Magnitude of The Force Supported at Point A

In summary, the magnitude of the force supported by the point at A under the action of the 2.9-kN load applied to the bracket is 2.10 kN. This is found by using the equations ƩMA=0, ƩFy=0, and ƩFx=0 to calculate the vertical and horizontal components of the force at point A. The final answer is obtained by using the Pythagorean theorem to find the magnitude of the force.
  • #1
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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  • #2
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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  • #3
Good catch Phantom, I fixed it. Doesn't change the final answer though.
 
  • #4
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

[itex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/itex]

[itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

Given what you're tying to do, are you sure you want to use sin(34) for that?
 
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  • #5
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.
 
  • #6
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:
 
  • #7
Gah! Typos are an enemy today. I did use 135*cos(34) and ended up with FA=2.09Kn.
 
  • #8
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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  • #9
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!
 

What is the magnitude of the force supported at point A?

The magnitude of the force supported at point A refers to the amount of force acting on an object or structure at a specific point. It is typically measured in units of Newtons (N) or pounds (lbs).

How do you calculate the magnitude of the force supported at point A?

The magnitude of the force supported at point A can be calculated using the equation F = ma, where F is the force, m is the mass of the object, and a is the acceleration. Alternatively, it can also be calculated using the equation F = kx, where k is the spring constant and x is the displacement from equilibrium.

What factors affect the magnitude of the force supported at point A?

The magnitude of the force supported at point A can be affected by a variety of factors, including the mass of the object, the acceleration or displacement, the strength of the supporting structure, and any external forces acting on the object.

Why is it important to determine the magnitude of the force supported at point A?

Determining the magnitude of the force supported at point A is important in understanding the stability and safety of an object or structure. It can also help in designing and engineering structures that can withstand the expected forces at that point.

How can the magnitude of the force supported at point A be measured?

The magnitude of the force supported at point A can be measured using various instruments such as force gauges, strain gauges, load cells, and accelerometers. These instruments can provide accurate measurements of the force acting on an object at a specific point.

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