# 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?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?

active| newest | oldestThe 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 = 600NAlso, 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.

looks likea 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 zeroandsum 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...

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.

Unless they want the forces expressed as values of each other

(B = 485.7nM + 4/7*Ay)

...or something.