The discussion revolves around calculating the combined moment of two forces about specified points (O, C, and D) given a force of 70 lb. The context involves understanding the principles of moments in mechanics, particularly in relation to forces and distances.
The discussion is active, with participants exploring different interpretations of the problem and clarifying concepts related to moments. Some guidance has been offered regarding the use of perpendicular distances and the application of the cross product, although no consensus has been reached on the specific calculations.
Participants note the difficulty in interpreting the problem due to unclear image references and question the assumptions about the distances involved. There is also mention of homework constraints regarding the points of interest for the moment calculations.
The image is a bit hard to read, but I assume the two forces are equal and opposite. M=Fd is right, where d is a certain distance. What distance? (It probably isn't the distance AB here.)J-dizzal said:
Which part of the question are you trying to answer here? It looks like you're trying find the moment about point B, which isn't listed in the question statement. It asks for moments about points O, C, and D.J-dizzal said:
haruspex said:
if the two forces are parallel then the distance AB is perpendicular because point A and point B are each on the axis of either forcegneill said:
Not collinear. If they were collinear there would be no moment.J-dizzal said:
It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.J-dizzal said:
yes sorry, they are not on the same axis non collinear.haruspex said:
i just tried using M=rxF and summed the two and got -2165.239## \hat k##haruspex said:
I thought the moment would be the same for any point? isn't that a properties of couple moments in 2d?gneill said:
well they are both points on the lines of application of the forces, i don't see how they are not the distance between the lines.haruspex said:
The distance between two parallel lines (or two skew lines ) is the shortest distance between points on them. If you pick two arbitrary points they are not necessarily as close as they could be, in fact they could be arbitrarily far apart.J-dizzal said:
Ok i can see that i was assuming incorrectly that the line AB is prependiharuspex said:
Ok, i was wrong to assume line AB was perpendicular to the couple. So would the best way to solve this would to be use ∑M = rxF from both the forces to point 0, then repeating that for the other two point?haruspex said:
Whichever way you do it will involve finding the perpendicular distance from a point to a line, so you might as well stick with your approach. You just need to do a bit of geometry to find that distance.J-dizzal said:
ok, but for future reference does the vector r in equation M= r x F have to be perpendicular to F?haruspex said:
No, the process of computing the cross product handles that. |r x F| = |r||F| sin(theta), where theta is the angle between the vectors.J-dizzal said:
well why didnt my calculations work for M = r Fharuspex said:
It'll take me a while to tease through your numbers, but using AB as the distance vector and taking the cross product I get 2520. Most likely you have a sign error somewhere.J-dizzal said:
ok thanks that is correct, i just did it the long way by summing the two moments together using ∑M = rF = 2520.227 unitsharuspex said:
so the quicker way is to cross rAB with F?haruspex said:
yes, i confirmed that M = r x F . and it works for both vectors as long as r and F are head to head and not tail to head.J-dizzal said:
It's certainly quicker.J-dizzal said:
indeed, i have an exam monday so i better remember this. thanks to all!haruspex said: