Moments & couples 2D (for now)

random101

Hey guys and girls, im new to this forum but been looking at it for a while, its a great read and always a challenge here or there, but for now i was hoping to get a little help

:)
1. The problem statement, all variables and given/known data

The direction of the two thrust vectors of an experimental aircraft can be independently changed from the conventional forward direction within limits. For the thrust configuration shown, determine the equivalent forces-couple system at point O. then replace this force-couple system by a single force and specify the point on the x-axis through which the line of action of this resultant passes. These results are vital to assessing the design performance.


2. Relevant equations




any help would be really really helpful cheers!!

random101

of the middle of the plane is Point O giving me a moment i can rotate on

lol now i can get somewhere :)

Dick

I hope you are getting it, because as you say, the diagram is not too clear.

random101

yeah i have got a much better diagram now

should find Mx but yeah im really not too sure how to go about it

random101

please help :(

PhanthomJay

As noted, your image wasn't too clear, and now I don't see it at all. It appears though that you should find the resultant of the 2 thrust force vectors (magnitude and direction), and the torque of those 2 vectors about point O (break the lower thrust vector into components for simplicity in calculating its torque), then place that resultant force and couple at O. To find the equivalent force through the x axis, I'd divide the couple by the y component of the resultant force.

random101

yeah thats what i have been attempting, ill give it another good shot when i get home :)

thanks :)

random101

im not to sure what i can do with that angle of thrust

at 15degrees

random101

Attached Thumbnails

 

random101

mm i think i have gone down the wrong road with this question should i be looking at it as a torsional question? that might make life easier

random101

bump?

PhanthomJay

Nice pic. As I mentioned earlier, the resultant force is the sum of the 2 thrust vectors, and the resultant moment about O (couple, or torque, if you wish), is the sum of the moments of those vectors about O. Try breaking each vector into its x and y components. The upper force has an x component only. The lower one has an x component of Tcos15 and a y component of T sin 15. Sum torques of those componnents about O, watch plus and minus signs. Show your work so that we may be of further assistance. Your result will be a function of T, which is not given.

random101

is the answer 2.69T

Mo = 0: T3+tsin15x10 - Tcos15x3

learningphysics

EDIT: Oops sorry! You're right. I was wrong. I'm still confused by the Mo=0.

PhanthomJay

Your picture seems to come and go quicker than the Chesire Cat. I think as I view it that you have 2 thrust vectors T, spaced 6 meters apart vertically, 3 m each from the x axis. One is horizontal, the other 15 degrees up from the horizontal. I don't see how yu can arrive at 2.69T as a resultant .....even if they were both horizontal, they'd add up to 2T, so surely the resultant must be less. You are just going to have to show your work if you need more help. Your moment looks good, but don't set it equal to zero, it is what it is. Then you still have to determine at what point on the x axis the resultant falls.

alex63

how could i solve this question?

Three forces acting in the XY-plane are applied to a body.
F1= i + 3j applied at (2,5)
F2= 3i - 3j applied at (0,11)
F3= i + 12j applied at (5,-4)

(a) Show that the moments of F1 and F1 about the point P(3,8) are both zero.

(b) Find the moment of F3 about P

(c) Write down the moment of F about P