Recent content by Juniper7

  1. J

    Static Equilibrium with a beam and cable

    I'm sorry, I haven't learned to sum moments yet. Or what moments are. Is there a way to do it without that?
  2. J

    Static Equilibrium and tension of cables

    Ok. So coordinates for each point would be in relation to (0.2,0) If I do it how you suggest: T1(0m) + T2(1.6m) = (45kg)(9.81m/s2)(0.91m) T2 = 250.8N Then can I use my earlier equation to solve for T1? T1(0.2m) + T2(1.8m) = (45kg)(9.81m/s2)(1.11m) 0.2T1 + (250.8N)(1.8m) =...
  3. J

    Static Equilibrium with a beam and cable

    Ok, So that would be like the force of friction? working in the opposite direction of the applied force?
  4. J

    Static Equilibrium and tension of cables

    Hm ok. So: T1(0.2m) + T2(1.8m) = (45kg)(9.81m/s2)(1.11m) T = 245N So that means each cord has a tension of 245N?
  5. J

    Static Equilibrium and tension of cables

    Homework Statement A 30Kg neon sign is suspended by two cables, as shown. Three neighbourhood cats (5.0 Kg each) find the sign a comfortable place. Calculate the tension in each cable with the cats are in the positions shown. Homework Equations xCM = m1x1 + m2x2 ...
  6. J

    Static Equilibrium with a beam and cable

    Homework Statement A uniform vertical beam of mass 40kg is acted on by a horizontal force of 520N at its top and is held, in the vertical position, by a cable as shown in the attached picture. a) Calculate the tension in the cable b) Determine the reaction forces acting on the beam by the...
  7. J

    Solving for Angular Acceleration & Speed of Rod in Freefall

    Oh, the centre of mass? so the radius would be half of the length, 0.5m, right?
  8. J

    Solving for Angular Acceleration & Speed of Rod in Freefall

    1) Would that be at the end of the rod, the opposite end of the pivot? 2) if the point of application is the end, then it would be 1.0 m and the radius should be 1.0m, right?
  9. J

    Rotational Energy/Speed of a System?

    Oh ok. So because the cord is attached to the block, they should both have the same velocities?
  10. J

    Solving for Angular Acceleration & Speed of Rod in Freefall

    Ok, so: τ = Iα τ = 1/3ml2α τ = rF F = mg τ = rmg rmg = 1/3ml2α (1.0m)(12kg)(9.81m/s2) = (1/3)(12kg)(1m)2α α = 29.4rad/s2 This seems rather high... would the radius of the rod be its length because it is pivoting on its end? I think i did something wrong... thanks for the help
  11. J

    Rotational Energy/Speed of a System?

    Are the velocities relevant to the speed the sphere turns at and the speed the pulley turns at?
  12. J

    Solving for Angular Acceleration & Speed of Rod in Freefall

    Homework Statement Consider a uniform rod of mass 12kg and length 1.0m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine, a) the angular acceleration of...
Back
Top