Recent content by RobertLelik

  1. R

    Forces in Equilibrium acting on wall climber

    I see now. Perfect! Thank you very much.
  2. R

    Forces in Equilibrium acting on wall climber

    I can see now that you can substitute (Ty + Ff) - Fg for ((Ty + (µ x Tx))-Fg. So that would mean (T(0.990) + µT(0.139)) - 650 N = 0 N But how do I carry on from there? Tx = T sin 8 or T(0.139) = Fn Ty = T cos 8 or T(0.990)
  3. R

    Forces in Equilibrium acting on wall climber

    The only equation to calculate force of friction that I know is Ff = µ x Fn It may be the substitutions you mentioned that I am not understanding. Could you please elaborate?
  4. R

    Forces in Equilibrium acting on wall climber

    This is all that is given for the problem. The black is original. The green was me.
  5. R

    Forces in Equilibrium acting on wall climber

    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color] Hello and thank you in advance for any help. I am currently studying physics and having difficulty working out a forces at equilibrium problem. With limited given variables, I am unable...