Magnitude of The Force Supported at Point A

AI Thread Summary
The discussion focuses on calculating the force supported at point A due to a 2.9-kN load on a bracket, neglecting friction. Initial calculations yielded a force at A of 1.672 kN, which was later adjusted to 2.09 kN after correcting the trigonometric functions used. Participants emphasized the importance of using the correct sine or cosine functions for the angles involved. The final force at A was confirmed to be 2.10 kN after rounding, highlighting the significance of precision in calculations. Overall, the thread illustrates common challenges in solving static equilibrium problems in physics.
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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