# How do I set up the equations for this?

Ryuk1990
I am to find the reactions at A. How would I set up the equations for Ay and Ax in this problem? It seems pretty confusing. Also, will there be a momentum in this problem? ## Answers and Replies

Homework Helper
This system is supposed to be in equilibrium, right? In that case, forces in both x and y directions must sum to 0 and torques must sum to 0. Write out those conditions in 3 equations, plug in the numbers, and you'll get the answer.

Ryuk1990
Well for forces in the X direction, it's just A (in the X direction) is 0, right?

For y direction, this is the equation I wrote down.

Ay + 300 - 100 - 200 = 0

Solving for the unknown, I got Ay = 0.

As for the momentum of A, here's what I wrote down. It doesn't exactly make sense to me.

-100(12) + 300(15) - 200(22) = 0

but...that comes out to -1100 = 0 which makes no sense? What's going on? Is the momentum of A just -1100?

Homework Helper
Gold Member
Well for forces in the X direction, it's just A (in the X direction) is 0, right?

For y direction, this is the equation I wrote down.

Ay + 300 - 100 - 200 = 0

Solving for the unknown, I got Ay = 0.

As for the momentum of A, here's what I wrote down. It doesn't exactly make sense to me.
You are using the wrong terminology;it's the moments (or torques) of all forces about A that must equal 0.
-100(12) + 300(15) - 200(22) = 0

but...that comes out to -1100 = 0 which makes no sense? What's going on? Is the momentum of A just -1100?
Since the support at A is fixed, it is capable of providing both forces and moment (or torque, or couple; these terms are sometimes used synonymously). So in your equation, you are missing the moment provided by the support at A.

Ryuk1990
You are using the wrong terminology;it's the moments (or torques) of all forces about A that must equal 0.Since the support at A is fixed, it is capable of providing both forces and moment (or torque, or couple; these terms are sometimes used synonymously). So in your equation, you are missing the moment provided by the support at A.

So the moment at A would be written like this?

A - 100(12) + 300(15) - 200(22) = 0

A = 1100

Also, was I right about the forces in the x and y directions?