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mechanics yet again watch

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    another past paper im stuck on!! ARGH!! exam on monday! plz help!!


    take a look at attachment
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    no1?
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    Which Mechanics is that? 2? They're some hard questions!
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    nop its a UCL past paper lol
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    7 isnt bad.
    Firstly, the string will become slack if the maximum extension on the way down is larger than a (because it will come back up the same height, greater than a, but the initial extension is only a).
    So the V^2 > ag comes from finding the velocity at which the particle will oscillate with amplitude a. By energy considerations, the particle initially has only KE and Elastic PE, then when it reaches the top it has zero KE, zero elastic PE (because the extension is 0) and all GPE, so
    1/2 mV^2 + 1/2k.a^2 = mag, where k is the spring constant. To find k, notice that initially, the string has extension a, and by equating forces
    mg = ka
    k = mg/a

    Therefore 1/2 mV^2 + 1/2mg/a * a^2 = mag
    1/2 V^2 + 1/2ag = ag
    V^2 = ag

    So for any larger V, the particle will go past the zero extension point, and the string will become slack.
    If V > ag, then when it reaches the point where the extension is 0, it will have velocity U and by considering energies, we have
    1/2mV^2 + 1/2mag = 1/2 m U^2 + mag
    V^2 + ag = U^2 + 2ag
    U^2 = V^2 - ag

    The ceiling is at a distance l from the point of zero extension, so the particle has velocity U from above. Then considering the energy change to GPE, when it hits the ceiling it will have mgl more GPE. Let the KE at the ceiling = 0 so we get the minumum value for V, then
    1/2 mU^2 = mgl
    1/2m(V^2 - ag) = mgl
    V^2 - ag = 2gl
    V^2 = g(2l + a)

    If V is larger than that, it will still have KE when it hits the ceiling, so
    1/2mU^2 = 1/2mW^2 + mgl
    U^2 = W^2 + 2gl
    W^2 = U^2 - 2gl

    If no energy is lost when the particle hits the ceiling, then it will just repeat the same thing over and over because no energy is lost in the system at all.
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    thanx ..ne1 got any idea bout others?
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    (2)
    We represent the results of the N trials by a string of N letters, with "F" representing failure and "S" success. There are NCm strings that have exactly m "S"s. Each such string has probability p^m q^(N - m). So P(exactly m successes) = NCm p^m q^(N - m).

    --

    P(man is at C on the evening of January N - 1)
    = P(camels are available on exactly 2 of the first (N - 1) days)
    = (N - 1)C2 p^2 (1 - p)^(N - 3)

    If the man is at C on the evening of January N - 1 then P(he reaches D on the evening of January N) = p. If, on the other hand, the man is not at C on the evening of January N - 1 then he can't reach D on the evening of January N.

    So

    P(man reaches D on the evening of January N)
    = p * P(man is at C on the evening of January N - 1)
    = (N - 1)C2 p^3 (1 - p)^(N - 3)

    --

    P(man reaches D on or before January 5)
    = P(camels are available for at least 3 of the first 5 days)
    = (1/2)^5 [5C3 + 5C4 + 5C5]
    = 1/2

    --

    (3)
    Let M and N be the normal reactions exerted by the ground on A and B.

    Let F and G be the rightwards and downwards components of the force that AB exerts on CD.

    See the attached diagram. The green arrows represent the forces that act on AB, and the red arrows those that act on CD.

    --

    Resolving horizontally for AB, F + W = mu M
    Resolving horizontally for CD, F + W = mu N

    Resolving vertically for AB, M + G = 2W
    Resolving vertically for CD, N = 2W + G

    Taking moments for AB about its midpoint,

    (W + M) cos(alpha) = (W + mu M) sin(alpha)

    --

    So

    M = N
    M = 2W + G
    2W - G = 2W + G
    G = 0

    N = M = 2W

    tan(alpha)
    = (W + M)/(W + mu M)
    = (3W)/(W + 2mu W)
    = 3/(1 + 2mu)

    alpha = arctan[3/(1 + 2mu)]

    --

    Horizontal component of the reaction at the joint
    = F
    = mu M - W
    = W(2mu - 1)

    Vertical component of the reaction at the joint
    = G
    = 0

    --

    Suppose that the reaction at the joint is zero. Then mu = 1/2 and alpha = arctan(3/2).

    --

    (4)
    Please see http://www.thestudentroom.co.uk/t81626.html for the answer to the first part.

    Let x and y be the angles shown in the first attached diagram. Then x = y. By considering two congruent triangles, we can then deduce the lengths shown in the second diagram. So x = pi/3.

    [corrected] Tension in middle part of rope = W exp(-(2/3)pi mu)

    P
    = Tension in middle part of rope * exp(-(2/3)pi mu)
    = W exp(-(4/3)pi mu)
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  1. File Type: pdf x statics 3.pdf (47.0 KB, 75 views)
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    For 3, the one with the rods in the shape of an X, I got close to the answer, but i think i might have misinterpreted the question because the last part suggests there is a non-trivial answer, which just doesnt make sense according to what i've done. Anyway, here:

    At A, there are 2 forces acting, the reaction, R from the ground and the frictional force, F. The total force downwards is 4w, so the reaction is 2w, and the friction is then 2uw in limiting equilibrium. The force at B is just W acting down.
    Let the rod AB lie at an angle of x to the horizontal. Take moments about the centre to get;
    0 = wcosx + Rcosx - Fsinx
    0 = (w+R) - Ftanx
    tanx = (w+R)/F = 3/2u
    :confused:
    Tbh, i cant see how their answer can work, because if u = 0, then clearly, both rods must be vertical, however, the angle from their answer would be 71 degrees, which makes no sense.

    The reaction at the centre can be found by taking moments about A. Giving
    2Rcos(2x-90) = Wcosx
    2RSin2x = Wcosx
    4Rsinxcosx = Wcosx
    R = W/4sinx

    Then the net reaction is clearly vertically upwards, so
    2Rcos(90-x) = 2Rsinx = 2sinxW/4sinx = W/2
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    (Original post by JamesF)
    The force at B is just W acting down.
    There is also a sideways force W at B.
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    5)
    Using F=ma,

    R = m.dv/dt
    -(av + bv²) = m.v.dv/ds
    -ds = [mv]/[av+bv²] dv
    -ds = [m]/[a + bv} dv
    -s = [m/b]ln(a+bv} + c

    at s=0, v = U,

    0 = [m/b]ln(a+bU) + c
    c = -[m/b]ln(a+bU)
    -s = [m/b]ln(a+bv) - [m/b]ln(a+bU)
    s = [m/b]ln{(a+bU)/(a+bv)} -----------------------------------------------(1)
    =====================

    s is max when particle comes to a rest, i.e. when v = 0.

    => s = [m/b]ln(1 + (b/a)U)
    ====================

    rearranging (1),


    ln{(a+bU)/(a+bv)} = bs/m
    k/(a+bv) = e^(bs/m), where k = a + bU
    ke^(-bs/m) = a + bv
    bv = ke^(-bs/m) - a
    b.ds/dt = ke^(-bs/m) - a
    b/[ke^(-bs/m) - a] ds = dt
    be^(bs/m)/[k - ae^(bs/m)] ds = dt

    integrating,

    -(m/a)ln(k - ae^(bs/m)) = t + d

    at t = 0, s = 0,

    -(m/a)ln(k - a) = d
    t = (m/a)ln{(k - a)/(k - ae^(bs/m))}
    ============================

    at max distance, s = [m/b]ln(1 + (b/a)U)

    when s = [m/b]ln(1 + (b/a)U),
    (bs/m) = ln(1 + (b/a)U)
    e^(bs/m) = 1 + (b/a)U
    ae^(bs/m) = a + bU
    k - ae^(bs/m) = k - a - bU
    k - ae^(bs/m) = a + bU - a - bU, substituting for k = a + bU
    k - ae^(bs/m) = 0

    giving,

    t = (m/a)ln{(k-a)/0}
    t = ∞ !!
    ======

    i.e. the mass will never actually reach the max distance in a finite time.
    ================================ =====================

    (Hmm, can someone give a better reasoning than the above. It doesn't seem very strong )


    Special case - a = 0
    ===============

    when a = 0,

    s = (m/b)ln(U/v)

    when v = ½U,

    s = (m/b)ln2
    =========

    from s = (m/b)ln(U/v)

    v = Ue^(-bs/m)
    ds/dt = Ue^(-bs/m)
    ∫ e^(bs/m) ds = ∫ U dt
    (m/b)e^(bs/m) = Ut + f

    at t = 0, s = 0, -> f = (m/b)

    t = (1/U){(m/b)(e^(bs/m) - 1)}
    t = (m/(bU))(e^(bs/m) - 1)

    when s = (m/b)ln2,
    bs/m = ln2
    e^(bs/m) = 2

    giving,

    t = (m/(bU))(2 - 1)
    t = m/(bU)
    ========
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    There was an error in my solution to Question 4, which I have now corrected.

    P equals W exp(-(4/3)pi mu), not W exp((4/3)pi mu).
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    thanx a lot guys just fingers crossd that i can pass tomorrow!!
 
 
 
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