How Do I Set Up Statics Forces and Moments Problems?

[akhmed966]

Homework Statement

In the attachment

Homework Equations

Sum of the forces in the X, Y, and Z
Sum of the Moments in the X, Y, and Z.

The Attempt at a Solution

I don't believe I need help solving the problem, I just need help setting it up.

I can find the vector of CD, which came out to 400i-100j-500k (I know I need to get the unit vector), and the tension in W is -(250*9.8)k. But after that I get lost. I tried to make a free body diagram, but I am not sure because the problem wants the force in B, so I think I have to break it down in components and what not, but I think it's the rod that throws me off.

How would I go about getting an equation for A and B, or would they just simply be Ax, Ay, Az, Bx, By, Bz and I simply take the sum of forces and moments? Like I said, I get lost after finding the vector of CD and tension in W.

 

[tiny-tim]

Hi akhmed966!

… I tried to make a free body diagram, but I am not sure because the problem wants the force in B, so I think I have to break it down in components and what not, but I think it's the rod that throws me off.

How would I go about getting an equation for A and B, or would they just simply be Ax, Ay, Az, Bx, By, Bz and I simply take the sum of forces and moments? Like I said, I get lost after finding the vector of CD and tension in W.

A free body diagram probably doesn't help when there are so many forces.

Yes, break it into Ax, Ay, Az, Bx, By, Bz and take the sum of forces and moments …

I don't think there's any short way of doing it (though you may be able to make it slightly shorter by eg taking moments about the most convenient point, and taking components perpendicular to one of the cables)

 

[akhmed966]

Ok, thanks.

I got three equations so far.

Sum of X = Tcd*.617+Ax+Bx = 0
Sum of Y = -Tcd*.154+Ay+By = 0
Sum of Z = -Tcd*.77+Az+Bz = 2450

7 unknowns and 3 equations. Not looking too good so far.

So, I was thinking I should take the moment about point A since that would reduce 3 unknowns, but I don't exactly remember how to do 3D moments with bars, I'm pretty sure I have to do the cross product, I just don't remember how to get the sum. How would I go about taking the moment at point A?

 

[tiny-tim]

Hi akhmed966!

Ok, thanks.

I got three equations so far.

Sum of X = Tcd*.617+Ax+Bx = 0
Sum of Y = -Tcd*.154+Ay+By = 0
Sum of Z = -Tcd*.77+Az+Bz = 2450

7 unknowns and 3 equations. Not looking too good so far.

Good point, but you've forgotten the extra information "rests against a smooth vertical wall at end B" … so the reaction at B is … ?

So, I was thinking I should take the moment about point A since that would reduce 3 unknowns, but I don't exactly remember how to do 3D moments with bars, I'm pretty sure I have to do the cross product, I just don't remember how to get the sum. How would I go about taking the moment at point A?

Yes, definitely take moments about somewhere, to get three extra equations (btw, taking moments about two points won't give you any extra information you can't get from the three linear equations) … personally, I'd go for D rather than A, since that eliminates two forces, and keeps things more symmetrical.

And the easiest way to do cross product is probably to use i x j = k etc.

btw, you might like to consider using components in the three directions along the bar, and the string CD, and perpendicular to both (instead of the usual x y and z) … I have a feeling that might make the equations simpler.

 

[akhmed966]

Hi akhmed966! Good point, but you've forgotten the extra information "rests against a smooth vertical wall at end B" … so the reaction at B is … ?

B? That's what he expects the answer to be in, but I have never handled a problem in that manner so I don't know how to modify the equation. Unless that is stating that it only has a force normal (N) opposite to that of the wall, so all I would need would be Bx (N = Bx) and not By and Bz?

Yes, definitely take moments about somewhere, to get three extra equations (btw, taking moments about two points won't give you any extra information you can't get from the three linear equations) … personally, I'd go for D rather than A, since that eliminates two forces, and keeps things more symmetrical.

If I take the moment about point D, wouldn't I not have anything usable data? If B is one variable, it would be like Az = 0, wouldn't it?

And the easiest way to do cross product is probably to use i x j = k etc.

btw, you might like to consider using components in the three directions along the bar, and the string CD, and perpendicular to both (instead of the usual x y and z) … I have a feeling that might make the equations simpler.

Yeah, that's the only method I actually use in 3d problems (or remember how), but it's like

|i j k|
|1 2 3| distance
|Tcd Tcd 3| tensions

And it would give me the moments in i, j, and k, and then I just sum up all the i, j, k moments and set them equal to zero, right? I remember doing that 2 months ago, when we started these problems.

I think I can solve this problem, but I just need to know if my assumption that Bx is all I need for point B, and if taking the moment about point D is valid or invalid.

Thanks again.

 

[tiny-tim]

Unless that is stating that it only has a force normal (N) opposite to that of the wall, so all I would need would be Bx (N = Bx) and not By and Bz?

That's right! There's no friction, so there can't be any force parallel to the wall, and so N = B_x.

If I take the moment about point D, wouldn't I not have anything usable data? If B is one variable, it would be like Az = 0, wouldn't it?

I don't understand. You'll have the reaction forces at both A and B.

Yeah, that's the only method I actually use in 3d problems (or remember how), but it's like
|i j k|
|1 2 3| distance
|Tcd Tcd 3| tensions

And it would give me the moments in i, j, and k, and then I just sum up all the i, j, k moments and set them equal to zero, right? I remember doing that 2 months ago, when we started these problems.

Yes, that's right (but I was actually not think ing about that matrix method, but the rather simpler method of writing r x A = (r_xi + r_yj + r_zk) x (A_xi + A_yj + A_zk))

I think I can solve this problem, but I just need to know if my assumption that Bx is all I need for point B, and if taking the moment about point D is valid or invalid.

Yes, just B_x. And you can always take moments about any point.

 

[akhmed966]

Oh, thank you so much.

I took the moment about D, and set the two remaining forces equal to zero. Sum of Moments about point D = r x Fb + r x Fa = 0, would that be right? Then I just gather the i, j, and k terms and set them each equal to 0, and have a total of 6 equation (3 from the sum of the x, y, and z).

Then I simply solve, would that be correct?

 

[tiny-tim]

I took the moment about D, and set the two remaining forces equal to zero. Sum of Moments about point D = r x Fb + r x Fa = 0, would that be right?

Yes, except you can't use the same r for both forces, can you?

Then I just gather the i, j, and k terms and set them each equal to 0, and have a total of 6 equation (3 from the sum of the x, y, and z).

Then I simply solve, would that be correct?

Yes.

 

[akhmed966]

Yes, except you can't use the same r for both forces, can you?


Yes.

No, thanks a ton.