
FBy=-1.41

Then I found FBX using trig functions and the angle between the vertical axis and FB:

(-1.41)tan(56)=-2.09

Then I summed all the forces to find the components of the force at point A:

ƩFy=0=-1.41+FAy →FAy=0.9Kn
check math!
ƩFx=0=-2.09+2.99+FAx →FAx=1.41
0 = -2.09 + 2.9 + FAx
Check typo and math!
 
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Good catch Phantom, I fixed it. Doesn't change the final answer though.
 
samccain93 said:

Homework Statement



Calculate the magnitude of the force supported by the point at A under the action of the 2.9-kN load applied to the bracket. Neglect friction in the slot.

Homework Equations



ƩMA=0
ƩFy=0
ƩFx=0

The Attempt at a Solution



I started by finding the moment around A so that I could solve for the vertical force at B:

ΔyAC=(135)sin(34)

MA=0=(2.9)(75.459)-FBy(155)
FBy=-1.41

\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}

\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}

Given what you're tying to do, are you sure you want to use sin(34) for that?
 
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I changed my calculation to 135*sin(34) and re-did all of my calculations to get a final answer of FA=2.09Kn, but that's still wrong.
 
samccain93 said:
I changed my calculation to 135*sin(34) and re-did all of my calculations to get a final answer of FA=2.09Kn, but that's still wrong.

What I meant to express is that you are using the sine function. Are you sure that the sine function is the appropriate function here? Maybe a cosine would be better? :wink:
 
Gah! Typos are an enemy today. I did use 135*cos(34) and ended up with FA=2.09Kn.
 
samccain93 said:
Gah! Typos are an enemy today. I did use 135*cos(34) and ended up with FA=2.09Kn.

Okay, 2.09 kN is the y-component of the force at point A. But there's still the x-component to figure out.
 
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The x-component of the force at A = 3.10-2.9 = 0.2

After Pythagorean theorem the force at A comes to 2.099Kn which the system says is incorrect.
 
  • #10
I figured it out, I just had to round to 2.10.

Thanks collinsmark!
 

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