Posers and Puzzles

Posers and Puzzles

  1. Joined
    15 Jun '06
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    16334
    14 Nov '07 00:38
    A 0.500 kg mass on a 3.20 meter string moves 2.5 radians/sec (1 revolution = 2 x pie x radians) in a horizontal circle. What angle does the string make with the vertical?
  2. Joined
    15 Jun '06
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    14 Nov '07 00:451 edit
    Originally posted by tomtom232
    A 0.500 kg mass on a 3.20 meter string moves 2.5 radians/sec (1 revolution = 2 x pie x radians) in a horizontal circle. What angle does the string make with the vertical?
    heres another one

    A sled weighing 200N rests on a 15 degree incline, held in place by static friction. The coefficient of static friction is 0.5. (a) What is the magnitude of the normal force on the sled? (b) Whatis the magnitude of the static friction on the sled? (c) The sled is now pulled up the incline at a constant speed by a child. The child weighs 500N and pulls on the rope with a constant force of 100N. The rope makes an angle of 30 degrees with the incline and has negligible weight. What is the magnitude of the kinetic fricition force on the sled? (d) What is the coefficient of kinetic friction between the sled and the incline?
  3. Joined
    09 Nov '07
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    140
    14 Nov '07 07:40
    first question is simple conical pendulum stuff:

    let tension=T, weight = mg, angular velocity = w, length of string l, angle x.
    resolve forces vertically:
    Tcos[x] = mg
    horizontally:
    Tsin[x] = m(w^2) x lsin[x]
    so T = m(w^2)l

    subsituting into the first equation:
    cos[x] = g/(w^2)l

    and so x = 1.1, about 61 degrees.
  4. Standard memberPBE6
    Bananarama
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    14 Nov '07 14:43
    Originally posted by etotheipi
    first question is simple conical pendulum stuff:

    let tension=T, weight = mg, angular velocity = w, length of string l, angle x.
    resolve forces vertically:
    Tcos[x] = mg
    horizontally:
    Tsin[x] = m(w^2) x lsin[x]
    so T = m(w^2)l

    subsituting into the first equation:
    cos[x] = g/(w^2)l

    and so x = 1.1, about 61 degrees.
    I agree, but x here is the angle the rope makes with the vertical. The "dip" is the additive inverse, 90 - 61 = 29 degrees.
  5. Joined
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    14 Nov '07 18:48
    Originally posted by PBE6
    I agree, but x here is the angle the rope makes with the vertical. The "dip" is the additive inverse, 90 - 61 = 29 degrees.
    Huh? 😕 Doesn't the question ask for the angle with the vertical?
  6. Joined
    09 Nov '07
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    14 Nov '07 19:01
    for the second one:

    let normal reaction force = R, frictional force up plane = F, coeff. of friction = m.

    a)
    resolving parallel to the plane:
    R = 200cos[15] = 193N

    b)
    limiting equilibrium implies F = mR
    so F = 97N

    c)
    the frictional force now acts down the plane, so
    resolving along the plane:
    F = 100cos[30] = 87N

    d)
    resolving parallel to the plane:
    R + 100sin[30] = 200
    so R = 150N
    limiting equilibrium implies F = mR
    so m = 0.58

    Last part im not so sure about -- can you use limiting equilibrium in this case??
  7. Standard memberPBE6
    Bananarama
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    14 Nov '07 22:10
    Originally posted by etotheipi
    Huh? 😕 Doesn't the question ask for the angle with the vertical?
    Sorry, you're right.
  8. Joined
    09 Nov '07
    Moves
    140
    14 Nov '07 22:23
    Wow, ignore my parts c and d from above as I have ignored a force 😲

    c)
    frictional force now acts down the plane
    resolving along plane:
    F + 200sin[15] = 100cos[30]
    so F = 35N

    d)
    resolving parallel to the plane:
    R + 100sin[30] = 200cos[15]
    so R = 143N
    limiting equilibrium implies F = mR
    so m = 0.24

    I hope 🙂
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