At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
ok so at at the required moment of time!.. you see only 4 buckets contribute to the momentum .. !!
yes. 2,3,4 but not one. taking moment around the axis, \(\Sigma M= (W_2+W_3+W_4) \times 1.6m \)
bucket 1 has force line distance to axis=0, the others have no force line.
shadow thats wrong.. you cannot take 1.6m ..moment is calculate as the force multiplied by the perpendicular distance from the centre!!
lol actually, yes, but to simplify calculations my teacher taught us a trick: to extend the force vectors and take the distance from the center. Yes, that's a lil' illegal, but it can simplify question sols like this one :P
no.. i ll upload pic wiat!
and what you are saying is the CORRECT way of taking the perpendicular distance.. its not illegal.. its perfect!
lol it is quite magical.
@Mashy: Ok, so for x and z, I find the perpendicular distance by the formula sin45x1.6, right? And then multiply both of them by 400? For y, I simply multiply 1.6 by 400 to get the moment about the centre? Finally, I add the three values to get the resultant moment. Am I right so far?
you're welcome :)