Free body diagram of pin support and cable

[whitejac]

Homework Statement

A) Draw a free body
B) determine the tension in rope A and the reactions in B.

Homework Equations

[/B]
Standard equilibrium equations

The Attempt at a Solution

My free body shows a collinear force T through the rope, the given downward force at 5ft, and then a Bx and By at the pin.
My confusion is this: the rope cannot exert a countering horizontal force because they're facing the same way. So the only thing is Bx, but cos(30)*600=513 and my book says Bx=493

 

[SteamKing]

Homework Statement

View attachment 89997
A) Draw a free body
B) determine the tension in rope A and the reactions in B.

Homework Equations

[/B]
Standard equilibrium equations

The Attempt at a Solution

My free body shows a collinear force T through the rope, the given downward force at 5ft, and then a Bx and By at the pin.
My confusion is this: the rope cannot exert a countering horizontal force because they're facing the same way. So the only thing is Bx, but cos(30)*600=513 and my book says Bx=493

You can't push on the rope.

The rope is in tension, and because the rope makes an angle to the vertical, there must be some non-zero horizontal component of the tension in the rope.

The horizontal reaction in the pin at B is what keeps everything in equilibrium horizontally.

 

[whitejac]

Hmm, I think I understand. The tension experiences equal and opposite reactions, much like the pin, but in this case then... I do not understand how we can find it. It sounds like a solution with too many variables for my unknowns. Whatever does or doesn't react at B will be absorbed by T. So how do I find the scaling?
So far I have:
[TSin(30) + By - 600Sin(30)]lb = 0
[TCos(30) + Bx - 600Cos(30)]lb = 0

My thought was to consider the moments - which could not be resisted by either side... persay, however at the point of the force it would exert equal and opposite moments at the ends, but the way I calulated it didn't seem to make much sense so I don't think it is the direction we're supposed to go...?
So 600Sin(30) * 5 = 600Sin(30) * 9, which doesn't make sense at all.

 

[SteamKing]

Hmm, I think I understand. The tension experiences equal and opposite reactions, much like the pin, but in this case then... I do not understand how we can find it. It sounds like a solution with 2 variables and 2 equations.

Most statically determinate beam problems are.

Instead of having simple supports at each end of the beam, you have a rope and a pinned connection.

In any event, you won't be able to solve this problem if you don't follow the steps laid out in the problem statement.

Once you have isolated the free body, you can then write the equations of static equilibrium and solve them.

 

[whitejac]

** I edited my last remark, showing you some of what I'm confused about. I can upload an image of how I view the diagram as well, if you think it'll be of benefit

 

[SteamKing]

Hmm, I think I understand. The tension experiences equal and opposite reactions, much like the pin, but in this case then... I do not understand how we can find it. It sounds like a solution with too many variables for my unknowns. Whatever does or doesn't react at B will be absorbed by T. So how do I find the scaling?
So far I have:
[TSin(30) + By - 600Sin(30)]lb = 0
[TCos(30) + Bx - 600Cos(30)]lb = 0

First, let's get the correct force equations. The angles for the rope and the 600-lb. load are measured from the vertical, not the horizontal. If you want to calculate the horizontal component of the tension or the applied load, which trig function do you use? For calculating the vertical components of each? If it's not obvious, draw a triangle and label the opposite side, the adjacent side, and the hypotenuse.

My thought was to consider the moments - which could not be resisted by either side... persay, however at the point of the force it would exert equal and opposite moments at the ends, but the way I calulated it didn't seem to make much sense so I don't think it is the direction we're supposed to go...?
So 600Sin(30) * 5 = 600Sin(30) * 9, which doesn't make sense at all.

Since you are worried about having too many unknowns and not enough equations of statics with which to solve them, why don't you write your moment equation about the pin location at B? Here, the only forces creating a moment acting on the beam are the vertical components of the rope tension and the 600-lb. load. But, before you write this moment equation, make sure you are using the correct force components (See comment above).

 

[whitejac]

Okay, and I apologize for sounding so inept at this. Free Bodies are my weakest skill in engineering. I know they ought to be intuitive, but for some reason I always get my arrows mixed up.

Here's my understanding of the forces. I'm not saying it's correct, but here's what I see:

I drew it like this because the moment is lifting the bar, and therefore countering the downward force. The horizontal component of T is because you cannot push a rope, as you said. So something is keeping it taught, you the rope must resist it. So at the wall or whatever, there is a force to the left and at the bar there is a force to the right (this is how tension is pictured in my book).

However, I am not sure that this is correct. The math implies something differently when I examine the unit circle. Because the angle of T is 30 ( + 90) degrees, the sine(θ) will make it negative while the cosine(θ) will remain positive. So there's a disconnect between what my mind sees and what I think the mathematics demands...

So:
∑Fy = By + Tcos(30) - 600cos(30)lb
∑Fx = Bx - Tsin(30) - 600sin(30)lb
∑Mb = 600cos(30)*9 ft-lb - Tcos(30) * 14

Solving for T = (600 * 9) / 14 = 386lb
Solving for Bx = (386 / 2) + (600 / 2) = 493 lb
Then, simply, By = 185

These are the answers in my book, so I know they're correct, but I can't seem to mentally visualize the reactions. I can mix arrows and algebra until I reach a desired result, but is it really supposed to be this difficult to see up front what should be common intuition?

 

[SteamKing]

Okay, and I apologize for sounding so inept at this. Free Bodies are my weakest skill in engineering. I know they ought to be intuitive, but for some reason I always get my arrows mixed up.

Here's my understanding of the forces. I'm not saying it's correct, but here's what I see:
View attachment 90053

I drew it like this because the moment is lifting the bar, and therefore countering the downward force. The horizontal component of T is because you cannot push a rope, as you said. So something is keeping it taught, you the rope must resist it. So at the wall or whatever, there is a force to the left and at the bar there is a force to the right (this is how tension is pictured in my book).

However, I am not sure that this is correct. The math implies something differently when I examine the unit circle. Because the angle of T is 30 ( + 90) degrees, the sine(θ) will make it negative while the cosine(θ) will remain positive. So there's a disconnect between what my mind sees and what I think the mathematics demands...

So:
∑Fy = By + Tcos(30) - 600cos(30)lb
∑Fx = Bx - Tsin(30) - 600sin(30)lb
∑Mb = 600cos(30)*9 ft-lb - Tcos(30) * 14

Solving for T = (600 * 9) / 14 = 386lb
Solving for Bx = (386 / 2) + (600 / 2) = 493 lb
Then, simply, By = 185

These are the answers in my book, so I know they're correct, but I can't seem to mentally visualize the reactions. I can mix arrows and algebra until I reach a desired result, but is it really supposed to be this difficult to see up front what should be common intuition?

The way you have drawn the components of the tension in the rope doesn't match the direction of the rope; specifically, the horizontal component of the tension is pointing in the opposite direction of how it should be pointing. When you combine Tx and Ty, you should obtain the magnitude and direction of T.

When starting a problem like this, it is a must to establish a consistent set of positive force and moment orientations. If your calculations result in a negative value, then the assumed orientation of the force or moment was opposite of what it should be.

We work out calculations for these kinds of problems because relying solely on intuition can sometimes lead us astray. This is not to say that you won't develop a better feel for how the calculations will turn out after working more problems and developing some experience dealing with them.

 

[whitejac]

When you combine Tx and Ty, you should obtain the magnitude and direction of T.

That makes a lot of sense (And I'm a little embarassed I broke down a vector without considering the fact it should add back up)...

Thank you for helping me with this problem. I can definitely see now that intuition doesn't lead us down the proper path, this problem is the perfect example haha. Every video and professor I've seen doing these types of things are so organized and adament about their consistent methodology. That sounds like the most surefire way to take deliberately and not let anything slip through your fingers.

 

[RETAJ8282]

Okay, and I apologize for sounding so inept at this. Free Bodies are my weakest skill in engineering. I know they ought to be intuitive, but for some reason I always get my arrows mixed up.

Here's my understanding of the forces. I'm not saying it's correct, but here's what I see:
View attachment 90053

I drew it like this because the moment is lifting the bar, and therefore countering the downward force. The horizontal component of T is because you cannot push a rope, as you said. So something is keeping it taught, you the rope must resist it. So at the wall or whatever, there is a force to the left and at the bar there is a force to the right (this is how tension is pictured in my book).

However, I am not sure that this is correct. The math implies something differently when I examine the unit circle. Because the angle of T is 30 ( + 90) degrees, the sine(θ) will make it negative while the cosine(θ) will remain positive. So there's a disconnect between what my mind sees and what I think the mathematics demands...

So:
∑Fy = By + Tcos(30) - 600cos(30)lb
∑Fx = Bx - Tsin(30) - 600sin(30)lb
∑Mb = 600cos(30)*9 ft-lb - Tcos(30) * 14

Solving for T = (600 * 9) / 14 = 386lb
Solving for Bx = (386 / 2) + (600 / 2) = 493 lb
Then, simply, By = 185

These are the answers in my book, so I know they're correct, but I can't seem to mentally visualize the reactions. I can mix arrows and algebra until I reach a desired result, but is it really supposed to be this difficult to see up front what should be common intuition?

Hello what is the source can you give me his name

 

[haruspex]

Hello what is the source can you give me his name

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