Recent content by leonkhasi

if i assume the angle θ like i mention then d=L/2/tan θ yes..but i guess i won't arrive at the solution so i must take θ as angle between hypotenuse and horizontal and d=tanθ.L/2. am i right?

i assume θ is the angle between the sag d and the hypotenuse..

Hello there is w1.g=2t1cosθ? and wat about should t2=w=t1? where t1 and t2 are tensions and should t1=t2 for tension tobe same in the system since force is transmitted on the same rope.

do we have to show the equilibrium of the forces at the point at the center ?

is d=tanθ⋅L/2? how do i determine how θ changes when i do not know basedwhich of the weights are heavier?

I'm just guessing how to arrive at the solution but i think its nowhere at all close to the solution :wideeyed:

[for d →0 i must say w is maximum.or i must say w1 is zero.

ouch I'm stuck.:frown:

i understand that w=1/2w1 is it not?

Homework Statement A pulley to which is attached w1 rides on a wire that is attached to a support at the left and that passes over a pulley on the right to weight w.the horizontal distance between the left support and the pulley is L.Express the sag d at the center in terms of w,w1 and L. The...