**Assume theta = 30 degrees**

We are encouraging you to communicate with us and with your colleagues in the class through the threaded discussions on the course blog. If you have questions on this homework, please ask here.

**Assume theta = 30 degrees**

We are encouraging you to communicate with us and with your colleagues in the class through the threaded discussions on the course blog. If you have questions on this homework, please ask here.

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Do we need theta to solve this problem?

yes, assume theta=30 degrees

Should we write the answer in terms of theta?

Assume theta=30 degrees

I was also wondering how to give answers in terms of theta. Thank you for updating us.

Do we need to write our answers in vector form?

Yes

No. For 24.B's problem statement, the "Find" section does not ask for the answers in vector form. However, 24.A does ask for answers in vector form.

Are there any force couples at C due to the pin? (i.e. Cx or Cy)

I think so. I am pretty sure that the pin resists motion in the x and y directions.

Because AC is a multi-force member, there should be Cx and Cy acting on AC at C. If, for example, you choose Cx and Cy to act on AC in the positive x and positive y directions, respectively, then Cx and Cy should be acting on the other multi-force member, CE, in the negative x and y directions, respectively, by Newton's third.

Is the reaction at A the normal force from the roller? Or is there an Ax and an Ay?

I am unsure about this as well, when I originally did the problem though I treated it as a roller and only gave it one force Na.

There is only one reaction force, the normal force at A, because there is a roller with no friction so there will not be a reaction force parallel to the surface. However, the normal force vector will have x and y components.

There should be a normal force at A from the roller. You can also right it in Ax and Ay form but there would be a relation between Ax and Ay due to theta is 30 degree.

Yes there is a norm

Yes there is a normal force at A with x and y components but it does not have frction so its parallel to the surface

I was confused about this too. Be sure to add the reaction forces at C as well.

Would there be a reaction force at E normal to the surface acting on the CE member?

Since it is a fixed support at E, there will be a reaction in the y and x direction. There will also be a reaction moment at E.

Also, should we assume the thickness of the member is negligible?

Hi Kate,

Yes. Unless specified, I don’t think we need to take into account the thickness and thus weight of the bar. We can simply assume that the forces applied are much larger than the weight of the bar itself. At least that’s what I’ll be doing in my calculations.

Hi Kate, we don't have to necessarily assume that. This problem can be solved with the given parameters.

Do moments still equal 0 on a secure member like CE?

Hi Mike,

Because CE is a fixed support, it has to produce a reaction moment to prevent rotation. You still set the moment equation equal to 0 if taking it at E, but now you have to account for the reaction moment in the equation as well.

Well, the moment equation would be set equal to zero regardless of where the moment is taken at. E does produce a reaction moment counter to the reaction on the system at A, that is true, but it counters the remaining moments throughout the system regardless of where they are calculated from.

As long as it does not rotate, the moment about any point should be 0.

Hi Mike,

It is important to remember that for fixed support, there exists three reactions - reaction in x, the reaction in y, and the moment reaction about the fixed point.

I think you need theta to solve the problem, but it isn't listed. Am I missing something?

Actually I just saw the update, I'm all set.

theta = 30 deg

Is the Y component of the Normal force at A found with COS(30) and the X component is found by using SIN(30)?

Wall and horizontal forms angel theta, so normal and vertical forms angle theta. So yes, you are correct.

Hi Noah,

Yes, you are correct that cos(30) is for the y component and sin(30) is for the x component in this problem. Finding the x and y components for reaction forces on an incline can be tricky sometimes!

Ah ok, I just came to check for theta and am glad to see it is here!

When writing moment equations with the two different arm segments, do we include the applied force on the other arm? For example, if finding the moment about A, do we need to include force F, even though it's on the other arm segment?

Hi Mehal,

I split my structure into two parts. The first part is the entire structure from A to C, and the second part is the rest of the structure that is C to E. Because it is split into two, you do not need to include the applied forces acting on the opposite section of the structure in your moment equation.

What are some examples of assumptions we can use for multiforce members?

Hi Josh, I'm not sure if I have them all covered, but I have simply been putting constant forces, in static equilibrium, and 2 force members (if applicable). Are there any others that I could potentially be lacking?

Do we need to give our answers in vector form like part A, or will the magnitude of the reactions suffice?

In 24.A it specially says to write in vector form, however it doesn't specially say that for 24.B. Personally, I wrote my answers for both problems in vector form.

Can we use the whole system without breaking it up as one of our FBDs?

When taking the moment about E, do you include the x and y components of Na separately in the equation? Or do you just keep it as Na?

Or do you calculate the moment about E only using the FBD for CE and therefore not need Na in the moment equation?

You calculate the moment about point E using the FDB for CE. Therefore, you would use the y component of Na when doing this.

Do we have to write our answers in vector form?

I would, if only to make sure there's no confusion about the directions of the forces.

I assume that there is Ax and Ay at end A; Ey Ex and Me at end E. Is these correct assumptions for the first FBD ?

That's what I have. However, I'd be careful about how you calculate your moment. The pin joint at C changes how force P will affect the moment at point E.

Is there a couple at point E since it's built into the wall?

Remember there is is a Moment on the wall at point E, because it is a Fixed point. And the answer should be in terms of theta. The best way to do this problem is draw the individual free body diagrams for each component, and solve for the individual equilibrium equations. Just a hint

When trying to solve for the reaction forces at C, does it matter at which side we start the problem from, from side A or side E?

If you draw two FBD's, one for each side of the joint, and the Cx reactions are equal and opposite, do you have to do a summation of Fx for the right FBD in your work to show that the Cx of the left will be equal to Ex, or can you just write that Cx of the right is equal to Ex.

From the FBD, Cx will be equal to Ex because those are the only horizontal forces acting on the other member. If you've already solved for Cx in the first member, you can just indicate a summation of Fx that Cx is equal to Ex.

Do we need to write our answers in vector notation? For example, the reaction at A can we just give what the normal force is or do we have to deconstruct it to its x and y-components?

I know that H24.A wants the answer in vector form. Does H24.B also need to be in vector form?

I'm assuming that the reactions at A for 24B go with the angle theta instead of with the x/y plane? Just making sure

Hello Andrew. Yes, the angle theta is 30*. They forgot to mention it in the problem, but it's written up top on this page.

Would the unit for a moment in this problem be kip ft?