What is the resulting force between the beam CB and E?

[ClearWhey]

Homework Statement

What is the resulting force between the beam CB and E? Look at the first picture!

Relevant Equations

Look below on picture two for my equations.

I assumed that the vertical force at point B would be the same as the force between beam CB and E because of Newtons law. Did I assume wrong? Look below for my calculations and answer which I got to 250N. I don’t know the correct answer but assume someone here can tell me if I solved the problem correctly?

Thanks in advance for the all the help!

 

[TSny]

Welcome to PF!

Looks like you assumed that point A is vertically above point B. That is, you did not take into account the 30 unit horizontal displacement between A and B.

 

[ClearWhey]

Welcome to PF!

View attachment 269480

Looks like you assumed that point A is vertically above point B. That is, you did not take into account the 30 unit horizontal displacement between A and B.

Thank you and thanks for trying to help me! What should I do? Have Not encountered a problem like this before. Btw did I assume right that by solving the vertical force at B I will get the force on E or is that wrong too?

 

[haruspex]

Thank you and thanks for trying to help me! What should I do? Have Not encountered a problem like this before. Btw did I assume right that by solving the vertical force at B I will get the force on E or is that wrong too?

@TSny is pointing out that the vertical component of compression Q acting on A also has a moment about B.

 

[TSny]

Thank you and thanks for trying to help me! What should I do? Have Not encountered a problem like this before.

What if you find the horizontal and vertical components of the force $\vec Q$ at point A? Can you find the moment of each of these components about point B?

Btw did I assume right that by solving the vertical force at B I will get the force on E or is that wrong too?

Yes. You got this by considering moments about C. Looks right to me.

 

[ClearWhey]

What if you find the horizontal and vertical components of the force $\vec Q$ at point A? Can you find the moment of each of these components about point B?

Yes. You got this by considering moments about C. Looks right to me.

English is not my native language so I have some difficulties understanding your sentence so sorry if I’m missunderstanding. But I take it you mean I should get the vertical and horizontal forces at A? I did that if you look at the next to last equation, took Q which I got from previous equations and times cos(25) to get the vertical force at point A.

 

[haruspex]

English is not my native language so I have some difficulties understanding your sentence so sorry if I’m missunderstanding. But I take it you mean I should get the vertical and horizontal forces at A? I did that if you look at the next to last equation, took Q which I got from previous equations and times cos(25) to get the vertical force at point A.

Yes, but you did not include it your sum of moments about B.

 

[TSny]

When you took moments (torques) about point B, you did not correctly find the total moment (torque) due to force Q. The horizontal component of Q has a nonzero moment about A and the vertical component of Q also has a nonzero moment about A. The part of your work that I circled in orange in post #2 is the moment due to the horizontal component of Q. You need to include the moment due to the vertical component of Q.

 

[ClearWhey]

When you took moments (torques) about point B, you did not correctly find the total moment (torque) due to force Q. The horizontal component of Q has a nonzero moment about A and the vertical component of Q also has a nonzero moment about A. The part of your work that I circled in orange in post #2 is the moment due to the horizontal component of Q. You need to include the moment due to the vertical component of Q.

From what I learned from the book because the force 50N is horizontal I only need to add/subtract the horizontal force from Q if I choose to take the moment from point B. But now you say I need to add the vertical force from point A too? Why? And do I just add the vertical force to the equation?

 

[haruspex]

From what I learned from the book because the force 50N is horizontal I only need to add/subtract the horizontal force from Q

Then you need to unlearn that because you must have misunderstood.
Three forces act on that member: the horizontal 50N, the force Q acting along DA and the reaction force at B.
The force at B has no moment about B; the 50N is purely horizontal, so is easy.
Because you know the horizontal and vertical displacements of A from B, the simplest way to deal with Q is to split it into horizontal and vertical components and express the moment each has about B. The horizontal component has a clockwise moment but the vertical component has an anticlockwise moment. Indeed, since the line DA projected passes very close to B, the two moments must almost cancel.

 

[ClearWhey]

Then you need to unlearn that because you must have misunderstood.
Three forces act on that member: the horizontal 50N, the force Q acting along DA and the reaction force at B.
The force at B has no moment about B; the 50N is purely horizontal, so is easy.
Because you know the horizontal and vertical displacements of A from B, the simplest way to deal with Q is to split it into horizontal and vertical components and express the moment each has about B. The horizontal component has a clockwise moment but the vertical component has an anticlockwise moment. Indeed, since the line DA projected passes very close to B, the two moments must almost cancel.

Alright I understand what you mean now and got the answer to 877N, confirmed with a friend and he got the same.

Now I understand you need to take into account the vertical and horizontal forces on a force that is skewed (can’t find another word to describe force Q).

Can you please explain to me why in this problem my teacher only took into account the vertical force on N and not the horizontal too when she took moment on A?

The problem was to find the forces at point A and point B.
Thanks for taking your time helping me!

 

[haruspex]

I assume you mean the force acting up at 45 degrees at point D.
If you resolve that into horizontal and vertical components, the horizontal component acts along ADC, so has no moment about A.
There are other ways of calculating moments. Here, instead of resolving N, we could just multiply N by its lever arm, i.e. the perpendicular distance from its line of action to A. That gives $N (\frac a{\sqrt 2})$, which is the same result.

 

[ClearWhey]

I assume you mean the force acting up at 45 degrees at point D.
If you resolve that into horizontal and vertical components, the horizontal component acts along ADC, so has no moment about A.
There are other ways of calculating moments. Here, instead of resolving N, we could just multiply N by its lever arm, i.e. the perpendicular distance from its line of action to A. That gives $N (\frac a{\sqrt 2})$, which is the same result.

Ohh now I get it, if the force Q had been straight above point B then I would ’t need to take the horizontal force into account? Did I understand it right?

 

[TSny]

Ohh now I get it, if the force Q had been straight above point B then I would ’t need to take the horizontal force into account? Did I understand it right?

You wouldn't need to take the vertical component of Q into account.

 

[ClearWhey]

You wouldn't need to take the vertical component of Q into account.

Yes the vertical component, wrote wrong. Thanks for all the help!