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1990 Paper 3 Q14
D: 1700.0 B: 1484.0
moments rigid body cube 3D vectors couple force system equilibrium line of action work-energy cross product

The edges \(OA,OB,OC\) of a rigid cube are taken as coordinate axes and \(O',A',B',C'\) are the vertices diagonally opposite \(O,A,B,C,\) respectively. The four forces acting on the cube are \[ \begin{pmatrix}\alpha\\ \beta\\ \gamma \end{pmatrix}\mbox{ at }O\ (0,0,0),\ \begin{pmatrix}\lambda\\ 0\\ 1 \end{pmatrix}\mbox{ at }O'\ (a,a,a),\ \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix}\mbox{ at }B\ (0,a,0),\ \mbox{ and }\begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix}\mbox{ at }B'\ (a,0,a). \] The moment of the system about \(O\) is zero: find \(\lambda,\mu\) and \(\nu\).

  1. Given that \(\alpha=\beta=\gamma=0,\) find the system consisting of a single force at \(B\) together with a couple which is equivalent to the given system.
  2. Given that \(\alpha=2,\beta=3\) and \(\gamma=2,\) find the equation of the locus about each point of which the moment of the system is zero. Find the number of units of work done on the cube when it moves (without rotation) a distance in the direction of this line under the action of the given forces only.

Solution: \begin{align*} &&\mathbf{M} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix}a\\ a \\ a \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \times \begin{pmatrix} a \\ 0 \\ a \end{pmatrix} \\ &&&= \begin{pmatrix} -a \\ -a(\lambda -1) \\ \lambda a \end{pmatrix} + \begin{pmatrix} -2a \\ 0 \\ -a \end{pmatrix} + \begin{pmatrix} \mu a \\ -a(1-\nu) \\ -a \mu \end{pmatrix} \\ &&&=a \begin{pmatrix} \mu - 3 \\ \nu - \lambda \\ \lambda-1-\mu \end{pmatrix} \\ \Rightarrow && \mu &= 3, \lambda = 4, \nu = 4 \end{align*}

  1. To find the force we add all vectors: \begin{align*} \mathbf{F} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \\ &= \begin{pmatrix}4\\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ 3 \\ 4 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 3 \\ 7 \end{pmatrix} \end{align*} Since the moment about \(O\) is \(0\), we have the moment about \(B\) is: \begin{align*} \mathbf{M} &= \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 7\end{pmatrix} \\ &= \begin{pmatrix} 7a \\ 0 \\ -4a\end{pmatrix} \end{align*}
  2. \begin{align*} \mathbf{0} &= \mathbf{r} \times \begin{pmatrix} 4 + 2 \\ 3+3 \\ 7+2 \end{pmatrix} \\ &= \mathbf{r} \times \begin{pmatrix} 6 \\ 6 \\ 9 \end{pmatrix} \\ \end{align*} Therefore \(\mathbf{r} = t\begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}\) (ie a line) \begin{align*} \text{Work done} &= \text{Force} \cdot \text{distance} \end{align*} Since they are parallel, it's just the magnitude of the force, which is \(3\sqrt{2^2+2^2+3^2} = 3\sqrt{17}\)

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1990 Paper 3 Q15
D: 1700.0 B: 1482.6
probability geometric distribution marginal distribution independence joint distribution expected value summation discrete random variables

An unbiased twelve-sided die has its faces marked \(A,A,A,B,B,B,B,B,B,B,B,B.\) In a series of throws of the die the first \(M\) throws show \(A,\) the next \(N\) throws show \(B\) and the \((M+N+1)\)th throw shows \(A\). Write down the probability that \(M=m\) and \(N=n\), where \(m\geqslant0\) and \(n\geqslant1.\) Find

  1. the marginal distributions of \(M\) and \(N\),
  2. the mean values of \(M\) and \(N\).
Investigate whether \(M\) and \(N\) are independent. Find the probability that \(N\) is greater than a given integer \(k\), where \(k\geqslant1,\) and find \(\mathrm{P}(N > M).\) Find also \(\mathrm{P}(N=M)\) and show that \(\mathrm{P}(N < M)=\frac{1}{52}.\)

Solution: \begin{align*} \mathbb{P}(M = m, N = n) &= \left ( \frac{3}{12} \right)^m \left ( \frac{9}{12} \right)^n \frac{3}{12} \\ &= \frac{3^n}{4^{m+n+1}} \end{align*}

  1. \begin{align*} \mathbb{P}(M = m) &= \sum_{n = 1}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{n = 1}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{1}{4^{m+1}} \sum_{n = 1}^{\infty} \left ( \frac34\right)^n \\ &= \frac{1}{4^{m+1}} \frac{3/4}{1/4} \\ &= \frac{3}{4^{m+1}} \\ \\ \mathbb{P}(N = n) &= \sum_{m = 0}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{m = 0}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{3^n}{4^{n+1}} \sum_{m = 0}^{\infty} \left ( \frac14\right)^n \\ &= \frac{3^n}{4^{n+1}} \frac{1}{3/4} \\ &= \frac{3^{n-1}}{4^{n}} \\ \end{align*}
  2. \(M+1 \sim Geo(\frac34) \Rightarrow \mathbb{E}(M) = \frac43 -1 = \frac13\) \(N \sim Geo(\frac14) \Rightarrow \mathbb{E}(N) = 4\)
\(M,N\) are independent since \(\mathbb{P}(M = m, N =n ) = \mathbb{P}(M=m)\mathbb{P}(N=n)\) \begin{align*} \mathbb{P}(N > k) &= \sum_{n=k+1}^{\infty} \mathbb{P}(N = n) \\ &= \sum_{n=k+1}^{\infty} \frac{3^{n-1}}{4^{n}} \\ &= \frac{3^k}{4^{k+1}} \sum_{n = 0}^{\infty} \left ( \frac34\right)^n \\ &= \frac{3^k}{4^{k+1}} \frac{1}{1/4} \\ &= \frac{3^k}{4^k} \end{align*} \begin{align*} \mathbb{P}(N > M) &= \sum_{m=0}^{\infty} \mathbb{P}(N > m) \mathbb{P}(M = m) \\ &= \sum_{m=0}^{\infty} \left (\frac34 \right)^m \frac{3}{4^{m+1}}\\ &=\sum_{m=0}^{\infty} \frac{3^{m+1}}{4^{2m+1}}\\ &= \frac{3}{4} \frac{1}{13/16} \\ &= \frac{12}{13} \\ \\ \mathbb{P}(N=M) &= \sum_{m=1}^{\infty} \mathbb{P}(N=m, M=m) \\ &= \sum_{m=1}^{\infty} \frac{3^m}{4^{2m+1}} \\ &= \frac{3}{64} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m \\ &= \frac{3}{64} \frac{1}{13/16} \\ &= \frac{3}{52}\\ \\ \mathbb{P}(N < M) &= 1 - \frac34 - \frac3{52} \\ &= 1 - \frac{48}{52} - \frac{3}{52} \\ &= 1 - \frac{51}{52} \\ &= \frac{1}{52} \end{align*}

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1990 Paper 3 Q16
D: 1700.0 B: 1484.0
probability conditional probability geometric probability triangle inequality obtuse triangle uniform distribution bivariate distribution integration

  1. A rod of unit length is cut into pieces of length \(X\) and \(1-X\); the latter is then cut in half. The random variable \(X\) is uniformly distributed over \([0,1].\) For some values of \(X\) a triangle can be formed from the three pieces of the rod. Show that the conditional probability that, if a triangle can be formed, it will be obtuse-angled is \(3-2\sqrt{2.}\)
  2. The bivariate distribution of the random variables \(X\) and \(Y\) is uniform over the triangle with vertices \((1,0),(1,1)\) and \((0,1).\) A pair of values \(x,y\) is chosen at random from this distribution and a (perhaps degenerate) triangle \(ABC\) is constructed with \(BC=x\) and \(CA=y\) and \(AB=2-x-y.\) Show that the construction is always possible and that \(\angle ABC\) is obtuse if and only if \[ y>\frac{x^{2}-2x+2}{2-x}. \] Deduce that the probability that \(\angle ABC\) is obtuse is \(3-4\ln2.\)

Solution:

  1. TikZ diagram
    The construction is possible if \(x + y > 2-x-y \Rightarrow x+y > 1\) (which is as the triangle is above the diagonal line), and \(x + (2-x-y) > y \Rightarrow 1 > y\) (true as the triangle is below the horizontal line) and \(y + (2-x-y) > x \Rightarrow 1 > x\) (true as the triangle is left of the vertical arrow). By the cosine rule: \begin{align*} && y^2 &= x^2 + (2-x-y)^2 - 2 x (2-x-y) \cos \angle ABC \\ \Rightarrow && \cos \angle ABC &= \frac{x^2+(2-x-y)^2 - y^2}{2x(2-x-y)} \\ &&&= \frac{4+2x^2-4x-4y+2xy}{2x(2-x-y)} \\ \underbrace{\Rightarrow}_{\cos \angle ABC < 0} && 0 &> 4+2x^2-4x-4y+2xy \\ \Rightarrow && 0 &> 2x^2-4x+4 - 2(x-2)y \\ \Rightarrow && y &> \frac{x^2-2x+2}{2-x} \\ &&&= -x + \frac{2}{2-x} \end{align*}
    TikZ diagram
    Therefore the area we want is: \begin{align*} A &= 1 - \int_0^1 \left ( -x + \frac{2}{2-x} \right)\d x \\ &= 1 - \left [-\frac12 x^2 - 2 \ln(2-x) \right]_0^1 \\ &= 1 + \frac12 -2 \ln 2 \\ &= \frac32 - 2 \ln 2 \end{align*} Therefore the relative area is: \(\frac{\frac32 - 2 \ln 2}{1/2} = 3 - 4 \ln 2\)

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