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- Thread starter Ryuk1990
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hotvette

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hotvette

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PhanthomJay

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neither pins nor rollers can supply moments. But you can sum moments (or torques) about any point and set that sum equal to zero for equilibrium. To solve for the reactions, use the three equilbrium equations: sum of F_x =0, sum of F_y = 0, and sum of moments about any point = 0. Note that rolller supports by definition cannot provide reaction forces in the direction along which they are free to slide. Try summing moments about C for starters. You needn't worry about member lengths or angles at this stage of the problem.

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For F_x, I got C_x = 0.

For F_y, I wrote out C - 1000 - 2000 = 0

and therefore C_y = 3000.

For moment about C, I wrote C - 1000(12) - 2000(24) = 0

and so C = 60,000.

Is this right so far?

If it is, how would I go about setting the equations for E?

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PhanthomJay

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well yes, since the sum of forces in the x direction must be zero, and E_x is zero by definition, then C_x must be 0.Well here's what I did.

For F_x, I got C_x = 0.

. No! It' sum of forces in y direction = 0. Where's the E_y force? C_y + E_y -1000 -2000 = 0. You have two unknowns here, so you can't solve this equation yet.For F_y, I wrote out C - 1000 - 2000 = 0

and therefore C_y = 3000.

you are taking moments about C, so the c force doesm't produce any moments about c. But the E_y force does. What's the moment about C from from the E_y force? Then you can solve for E_y, and then go back into the other equation to solve for C_y.For moment about C, I wrote C - 1000(12) - 2000(24) = 0

and so C = 60,000.

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you are taking moments about C, so the c force doesm't produce any moments about c. But the E_y force does. What's the moment about C from from the E_y force? Then you can solve for E_y, and then go back into the other equation to solve for C_y.

So for the moment about C, would it be E_y(8) - 1000(12) - 2000(24) = 0?

Or should the distance for E_y be the diagonal distance of the beam which is 10?

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PhanthomJay

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neither is correct. You have to go back to the basic definition of a moment. The magnitude of the moment of a force about a point is equal to the force times the perpendicular distance from the line of action of that force to that point. The perpendicular distance from the line of action of the force, E_y, to the point C, is not 8 ft., nor is it 10 feet. Please study up on torques or moments. There is an alternate way to calculate moments that might confuse you at this time. Echoing hotvette's comment above, nothing discussed in class at all, or are you in a self help course? Do you ultimately need to find imternal member forces, or are you solely studying reaction forces? PS: Watch your plus and minus signs throughout.So for the moment about C, would it be E_y(8) - 1000(12) - 2000(24) = 0?

Or should the distance for E_y be the diagonal distance of the beam which is 10?

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neither is correct. You have to go back to the basic definition of a moment. The magnitude of the moment of a force about a point is equal to the force times the perpendicular distance from the line of action of that force to that point. The perpendicular distance from the line of action of the force, E_y, to the point C, is not 8 ft., nor is it 10 feet. Please study up on torques or moments. There is an alternate way to calculate moments that might confuse you at this time. Echoing hotvette's comment above, nothing discussed in class at all, or are you in a self help course? Do you ultimately need to find imternal member forces, or are you solely studying reaction forces? PS: Watch your plus and minus signs throughout.

It's an engineering class but we've never talked about truss problems which is why I'm confused. We've only talked about very basic problems nothing like this.

So the perpendicular distance of the force of E_y to point C is 6 feet?

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hotvette

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So the perpendicular distance of the force of E_y to point C is 6 feet?

Correct.

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Now I'm going to attempt to find the equations for the reactions at E. Is E_x essentially just zero as well?

As for y direction, I wrote down E_y + C_y - 1000 - 2000 = 0

For the moment about E, I wrote down Cy(6) + 1000(6) + 2000(18) = 0

Is this right?

- #12

hotvette

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Nope. Show us how you arrived at Ey. Also, no need to take moment about E. Taking a moment about C and summing vertical forces is enough to solve the entire problem. Also, correct about Ex = 0 (rolling joint can't support a force in x)

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I made a mistake on the calculator. I got 10,000 for Ey and 7000 downward for Cy.

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Also, no need to take moment about E. Taking a moment about C and summing vertical forces is enough to solve the entire problem.

But the problem asks to find reactions at both C and E. So shouldn't I write the moment about E anyway?

- #15

hotvette

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But the problem asks to find reactions at both C and E. So shouldn't I write the moment about E anyway?

You already have the reactions at C and E, but go ahead. It's a great way to verify your answer.

- #16

hotvette

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[soapbox]

1. Always choose, draw, and label a coordinate system. It is surprising how helpful this can be for keeping track of signs, especially in more complicated problems. May seem trivial, but is worth it in the long run to get into the habit of doing it every time. Homework can (and should in my opinion) be marked down if a coordinate system isn't present.

2. Replace the reaction joints with equivalent force vectors (i.e. draw a free body diagram). This makes it easier to sum forces and moments. It also makes one really think about what forces / moments the joint can support. Same comment about homework being marked down.

[/soapbox]

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I agree hotvette, but how would you draw a coordinate system or FBD for a truss problem like this?

- #18

hotvette

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but how would you draw a coordinate system or FBD for a truss problem like this?

If I were solving this problem, I'd draw the figure, add a small coordinate axis (doesn't matter where) with +x to the right and +y upward, and instead of the reaction joints, I'd draw a vertical vector at E (with label Ey), and at C I'd draw a vertical vector (with label Cy) and horizontal vector (with label Cx). It actually doesn't matter whether the reaction forces are drawn in the postive or negative direction as long as you are consistent in treatment when summing forces and moments. If a reaction force turns out to have a negative value, it just means you guessed the wrong direction when you drew it.

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