How to Calculate Reactions at Point A Using the Laws of Sine and Cosine?

[drkidd22]

Homework Statement

See attached image file

Homework Equations

law sine
law cosine

The Attempt at a Solution

Find the reactions about two points from the applied force at point B

Bx = 200Cos60 = 100N
By = 200Sin60 = 173.20N

May = -173N(.4m) + CyN(.15m) = 0
May = -461.86Nm

Max = -100N(.3) = -30Nm

Ma = -491.86Nm

I tried solving to find reaction at a, but not 100% if my result is correct. Any help would be appreciated

 

[PhanthomJay]

Homework Statement

See attached image file

Homework Equations

law sine
law cosine

these are not relevant; what are the equilibrium equations?

The Attempt at a Solution

Find the reactions about two points from the applied force at point B

Bx = 200Cos60 = 100N
By = 200Sin60 = 173.20N

yes, direction?

May = -173N(.4m) + CyN(.15m) = 0
May = -461.86Nm

Are you trying to solve for Cy (in Newtons) by summing moments about A? If so, your equiibrium equation and units are incorrect.

Max = -100N(.3) = -30Nm

What point are you summing moments about?

Ma = -491.86Nm

I tried solving to find reaction at a, but not 100% if my result is correct. Any help would be appreciated

You must first note that the support at A can provide only a vertical force, since it is a roller joint free to slide horizontally. The pinned support at C can provide both a horizontal and a vertical force. You have 3 unknowns, Ay, Cx, and Cy, and 3 equilibrium equations. Try summing forces in the x direction first, then try again to sum moments about A. Do not forget to include the moments from all force or force components.

 

[Redbelly98]

EDIT: PhanthomJay beat my response. Gotta learn to type faster

Homework Equations

law sine
law cosine

Those are not relevant at all here. Instead, something involving the sum of the forces and sum of the moments would apply.

The Attempt at a Solution

Find the reactions about two points from the applied force at point B

Bx = 200Cos60 = 100N
By = 200Sin60 = 173.20N

Okay, though By will point downward and that is usually indicated as a negative value.

May = -173N(.4m) + CyN(.15m) = 0
May = -461.86Nm

Is this supposed to be the torque about point A? It's not correct, for one thing you are saying it is both 0 and -462 Nm.

Before trying again, make a list of all x and y forces that act at points A and C. (You already have done this for point B.)

Max = -100N(.3) = -30Nm

Ma = -491.86Nm

I tried solving to find reaction at a, but not 100% if my result is correct. Any help would be appreciated

 

[drkidd22]

So Cx + Bx = 0
Cx + 100 = 0
Cx = -100N?

 

[Redbelly98]

Yes, that's right.
Next do the same thing for:

• the y-forces
• the moments about a point of your choice.

(Hint: either points A or C are good choices for setting up the moment equation.)

 

[drkidd22]

I'm confused trying to find Cy and Ay.
But this is what I tried doing.0 = Fay + FBy
Fay -173.2 = 0
Fay = 173.2

Fay + Fby + Fcy = 0
Fcy = 0?

or is it

250mm(-173.2N) = Fay(150mm)
Fay = 288.66N (I think this is the right way)

 

[PhanthomJay]

I'm confused trying to find Cy and Ay.
But this is what I tried doing.


0 = Fay + FBy

No

Fay + Fby + Fcy = 0

Yes

or is it

250mm(-173.2N) = Fay(150mm)
Fay = 288.66N (I think this is the right way)

No. If you are summing torques (moments) about C (are you?), you must include the torque from Bx (the horizontal component of the force at B). Solve for Ay, then use your other equation to solve for Cy.

 

[drkidd22]

I'm really frustrated with this and I give up, If someone explains this to me I will unlock whatever it is that's blocking mi mind to get through it. Can someone please explain it to me.

 

[PhanthomJay]

You have already calculated Cx correctly, by noting that the sum of the x forces must equal zero, for equilibrium . Since Ax = 0 due to the roller support at A, Cx is 100N acting to the left on the bracket. That is to say,
Ax +Bx +Cx =0
0 + 100 + Cx =0
Cx =-100N

Now you look at forces in the y direction. They also must sum to zero.
Ay +By + Cy =0
You know the value of By, but you don't know Ay or Cy yet, so you need one more equation of equilibrium, which is
"Sum of Moments of all Forces about any Point =0."

If you choose point C as your point about which to sum moments, then calculate the moment about C from By (which you did), the moment from Bx (which you did not do), and the moment from Ay (which you did, in terms of Ay, an unknown), and add them all up, set them equal to zero, and solve for Ay. When you sum moments, watch your plus and minus signs; if you choose a clockwise moment as minus, then a counterclockwise moment is plus. Note that the forces at C do not contribute any moment when you sum moments about C. Once you solve for Ay, then you can get Cy from your Ay +By + Cy =0 equation.

 

[drkidd22]

Ok I drank some cheap wine and my light bulb went off. I feel stupid now, but I got it figured out.

Ma = 0
(-100Nx300mm)+Cy(150mm)-(173.21Nx400mm)
= Cy = 661.89N

Mc = 0
(Ay150mm) - (100Nx300mm)-(173.21Nx250mm)
Ay = 488.7N

Thanks to all for your help

 

[PhanthomJay]

Always check your answer by summing forces in the y direction
Ay + By + Cy =0
It's a good check for the proper direction of the y forces.
Then celebrate with a tall glass of Cabernet Sauvignon.
And almost never ever give up.