OT - Any Math/static/engineer guys out there???

[Misfit6669]

I am having a brain fart and missing something on this problem. Not sure if Brandon is still around, he has saved my arse in the past but this is open for anybody.

I am not seeing how to find Theta or B, it seems like I need for info. If I pick a theta like 45 then I am fine to figure out B to be 692 lb which is a valid answer to get a mag of 1200 lb.

BOB (back of book) has an answer of 744 lb and a theta of 23.8 but I don't know how they got that without having more info. If I use 23.8 as theta then I also get 744 but again, not sure how they got it. I keep thinking that they just want me to pick an angle and find B but none of the other questions are like that. I have done the 10 problems before and after this one and got the right answers but I am stupid on this one.

WHAT AM I MISSING???

Any thoughts would be appreciated big time.

 

[murfman]

I'd say that in order to get 600# on B you would have to have at least 30*. My logic here is that you have A and 60* at the bottom angle gives you 600#. If you take the angle of B or Theta, at 30*, that gives you 60* at the bottom and hence 600# of force.

I am not a mechanical engineer, but I did stay at a Holiday Inn Express last night!!!

 

[Misfit6669]

Thanks Murf, a guy in the lounge was able to help me out. I wasn't getting the 300 along the x-axis on the left so I couldn't use tan theta = 300/680 which gets me to 23.8 which in turn gets me to 743 lb which are both good answers.

And Paul, who's the avatar?

 

[murfman]

Brian, that is Okla---homa ;=) Robert just loves her!!!!

 

[Misfit6669]

Nice one, do you have a better pic of her? I would to see how nice her, ughh... eyes look, yeah that's it, her eyes.

 

[Darknes]

Ah, I love these problems!

The total vertical force is 1200 pounds, so do a summation of the A and B component forces in the vertical direction:

1200 = 600*sin60 + T*cos(theta)

You need another equation (2 var's), so do a summation of the forces in the horizontal direction, which should equal zero:

0 = 600*cos60 - T*sin(theta)


Solve the two simultaneously, and get the answer:
T = 744
(theta) = 23.8

 

[Darknes]

I see you already figured it out.

 

[murfman]

So dark, in my non engineering logic, how close was I?

 

[Darknes]

<a href=showthread.php?s=&postid=10776783#post10776783 target=_blank>Originally posted</a> by murfman
So dark, in my non engineering logic, how close was I?

Your logic was close, but you misunderstood the problem.

The total vertical force should be 1200. However, the 600 pounds in rope A is not all vertical. Part of the force is pulling up, and part is pulling to the right.

Now, if you ignore the 1200 and balance out the ropes, your logic was dead on.

 

[reefkoi]

hmmmm I'd throw a chain around it and pull it out with the 9,000lbs of force my excavator produces LOL
Sorry stupid un-educated ditch digger mentality, ignore me.
Chris

 

[Misfit6669]

Ok guys, got a few more for you, here is your chance Mutt


<a href="http://picasaweb.google.com/misfit69/MyDocuments/photo#5119764230443451490"><img src="http://lh4.google.com/misfit69/Rw0OcSFVRGI/AAAAAAAAKcA/UKZ0qK6BxUw/s400/64.jpg" /></a>

<a href="http://picasaweb.google.com/misfit69/MyDocuments/photo#5119764277688091762"><img src="http://lh3.google.com/misfit69/Rw0OfCFVRHI/AAAAAAAAKcI/OYuBEnn7JXM/s400/134.jpg" /></a>

 

[murfman]

hey Brian, we hang out over at www.thescmas.com now.

 

[Darknes]

For the second problem, you know the sum of forces is -120 kN in the z-direction. You want to find the moment that each force exerts on the origin. It helps to tabulate the data:

distance: force: moment:
10x -20z 120y
4x + 3y -50z 200y - 150x
11y -20z -220x
10x + 13y -50z 500y - 650x

Sum of the moments: -1020x + 820y

Now to find the position:
(Xx+Yy)*-120z = -1020x + 580y
120Xy - 120Yx = -1020x + 580y

Split into 2 equations:
120X = 580; X= 4.83
-120Y = -1020; Y= 8.5

So, the force is -120kN at position X=4.83m, Y=8.5m

(Hope these numbers are correct )