Recent content by wah31

I tried another approach that I believe is the right (see attached file): For g’ = 0 Tension in the middle of the cable T0 H = (g’ / 8d) ∙ (S2 – 4d2)=0 V = G (the cable doesn't support any of it's own weight) T0 = G Tension at the end of the cable T1 H = (g’ / 8d) ∙ (S2 – 4d2)=0 V = G T1 =...

Yes you are right :check: Thank you.. it's cool :oldbiggrin:

I'm not convinced either I thought about it..

You just have to validate or not thes values (T1, P, T2, T3, T4, T5)

Homework Statement Man standing on scale and pulling on a rope connected to two pulleys Homework Equations The mass of the man is: m=100 kg The weight of the pulleys, cables, balance, and the frictions are not taken into account g=10m x s-2 3. The Attempt at a Solution the force that the...

Now if I summarize we get: For g’ = 0 Tension in the middle of the cable T0 Th = (g’ / 8d) ∙ (S2 – 4d2)=0 Tv = g’ ∙ S / 2 =0 T0 = 0 Tension at the end of the cable T1 Th = (g’ / 8d) ∙ (S2 – 4d2)=0 Tv = G T1 = (Th2 + Tv2)0.5 = G For g‘ ≠ 0 Tension in the middle of the cable T0 Th =...

It is the same .. You mean that T1 is equal to G?

That’s right, g’ is the cable weight per unit length I think G is there to balance the cable weight i.e. the vertical component of tension Tv These equations come from the catenary principle, have a look below: http://home.earthlink.net/~w6rmk/math/catenary.htm

Mod note: Moved from technical forum, so missing the HW template. Hello everybody and happy new year :partytime: Here is a solution for the following case. Do you agree with ? Thank you in advance for your answer The question is to give the values of T0 T1 and d in terms of g', L and G in...