Statics Problem Help?
I need help with this statics question, because it seems easy enough but I know I am not doing something correctly. Look at the picture and use it along with the question, which says;
Four forces and a couple act on the beam. The vector sum of the forces is zero, and the sum of
the moments about the left end of the beam is zero. What are the forces Ax, Ay, and B?
Ax is of no importance, it can be anything because it does not affect the moment because it line of action intersects the point of rotation, so i don't care about it. So it would seem to me that the 800N force multiplied by the 4m, gives 3200Nm plus the 200Nm from the couple makes 3400Nm, and since it has to be counteracted by B, it would be divided by 7m which equals 485.7Nm. But what is Ay? It seems to me it can be anything, as long as B is adjusted accordingly to account for it, but I think I need an actual value.
Is Ay suppose to be part of the couple, because if it is it would have a value of 25N which would factor into the 800N because that force lies on the same place where the other portion of the couple would be. So would 25N be Ay because it says in the question that 4 forces (800N, Ax, Ay, B) AND A COUPLE, which it makes me think that Ay isn't part of the couple?
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The problem says that the vector sum of the forces is zero.
The sum of the forces in the x-direction is: Ax + 0 + 0 +0 = 0. Thus Ax=0.
The sum of the forces in the y-direction is: Ay+ B -800N =0
Then there's the sum of your moments:
-4m*Ay - 200 N*m - 3200 N*m + 7m*B = 0
So you've got two equations with two unknowns:
Ay + B - 800 = 0
-4*Ay + 7*B -3400 =0
And the solution to this is Ay = 200N, B = 600N
The sum of the forces in the x-direction is: Ax + 0 + 0 +0 = 0. Thus Ax=0.
The sum of the forces in the y-direction is: Ay+ B -800N =0
Then there's the sum of your moments:
-4m*Ay - 200 N*m - 3200 N*m + 7m*B = 0
So you've got two equations with two unknowns:
Ay + B - 800 = 0
-4*Ay + 7*B -3400 =0
And the solution to this is Ay = 200N, B = 600N
Oct 1, 2011. 3:56 PMWesley666 (author)
says:
I am going to try this out, because I these seems sound, but I definitely need to write it out for myself.
Also, if a force acts with a line of action that intersects the point of rotation, it won't cause a moment, no matter how strong or weak the force is, it will only cause a translation, so Ax can be any value, that I am 100% sure of though. If you take a bottle and loosen the cap, and push on the cap with your finger so that your finger would be perpendicular to the tangent at that spot, nothing (Rotationally) will happen, no matter how hard you push. If you move your finger so that it is pushing the same direction, but is now not perpendicular to the tangent at that spot, the cap will turn.
Also, if a force acts with a line of action that intersects the point of rotation, it won't cause a moment, no matter how strong or weak the force is, it will only cause a translation, so Ax can be any value, that I am 100% sure of though. If you take a bottle and loosen the cap, and push on the cap with your finger so that your finger would be perpendicular to the tangent at that spot, nothing (Rotationally) will happen, no matter how hard you push. If you move your finger so that it is pushing the same direction, but is now not perpendicular to the tangent at that spot, the cap will turn.
Oct 1, 2011. 3:59 PMWesley666 (author)
says:
But then when I read it again and it says the sum of the forces is zero then yes it would be zero, in this case, but if the question said that the forces that caused moments were equal to zero, then it could be any value, because, like I said below, it doesn't cause a moment. So Ax=0. I overlooked Ax too quickly, and what was being asked.
The picture of this thing kind of looks like a board bolted to a wall, with the bolt stuck through a hole 4m from the left end, but I don't think that's what it is.
Rather, I think this is a board floating in space, with these four mysterious forces, and also one mysterious pure moment acting on that point where I used to think the bolt was.
Moreover I think these conditions of the vector sum of the forces equal to zero and sum of the moments equal to zero, are the conditions necessary to keep that board just floating in space there, and have it not rotate.
I'll admit that a board floating in space surrounded by mysterious forces (plus one mysterious torque) is kind of a weird thing, but uh... if it seems abstract, I think that's because it's a homework problem. I honestly think the point here was just to get you to write those three equations, and then solve them.
This problem would be more "fun" if that "beam" were drawn as an airplane, instead of what looks like a board bolted to a wall. Erm... Maybe...
Rather, I think this is a board floating in space, with these four mysterious forces, and also one mysterious pure moment acting on that point where I used to think the bolt was.
Moreover I think these conditions of the vector sum of the forces equal to zero and sum of the moments equal to zero, are the conditions necessary to keep that board just floating in space there, and have it not rotate.
I'll admit that a board floating in space surrounded by mysterious forces (plus one mysterious torque) is kind of a weird thing, but uh... if it seems abstract, I think that's because it's a homework problem. I honestly think the point here was just to get you to write those three equations, and then solve them.
This problem would be more "fun" if that "beam" were drawn as an airplane, instead of what looks like a board bolted to a wall. Erm... Maybe...
Oct 1, 2011. 2:22 PMFoolishSage
says:
It seems to me you are trying to solve 1 equation (sum of all forces is zero) with 2 unknown variables (A and B) and therefore the problem is unsolvable, but there are two equations.
1. vectors add to zero
2. moments on the left add to zero
Figure out what is meant and then you have two equations with two variables leading to a single solution. If you have trouble interpreting the question try asking your teacher.
1. vectors add to zero
2. moments on the left add to zero
Figure out what is meant and then you have two equations with two variables leading to a single solution. If you have trouble interpreting the question try asking your teacher.
Oct 1, 2011. 3:44 PMWesley666 (author)
says:
This is like Jack O Lopez's comment above, and I think it is the right way to go, so I will try it out.
Oct 2, 2011. 2:25 AMFoolishSage
says:
I was trying to give a hint without solving it for you but it seems jack already gave you the answer :p
Oct 2, 2011. 6:24 PMWesley666 (author)
says:
Yes, I know why you weren't giving away the answer, just the method for solving it. But thank you for helping though! :D
I think you are right...
Unless they want the forces expressed as values of each other
(B = 485.7nM + 4/7*Ay)
...or something.
Unless they want the forces expressed as values of each other
(B = 485.7nM + 4/7*Ay)
...or something.
Oct 1, 2011. 3:46 PMWesley666 (author)
says:
We rarely get questions that ask for answers expressed as values of each other, but I did think of it. I am going to try some of the above suggestions, and if none work out forward and backwards, then I will use this as a last chance answer.
Oct 1, 2011. 3:43 PMWesley666 (author)
says:
Yes, I know it is, but I want to know if Ay is one part of the couple, because if it is, then its magnitude would be 25N.
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