- #1

Ryuk1990

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_{y}and A

_{x}in this problem? It seems pretty confusing. Also, will there be a momentum in this problem?

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- Thread starter Ryuk1990
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- #1

Ryuk1990

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- #2

ideasrule

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- #3

Ryuk1990

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

A

Solving for the unknown, I got A

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

PhanthomJay

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You are using the wrong terminology;it's theWell 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.

A_{y}+ 300 - 100 - 200 = 0

Solving for the unknown, I got A_{y}= 0.

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

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.-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?

- #5

Ryuk1990

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You are using the wrong terminology;it's themoments(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?

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