Recent content by Ithilrandir

  1. I

    Difference in tension introduced by friction

    Right. Thanks, I tend to overlook simplifying things involving small values unless it's an angle.
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    Difference in tension introduced by friction

    OK I did a force body diagram, with T to the right, F to the left, N upwards and T + ΔT at an angle Δθ to the horizontal. Using small angle approximation, (T + ΔT)Δθ = N, μ(T + ΔT)Δθ = F (T + ΔT)cosΔθ = T - F, small angle approx, T + ΔT = T - F ΔT = F μ(T + ΔT)Δθ = ΔT If the μΔTΔθ part...
  3. I

    Difference in tension introduced by friction

    Right I see. The forces on the segment should be the friction, the normal force and the tension from both sides.
  4. I

    Difference in tension introduced by friction

    I can't see anything wrong with the formula as the force experienced by the cord is by turning a small degree against the post.
  5. I

    Difference in tension introduced by friction

    I have no clue how to do this, so I did my best guess of it. Friction is μN, N being the normal force. The normal force when it is in contract with the pole should be V2/R, R being the radius of the pole. So ΔT = μV2/R The answer provided is μTΔθ
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    Minimum length x for no slipping

    I manged to get the answer after pouring fixing some mistakes in my algebra with the new equation.
  7. I

    Minimum length x for no slipping

    Ah I see. I should've placed the friction where I had placed the normal force. I will test it tmr after work.
  8. I

    Minimum length x for no slipping

    The two frictional forces should be in the same direction, so isn't the net effect just 2F x 15?
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    Minimum length x for no slipping

    This doesn't seem right in calculations as I get 10.716 for X.
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    Minimum length x for no slipping

    My bad. 15F + 50(X-15) = 22N. Is this right?
  11. I

    Minimum length x for no slipping

    The Friction acts clockwise, the normal force acts anticlockwise. So 15F + 50W = 22N.
  12. I

    Minimum length x for no slipping

    The D was a typo, I had meant X. I have F acting on the inner side of the support acting upwards. The upper N is on the left side acting left, the lower N is on the right side acting right.
  13. I

    Minimum length x for no slipping

    I'm letting the weight of the hanger be W. Since there is no slipping, the total frictional force will be = total weight. When the load of 50W is placed at X, there'll be a normal force at the left end of the pole on top to the left, and another normal force at the right end of the pole at the...
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    Find the distance x so that the two frictional forces are equal

    I'm thinking it's because the net movement is from the centre of mass, motions from other points of reference will essentially have an added rotational element.
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    Find the distance x so that the two frictional forces are equal

    I didn't make use of angular acceleration in my calculations, so no. I just took into account of friction when I calculated the moment about point P.