Statics: forces on a beam and reactions

[laekoth]

Homework Statement

a horizontal beam has a pin at 0m(point A), a 90N (-90i) force at .2m(point B), a pin at .4m(point C) and a 180N (-127.3i + -127.3j)at (45 degrees) force at .6m(point D). the pin at point A is attached to a vertical wall with rollers, the pin at point C can't move. The beam is in equilibrium.

Homework Equations

sum of forces in the x direction = 0
sum of forces in the x direction = 0
Moment about any point = 0

The Attempt at a Solution

The part of the problem that stumps is is that the y component of the force at point D is inline with two reaction forces at point A and point C and I don't see how to solve for each one alone.

Also, if the Moment of A = 0, then -90(.2) + Rbx(.4) + -sin45(180)(.6) = 0 and forces in x = 0 so 90 + sin45(180) - Rbx = 0 yet these both give different answers, so I assume I have a concept error here somewhere.

 

[PhanthomJay]

Homework Statement

a horizontal beam has a pin at 0m(point A), a 90N (-90i) force at .2m(point B), a pin at .4m(point C) and a 180N (-127.3i + -127.3j)at (45 degrees) force at .6m(point D). the pin at point A is attached to a vertical wall with rollers, the pin at point C can't move. The beam is in equilibrium.

I assume that the pin at A is attached to the top of the wall with rollers, or else it would be unstable.

Homework Equations

sum of forces in the x direction = 0
sum of forces in the Y [/color]direction = 0
Moment about any point = 0

yes...

The Attempt at a Solution

The part of the problem that stumps is is that the y component of the force at point D is inline with two reaction forces at point A and point C and I don't see how to solve for each one alone.

use your 3rd relevant equation

Also, if the Moment of A = 0, then -90(.2) + Rbx(.4) + -sin45(180)(.6) = 0

This is not right, if you are summing moments about A, then you want to look at moments due to vertical forces, and you don't mean Rbx, you mean Rcy; and get rid of that -90(.2) term

and forces in x = 0 so 90 + sin45(180) - Rbx = 0 yet these both give different answers, so I assume I have a concept error here somewhere.

You mean Rcx here.

 

[laekoth]

I had renamed the points to explain the question on here, and copied over my work incorrectly.

To clarify I attempted to draw the problem:

From how I understand the problem these should be correct:

Moment at A: -F1(.2) + Rcy(.4) + -F2y(.6) = -90(.2) + Rcy(.4) + -sin45(180)(.6) = 0
I could use any point to compute the moment at, I just chose A. I don't see how i can get rid of the -90(.2) term, can you explain?

forces in y: -F1 + -F2y + Rcy = -90 + -sin45(180) + Rcy = 0
forces in x: Ra + Rcx - F2x = Rax + Rcx - cos45(180) = 0

Just looking at the problem, it doesn't look like it *can* be in equilibrium. The problem is basic, and I haven't had a difficulties with similar problems.

 

[PhanthomJay]

In the OP, you noted that the force at B was -90i N. That is a horizontal 90 N force acting left. You have shown it as -90j N, as a vertical force acting down. Which one is it?

Then as I tried to explain, the beam cannot be in equilibrium as drawn and described, because the roller at A cannot take forces in the vertical direction, and thus, the beam wil rotate. The support at A must be on top of the wall, with no Ax reaction due to slippage in the x direction.

 

[laekoth]

The drawing is correct, it's -90j.

If i could scan the picture, i would. I guess I'm not crazy thinking the beam can't be in equilibrium, it's good to have a second opinion on that.

Edit: professor confirmed the problem was intentionally unsolvable.