Problems

Two uniform ladders \(AB\) and \(BC\) of equal length are hinged smoothly at \(B\). The weight of \(AB\) is \(W\) and the weight of \(BC\) is \(4W \). The ladders stand on rough horizontal ground with \(\angle ABC=60^\circ\,\). The coefficient of friction between each ladder and the ground is \(\mu\). A decorator of weight \(7W\) begins to climb the ladder \(AB\) slowly. When she has climbed up \(\frac13\) of the ladder, one of the ladders slips. Which ladder slips, and what is the value of \(\mu\)?

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Solution:

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\begin{align*} \text{N2}(\rightarrow): && F_A - F_C &= 0\\ && F_A &= F_C \\ \text{N2}(\uparrow): && R_A + R_C - 7W - W - 4W &= 0\\ && R_A + R_C &= 12W \\ \overset{\curvearrowright}{A}: && \frac{1}{6}7W + \frac{1}{4}W + \frac{3}{4}4W - R_C &= 0 \\ \Rightarrow && R_C &= \frac{53}{12}W\\ \Rightarrow && R_A = 12W - \frac{53}{12}W &= \frac{91}{12}W \\ \overset{\curvearrowleft}{B}(AB): && \frac{1}{2}W + \frac{2}{3}7W - R_A+\sqrt{3}F_A &= 0 \\ \Rightarrow && F_A = \frac{1}{\sqrt{3}} \l \frac{91}{12}-\frac12-\frac{14}3\r W &= \frac{29}{12\sqrt{3}}W \end{align*} We know that the system is about to slip, so equality holds in one of \(F_A \leq \mu R_A\) or \(F_C \leq \mu R_C\). Since \(F_A = F_C\), we know it must occur for whichever of \(\mu R_A\) and \(\mu R_C\) is smaller. Since \(R_C\) is much smaller, this must be the ladder about to slip \(BC\) and \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]

In a certain factory, microchips are made by two machines. Machine A makes a proportion~\(\lambda\) of the chips, where \(0 < \lambda < 1\), and machine B makes the rest. A proportion \(p\) of the chips made by machine A are perfect, and a proportion \(q\) of those made by machine B are perfect, where \(0 < p < 1\) and \(0 < q < 1\). The chips are sorted into two groups: group 1 contains those that are perfect and group 2 contains those that are imperfect. In a large random sample taken from group 1, it is found that \(\frac 2 5\) were made by machine A. Show that \(\lambda\) can estimated as \[ {2q \over 3p + 2q}\;. \] Subsequently, it is discovered that the sorting process is faulty: there is a probability of \(\frac 14\) that a perfect chip is assigned to group 2 and a probability of \(\frac 14\) that an imperfect chip is assigned to group 1. Taking into account this additional information, obtain a new estimate of \(\lambda\,\).

1. Three real numbers are drawn independently from the continuous rectangular distribution on \([ 0, 1 ]\,\). The random variable \(X\) is the maximum of the three numbers. Show that the probability that \(X \le 0.8\) is \(0.512\,\), and calculate the expectation of \(X\).
2. \(N\) real numbers are drawn independently from a continuous rectangular distribution on \([ 0, a ]\,\). The random variable \(X\) is the maximum of the \(N\) numbers. A hypothesis test with a significance level of 5\% is carried out using the value, \(x\), of \(X \). The null hypothesis is that \(a=1\) and the alternative hypothesis is that \(a<1 \,\). The form of the test is such that \(H_0\) is rejected if \(x < c\,\), for some chosen number \(c\,\). Using the approximation \(2^{10} \approx 10^3\,\), determine the smallest integer value of \(N\) such that if \(x \le 0.8\) the null hypothesis will be rejected. With this value of \(N\), write down the probability that the null hypothesis is rejected if \(a = 0.8\,\), and find the probability that the null hypothesis is rejected if \(a = 0.9\,\).

Three pirates are sharing out the contents of a treasure chest containing \(n\) gold coins and \(2\) lead coins. The first pirate takes out coins one at a time until he takes out one of the lead coins. The second pirate then takes out coins one at a time until she draws the second lead coin. The third pirate takes out all the gold coins remaining in the chest. Find:

1. the probability that the first pirate will have some gold coins;
2. the probability that the second pirate will have some gold coins;
3. the probability that all three pirates will have some gold coins.

Find all real values of \(x\) that satisfy:

1. \( \ds \sqrt{3x^2+1} + \sqrt{x} -2x-1=0 \;;\)
2. \( \ds \sqrt{3x^2+1} - 2\sqrt{x} +x-1=0 \;;\)
3. \( \ds \sqrt{3x^2+1} - 2\sqrt{x} -x+1=0 \;.\)

Prove that, if \(\vert \alpha\vert < 2\sqrt{2},\) then there is no value of \(x\) for which \begin{equation} x^2 -{\alpha}\vert x \vert + 2 < 0\;. \tag{\(*\)} \end{equation} Find the solution set of \((*)\) for \({\alpha}=3\,\). For \({\alpha} > 2\sqrt{2}\,\), the sum of the lengths of the intervals in which \(x\) satisfies \((*)\) is denoted by \(S\,\). Find \(S\) in terms of \({\alpha}\) and deduce that \(S < 2{\alpha}\,\). Sketch the graph of \(S\,\) against \(\alpha \,\).

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Solution: There are two cases to consider by they are equivalent to \(x^2 \pm \alpha x + 2 < 0\), which has no solution solutions if \(\Delta < 0\), ie if \(\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}\). If \(\alpha = 3\), we have \begin{align*} && 0 & > x^2-3x+2 \\ &&&= (x-2)(x-1) \\ \Rightarrow && x & \in (1,2) \\ \\ && 0 &> x^2+3x+2 \\ &&& = (x+2)(x+1) \\ \Rightarrow && x &\in (-2,-1) \end{align*} Both cases work here, so \(x \in (-2, -1) \cup (1,2)\). \begin{align*} && 0 &> x^2 \pm \alpha x + 2 \\ &&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4} \end{align*} The potential intervals therefore are \((\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})\) and \((\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})\). Neither of these intervals overlap with \(0\), since \(\alpha^2 > \alpha^2-8\), and their lengths are both \(\sqrt{\alpha^2-8}\), therefore \(S = 2\sqrt{\alpha^2-8} < 2\alpha\)

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The curve \(C\) has equation $$ y = x(x+1)(x-2)^4. $$ Determine the coordinates of all the stationary points of \(C\) and the nature of each. Sketch \(C\). In separate diagrams draw sketches of the curves whose equations are:

1. \( y^2 = x(x+1)(x-2)^4\;\);
2. \(y = x^2(x^2+1)(x^2-2)^4\,\).

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Solution: \begin{align*} && y &= x(x+1)(x-2)^4 \\ \Rightarrow && y' &= (x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\ &&&= (x-2)^3 \left ( (2x+1)(x-2)+4x(x+1) \right) \\ &&&= (x-2)^3 \left (2x^2-3x-2+4x^2+4x \right) \\ &&&=(x-2)^3(6x^2+x-2) \\ &&&=(x-2)^3(2x-1)(3x+2) \end{align*} Therefore there are stationary points at \((2,0), (\frac12, -\frac{625}{64}), (-\frac23, -\frac{4078}{81})\) \((0,2)\) is a minimum by considering the sign of \(y'\) either side. \( (-\frac23, \frac{2560}{729})\) is a minimum, since it's the first stationary point. \( (\frac12, \frac{243}{64})\) is a maximum since you can't have consecutive minima and the second derivative is clearly non-zero.

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1. An attempt is made to move a rod of length \(L\) from a corridor of width \(a\) into a corridor of width~\(b\), where \(a \ne b.\) The corridors meet at right angles, as shown in Figure 1 and the rod remains horizontal. Show that if the attempt is to be successful then $$ L \le a \cosec {\alpha} + b \sec {\alpha} \;, $$ where \({\alpha}\) satisfies $$ \tan^3\alpha =\frac a b \;. $$
2. An attempt is made to move a rectangular table-top, of width \(w\) and length \(l\), from one corridor to the other, as shown in the Figure 2. The table-top remains horizontal. Show that if the attempt is to be successful then $$ l\le a \cosec {\beta} + b \sec {\beta} -2w \cosec 2{\beta}, $$ where \({\beta}\) satisfies $$ w= \left(\frac {a -b \tan^3 \beta} {1 - \tan^2 \beta} \right) \cos \beta \;. $$

Evaluate \(\int_0^{{\pi}} x \sin x\,\d x\) and \(\int_0^{{\pi}} x \cos x\,\d x\;\). The function \(\f\) satisfies the equation \begin{equation} \f(t)=t + \int_0^{{\pi}} \f(x)\sin(x+t)\,\d x\;. \tag{\(*\)} \end{equation} Show that \[ \f(t)=t + A\sin t + B\cos t\;, \] where \(A= \int_0^{{\pi}}\,\f(x)\cos x\,\d x\;\) and \(B= \int_0^{{\pi}}\,\f(x)\sin x\,\d x\;\). Find \(A\) and \(B\) by substituting for \(\f(t)\) and \(\f(x)\) in \((*)\) and equating coefficients of \(\sin t\) and \(\cos t\,\).

The vectors \({\bf a}\) and \({\bf b}\) lie in the plane \(\Pi\,\). Given that \(\vert {\bf a} \vert= 1\) and \({\bf a}.{\bf b} = 3,\) find, in terms of \({\bf a}\) and \({\bf b}\,\), a vector \({\bf p}\) parallel to \({\bf a}\) and a vector \({\bf q}\) perpendicular to \({\bf a}\,\), both lying in the plane \(\Pi\,\), such that $${\bf p}+{\bf q}={\bf a}+{\bf b}\;.$$ The vector \({\bf c}\) is not parallel to the plane \(\Pi\) and is such that \({\bf a}.{\bf c} = -2\) and \({\bf b}.{\bf c} = 2\,\). Given that \(\vert {\bf b} \vert = 5\,\), find, in terms of \({\bf a}, {\bf b}\) and \({\bf c},\) vectors \({\bf P}\), \({\bf Q}\) and \({\bf R}\) such that \({\bf P}\) and \({\bf Q}\) are parallel to \({\bf p}\) and \({\bf q},\) respectively, \({\bf R}\) is perpendicular to the plane \(\Pi\) and $${\bf P} + {\bf Q} + {\bf R} = {\bf a}+{\bf b}+{\bf c}\;.$$