Magnitude of The Force Supported at Point A

[samccain93]

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

ƩM_A=0
ƩF_y=0
ƩF_x=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:

Δy_AC=(135)sin(34)

M_A=0=(2.9)(75.459)-F_By(155)
F_By=-1.41

Then I found F_BX using trig functions and the angle between the vertical axis and F_B:

(-1.41)tan(56)=-2.09

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

ƩF_y=0=-1.41+F_Ay
F_Ay=1.41Kn
ƩF_x=0=-2.09+2.99+F_Ax
F_Ax=0.9Kn

Then I found F_A using the Pythagorean theorem

F_A=√(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!

 

[PhanthomJay]

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

ƩM_A=0
ƩF_y=0
ƩF_x=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:

Δy_AC=(135)sin(34)

M_A=0=(2.9)(75.459)-F_By(155)
F_By=-1.41

Then I found F_BX using trig functions and the angle between the vertical axis and F_B:

(-1.41)tan(56)=-2.09

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

ƩF_y=0=-1.41+F_Ay →F_Ay=0.9Kn

check math!

ƩF_x=0=-2.09+2.99+F_Ax →F_Ax=1.41

0 = -2.09 + 2.9 + F_Ax
Check typo and math!

 

[samccain93]

Good catch Phantom, I fixed it. Doesn't change the final answer though.

 

[collinsmark]

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

ƩM_A=0
ƩF_y=0
ƩF_x=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:

Δy_AC=(135)sin(34)

M_A=0=(2.9)(75.459)-F_By(155)
F_By=-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?

 

[samccain93]

I changed my calculation to 135*sin(34) and re-did all of my calculations to get a final answer of F_A=2.09Kn, but that's still wrong.

 

[collinsmark]

I changed my calculation to 135*sin(34) and re-did all of my calculations to get a final answer of F_A=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?

 

[samccain93]

Gah! Typos are an enemy today. I did use 135*cos(34) and ended up with FA=2.09Kn.

 

[collinsmark]

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.

 

[samccain93]

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.

 

[samccain93]

I figured it out, I just had to round to 2.10.

Thanks collinsmark!