Determine the force in the bar CB

[ClearWhey]

Homework Statement

Determine the force in the bar CB. The own weights can be neglected. Look at the first picture for the question.

Relevant Equations

Look at the second picture for my equations and results.

I solved this with two methods as you can see down below in the picture. Which is the correct way?

I remember I learned in class you could use both ways but why am I getting different results?

 

[TSny]

Did you assume that F acts perpendicular to bar ACD in your first method?

 

[TSny]

For clarification, can you describe the type of connection or linkage at points A, B, and C.

 

[ClearWhey]

Did you assume that F acts perpendicular to bar ACD in your first method?

I actually didn't, just assumed the force would be that way if you removed the bar CB. But now that you write this I realize that method 1 only works if bar CB was perpendicular right?

Does it mean the second method is the right one?

For clarification, can you describe the type of connection or linkage at points A, B, and C.

Sorry but my english is not great. I have no idea what you mean with this. Thanks for helping me tho!

 

[TSny]

It's not clear to me what's going on at point C. I see two options.

Option 1: The long rod is resting against the rounded, frictionless end of the short rod at C. The other end of the short rod is fixed in place at B so that the short rod can't rotate about B.

Option 2: The rods are free to rotate about points A and B. The two rods are pinned together at C so that they are free to rotate about the pin. (It doesn't look like the two rods are pinned together in the figure.)

Since you are told to neglect the weights of both rods, I think it's probably option 2 that's intended. Your first solution would apply to option 1. Your second solution would apply to option 2.

 

[ClearWhey]

It's not clear to me what's going on at point C. I see two options.

Option 1: The long rod is resting against the rounded, frictionless end of the short rod at C. The other end of the short rod is fixed in place at B so that the short rod can't rotate about B.

Option 2: The rods are free to rotate about points A and B. The two rods are pinned together at C so that they are free to rotate about the pin. (It doesn't look like the two rods are pinned together in the figure.)

Since you are told to neglect the weights of both rods, I think it's probably option 2 that's intended. Your first solution would apply to option 1. Your second solution would apply to option 2.

Wow I didn't know it was this complicated haha. Thanks for clarifying it to me, now I understand how to think about these kind of questions. Thanks for the help!

 

[ClearWhey]

It's not clear to me what's going on at point C. I see two options.

Option 1: The long rod is resting against the rounded, frictionless end of the short rod at C. The other end of the short rod is fixed in place at B so that the short rod can't rotate about B.

Option 2: The rods are free to rotate about points A and B. The two rods are pinned together at C so that they are free to rotate about the pin. (It doesn't look like the two rods are pinned together in the figure.)

Since you are told to neglect the weights of both rods, I think it's probably option 2 that's intended. Your first solution would apply to option 1. Your second solution would apply to option 2.

I also have one more question about the way I solved it the first time. I can only remove the rod CB and write a force F like that if the rod CB was perpendicular right?

 

[TSny]

I also have one more question about the way I solved it the first time. I can only remove the rod CB and write a force F like that if the rod CB was perpendicular right?

I'm not sure what you mean by removing the rod CB. You are interested in the direction of the force F that rod CB exerts on rod AD. Suppose rod AD rests against the end of rod CB as shown below. This is option 1 in post #5.

If the surfaces are frictionless at the point of contact, can you deduce the direction of the force F? In the figure, I've exaggerated the fact that the two rods are not perpendicular to each other.

 

[ClearWhey]

I'm not sure what you mean by removing the rod CB. You are interested in the direction of the force F that rod CB exerts on rod AD. Suppose rod AD rests against the end of rod CB as shown below. This is option 1 in post #5.
View attachment 273143

If the surfaces are frictionless at the point of contact, can you deduce the direction of the force F? In the figure, I've exaggerated the fact that the two rods are not perpendicular to each other.

I don't know the word in english but by removing I mean when you draw the object and the forces (what I learned in class) you can sometimes remove certain objects and then draw the force it makes on the other object. Hope you understand what I mean.

If the rod CB was perpendicular for instance it would be easier to just remove it and then draw a force F that will impact rod AD. Then calculate force F from that instead of drawing the whole figure and calculate force F from point B right? Then you have to use cos or tan and so on...

 

[TSny]

I don't know the word in english but by removing I mean when you draw the object and the forces (what I learned in class) you can sometimes remove certain objects and then draw the force it makes on the other object. Hope you understand what I mean.

Ok. I think you are describing the idea of drawing a free body diagram for each rod.

If the rod CB was perpendicular for instance it would be easier to just remove it and then draw a force F that will impact rod AD. Then calculate force F from that instead of drawing the whole figure and calculate force F from point B right? Then you have to use cos or tan and so on...

If rod AD rests against rod BC and if there is no friction between the rods, then there is only a normal force between the surfaces at the point of contact. So F is the normal force that rod BC exerts on AD.

 

[ClearWhey]

Ok. I think you are describing the idea of drawing a free body diagram for each rod.

Yeah that is what I meant.

If rod AD rests against rod BC and if there is no friction between the rods, then there is only a normal force between the surfaces at the point of contact. So F is the normal force that rod BC exerts on AD.

Yes, like I thought. Thanks for the explanation!