Recent content by j123

-so the greatest tension is at the top of the chain. -so EFy is 6958N of boulder + 5831N of chain =12789N, right? (someone else said i should include the tension but i thought these were the only two forces and they are the T also)

-OK, (i thought the tension was these two forces) I am actually really fustrated and completely out of this. Can you explain? and am i even close to the end? i kinda gave up on this problem already, i really just want to know what I am doing wrong. (will check on this later today)

-so EFy is 6958N of boulder + 5831N of chain =12789N (isn't this divided by the two masses equal back to g?) -the max tension is 16326.8N for the chain -isn't the tension the greatest where the boulder pulls on the chain?

-the weight of the chain is w=mg so 595kg*9.8= 5831N x 2.8 times is = 16326.8N.? -where the boulder and the chain connects, i would say. -Newton's 2nd law: EFx=ma.x EFy=ma.y (how many steps do i apply? and how?, I think only EFy is only used since it's going up) so Fy/m = a.

Homework Statement A 710kg boulder is raised from a quarry 191m deep by a long uniform chain having a mass of 595kg. This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.80 times its weight without breaking. How do i find the maximum...