Fixed Support Moment Homework: Find Moment at Point A

[Kaushy]

Homework Statement

This is the diagram, and the question is written there too (sorry for blurry picture)

Homework Equations

My main question is that when you draw out the FBD, at point A, there is a force going up, and a force going to the right. Why isn't there a moment around A? I have had other questions where a beam is attached to a wall and I'll have the x, y, and moment at the connection. Is it because there is a ground at the bottom? Would there be a moment at A if it wasn't there?

The Attempt at a Solution

(left, right, clockwise positive)
Sum of all forces in X = 0
Sum of all forces in Y = 0
Sum of moments (from A) = 0, so is the formula: 0 = 4*10000sin30 - 4*Tsin80 (or 0 = MomentA + 4*10000sin30 - 4*Tsin80?)[/B]

 

[BvU]

Hello Kaushy,
There's no moment around A, otherwise the beam would start rotating!
Same as when you're standing on the floor, then there's no net force in the vertical direction, otherwise you would start accelerating vertically.
Which also means that, at A, the forces from beam and wall better add up to zero in both vertical and horizontal directions! (otherwise ...)

For the sum of moments around A I agree with the 4 * 10000 * sin 30##^\circ##, only I would call it negative (we usually have counterclockwise as the positive rotation direction).
[edit] sorry, that was correct too !
I'm not that happy with the other term: the perpendicular distance for the tension in the cable is not 4 * sin 80##^\circ##...

Do I understand correctly that you don't have to take the weight of the beam into account ?

--

 

[Kaushy]

When I am given this, seemingly similar problem (i have the solution), the formula for moment at point A is 0 = momentA - 20(5) - 20(5sin26.56) (counterclockwise positive). Why do we consider the moment as part of the equation in this and not the first one? (yes, assume the beam is weightless)

 

[SteamKing]

Homework Statement

View attachment 92051
This is the diagram, and the question is written there too (sorry for blurry picture)

Homework Equations

My main question is that when you draw out the FBD, at point A, there is a force going up, and a force going to the right. Why isn't there a moment around A? I have had other questions where a beam is attached to a wall and I'll have the x, y, and moment at the connection. Is it because there is a ground at the bottom? Would there be a moment at A if it wasn't there?

The Attempt at a Solution

(left, right, clockwise positive)
Sum of all forces in X = 0
Sum of all forces in Y = 0
Sum of moments (from A) = 0, so is the formula: 0 = 4*10000sin30 - 4*Tsin80 (or 0 = MomentA + 4*10000sin30 - 4*Tsin80?)[/B]

You should review the definition of a moment:

http://emweb.unl.edu/NEGAHBAN/EM223/note6/note6.htm

The moment arm is the perpendicular distance between the line of action of the force and the reference point about which the moment is calculated.

For the unknown cable tension in line BC, you should break this force up into its horizontal and vertical components. This will make your calculations much easier.

You should also pay careful attention to which angles and which trig functions you are using.
The moment of the tension T_BC about A is not equal to 4*T*sin 80°.

 

[SteamKing]

View attachment 92053
When I am given this, seemingly similar problem (i have the solution), the formula for moment at point A is 0 = momentA - 20(5) - 20(5sin26.56) (counterclockwise positive). Why do we consider the moment as part of the equation in this and not the first one? (yes, assume the beam is weightless)

Here, the beam is fixed to the wall at point A and is not free to rotate.

You should always draw a free body diagram of any beams you are analyzing and determine how the ends of the beam are fixed or not.

In the problem in the OP, the sketch is not clear on how the beam AB is connected to point A, but usually there is a pin connection, or a ball-and-socket joint which allows the beam AB to rotate up and down. Neither the pin connection of the ball-and-socket joint permit a reaction moment to develop.

 

[BvU]

View attachment 92053
When I am given this, seemingly similar problem (i have the solution), the formula for moment at point A is 0 = momentA - 20(5) - 20(5sin26.56) (counterclockwise positive). Why do we consider the moment as part of the equation in this and not the first one? (yes, assume the beam is weightless)

In this case the beam is attached to the wall and can't rotate: the wall exercises a torque on the beam.
In post #1 A is drawn as a pivot point, a hinge for example (at least, that's how I interpret the black dot).

 

[BvU]

The moment of the tension T_BC about A is not equal to 4*T*sin 80°.

Was my first reaction too, but on second thoughts I agreed with Kaushy. Bedtime for me, but I would appreciate some enlightenment !

 

[Kaushy]

I'm almost certain it is 4*T*sin80. Regardless, from my understanding, I'm not considering there to be a moment since the question in OP isn't actually a fixed point. I could not make that out from this diagram given. Thank you for both of your help.

 

[SteamKing]

I'm almost certain it is 4*T*sin80. Regardless, from my understanding, I'm not considering there to be a moment since the question in OP isn't actually a fixed point. I could not make that out from this diagram given. Thank you for both of your help.

Triangle ABC is not a right triangle, so using trig functions like you would with a right triangle to determine moment arms may not always be correct. This is why I recommended that you decompose the tension force in line BC into its horizontal and vertical components.

The horizontal component of T_BC will be perpendicular to the wall, thus forming a right triangle by definition. The moment due to the vertical component of T_BC can easily be computed by inspection.

The horizontal component of T_BC is T_BC ⋅ cos 20°. Its moment arm about A is 4 ⋅ sin 60° .

Likewise, the vertical component of T_BC can be computed and its moment arm about point A determined.

The sum of these two moments will be the moment of T_BC about point A. Be sure to account for the direction of these individual moments about A when summing them.

 

[BvU]

Triangle ABC is not a right triangle, so

Perhaps, but Angle ABC is definitely 80 degrees and that's all OP uses!

The horizontal component of TBC is TBC ⋅ cos 20°. Its moment arm about A is 4 ⋅ sin 60° .

Likewise, the vertical component of TBC can be computed and its moment arm about point A determined.

The sum of these two moments will be the moment of TBC about point A

Which then, according to our reasoning, should be the same as OP expression ...

 

[SteamKing]

Perhaps, but Angle ABC is definitely 80 degrees and that's all OP uses!
Which then, according to our reasoning, should be the same as OP expression ...

I've confirmed the OP's expression of the moment about A due to the tension T_BC.

I generally prefer using the vector definition M = r × F since it saves having to figure out all the components of r and F and then multiplying them together.

 

[BvU]

OK, so Kaushy is doing fine finding T from ##\sum\vec\tau = 0## .
Reaction forces at A also OK ?

 

[SteamKing]

OK, so Kaushy is doing fine finding T from ##\sum\vec\tau = 0## .
Reaction forces at A also OK ?

I haven't seen any calculations so far ...

 

[BvU]

Perhaps Kaushy has left us to quarrel amongst ourselves and moved on to greener pastures... (or the next exercise!)