How do I set up the equations for this?

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In summary, to find the reactions at A in this problem, you need to set up equations for the forces and moments in both the x and y directions. In order for the system to be in equilibrium, these forces and moments must sum to 0. The equations for the y direction can be written as Ay + 300 - 100 - 200 = 0, and solving for Ay gives a value of 0. However, you also need to consider the moment provided by the support at A, which was missing in your initial equation. The correct equation would be A - 100(12) + 300(15) - 200(22) = 0, and solving for A gives a value of 110
  • #1
Ryuk1990
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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?

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  • #2
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.
 
  • #3
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?
 
  • #4
Ryuk1990 said:
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.
 
  • #5
PhanthomJay said:
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?
 

1. How do I determine which equations to use for a specific problem?

The first step is to carefully read and understand the problem. Identify the known values, unknown values, and any other relevant information. Then, use your knowledge of the relevant concepts and equations to select the most appropriate equation(s) to solve the problem.

2. How do I set up equations for systems of equations?

For systems of equations, you will have multiple equations with multiple variables. Start by identifying the number of equations and variables in the system. Then, use techniques such as substitution or elimination to solve for the unknown variables.

3. How do I handle units when setting up equations?

Units are an important aspect of scientific equations. Make sure to keep track of the units for each value and use conversion factors if necessary. It is also important to ensure that all units are consistent throughout the equation.

4. How do I incorporate scientific notation into equations?

To incorporate scientific notation into equations, simply treat the numbers as you would with regular numbers. Use the rules of exponents and scientific notation to simplify and manipulate the equations as needed.

5. How do I check my equations for accuracy?

To check your equations for accuracy, plug in the known values and solve for the unknown values. Then, substitute those values back into the original equations to see if they satisfy the equation. Additionally, you can use dimensional analysis to ensure that the units on both sides of the equation are consistent.

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