Elastic string equilibrium

4 \(A\) B \(A \quad B\) \(\begin{array} { l l } B & \\ A & B \end{array}\) \(P B\) $$P \theta$$ \(\theta\) P \(\theta\) L \(P\) \(P\)

1. (i)
   P \(\theta\) \(P \quad \underline { \theta }\)
2. - \(\underline { \theta }\) \(5 \theta\) $$\begin{gathered} \\ \theta \end{gathered} \quad P$$
3. \(P\)

6 \(\begin{array} { c c c c c c c c } & P & & P & & & \theta & \\ \theta & P & & A & \theta & \theta & P & \\ \text { (i) } & & P & & & \theta & \\ \text { (ii) } & P & & & P & & & \\ \text { (iii) } & & & & & & \theta \\ & & & & & & \theta \end{array}\) 7 \includegraphics[max width=\textwidth, alt={}, center]{bc436b32-01f9-41dc-b2f7-ce49e18d3e6c-3_512_1095_1439_580}

2. \(B\) \(\begin{array} { l l l } A & P & A \end{array}\) AN PA

3 \includegraphics[max width=\textwidth, alt={}, center]{3e7472a8-df1e-45c4-81fb-e4397bddf5ad-2_202_972_1619_584} A light elastic string has natural length 0.8 m and modulus of elasticity 12 N . The ends of the string are attached to fixed points \(A\) and \(B\), which are at the same horizontal level and 0.96 m apart. A particle of weight \(W \mathrm {~N}\) is attached to the mid-point of the string and hangs in equilibrium at a point 0.14 m below \(A B\) (see diagram). Find \(W\).

1 \includegraphics[max width=\textwidth, alt={}, center]{ece63d46-5e56-4668-939a-9dbbcfc1a77a-2_248_1267_276_440} A light elastic string has natural length 0.6 m and modulus of elasticity \(\lambda \mathrm { N }\). The ends of the string are attached to fixed points \(A\) and \(B\), which are at the same horizontal level and 0.63 m apart. A particle \(P\) of mass 0.064 kg is attached to the mid-point of the string and hangs in equilibrium at a point 0.08 m below \(A B\) (see diagram). Find

1. the tension in the string,
2. the value of \(\lambda\).

1 A light elastic string has natural length 1.5 m and modulus of elasticity 60 N . The string is stretched between two fixed points \(A\) and \(B\), which are at the same horizontal level and 2 m apart.

1. Find the tension in the string. A particle of weight \(W \mathrm {~N}\) is now attached to the mid-point of the string and the particle is in equilibrium at a point 0.75 m vertically below the mid-point of \(A B\).
2. Find the value of \(W\).

2. \begin{figure}[h] \end{figure} A light elastic string \(A B\) has modulus of elasticity \(2 m g\) and natural length \(k a\), where \(k\) is a constant.
The end \(A\) of the elastic string is attached to a fixed point. The other end \(B\) is attached to a particle of mass \(m\). The particle is held in equilibrium, with the elastic string taut, by a force that acts in a direction that is perpendicular to the string. The line of action of the force and the elastic string lie in the same vertical plane. The string makes an angle \(\theta\) with the downward vertical at \(A\), as shown in Figure 2. Given that the length \(A B = \frac { 21 } { 10 } a\) and \(\tan \theta = \frac { 3 } { 4 }\), find the value of \(k\).

1. \begin{figure}[h] \end{figure} A light elastic string \(A B\) has natural length \(4 a\) and modulus of elasticity \(\lambda\). The end \(A\) is attached to a fixed point and the end \(B\) is attached to a particle of mass \(m\). The particle is held in equilibrium, with the string stretched, by a horizontal force of magnitude \(k m g\).
The line of action of the horizontal force lies in the vertical plane containing the elastic string.
The string \(A B\) makes an angle \(\alpha\) with the vertical, where \(\tan \alpha = \frac { 4 } { 3 }\) With the particle in this position, \(A B = 5 a\), as shown in Figure 1.

1. Show that \(\lambda = \frac { 20 m g } { 3 }\)
2. Find the value of \(k\).

6. A light elastic spring \(A B\) has natural length \(2 a\) and modulus of elasticity \(2 m n ^ { 2 } a\), where \(n\) is a constant. A particle \(P\) of mass \(m\) is attached to the end \(A\) of the spring. At time \(t = 0\), the spring, with \(P\) attached, lies at rest and unstretched on a smooth horizontal plane. The other end \(B\) of the spring is then pulled along the plane in the direction \(A B\) with constant acceleration \(f\). At time \(t\) the extension of the spring is \(x\).

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
2. Find \(x\) in terms of \(n , f\) and \(t\). Hence find
3. the maximum extension of the spring,
4. the speed of \(P\) when the spring first reaches its maximum extension.
   1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors due east and due north respectively]
   A man cycles at a constant speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on level ground and finds that when his velocity is \(u \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) the velocity of the wind appears to be \(v ( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\), where \(v\) is a positive constant. When the man cycles with velocity \(\frac { 1 } { 5 } u ( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\), the velocity of the wind appears to be \(w \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(w\) is a positive constant. Find, in terms of \(u\), the true velocity of the wind.

7. \begin{figure}[h] \end{figure} Figure 3 shows a framework \(A B C\), consisting of two uniform rods rigidly joined together at \(B\) so that \(\angle A B C = 90 ^ { \circ }\). The \(\operatorname { rod } A B\) has length \(2 a\) and mass \(4 m\), and the \(\operatorname { rod } B C\) has length \(a\) and mass \(2 m\). The framework is smoothly hinged at \(A\) to a fixed point, so that the framework can rotate in a fixed vertical plane. One end of a light elastic string, of natural length \(2 a\) and modulus of elasticity \(3 m g\), is attached to \(A\). The string passes through a small smooth ring \(R\) fixed at a distance \(2 a\) from \(A\), on the same horizontal level as \(A\) and in the same vertical plane as the framework. The other end of the string is attached to \(B\). The angle \(A R B\) is \(\theta\), where \(0 < \theta < \frac { \pi } { 2 }\).

1. Show that the potential energy \(V\) of the system is given by $$V = 8 a m g \sin 2 \theta + 5 a m g \cos 2 \theta + \text { constant }$$
2. Find the value of \(\theta\) for which the system is in equilibrium.
3. Determine the stability of this position of equilibrium.
   1. A smooth uniform sphere \(S\), of mass \(m\), is moving on a smooth horizontal plane when it collides obliquely with another smooth uniform sphere \(T\), of the same radius as \(S\) but of mass \(2 m\), which is at rest on the plane. Immediately before the collision the velocity of \(S\) makes an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\), with the line joining the centres of the spheres. Immediately after the collision the speed of \(T\) is \(V\). The coefficient of restitution between the spheres is \(\frac { 3 } { 4 }\).
   2. Find, in terms of \(V\), the speed of \(S\)
      1. immediately before the collision,
      2. immediately after the collision.
   3. Find the angle through which the direction of motion of \(S\) is deflected as a result of the collision.
      

\includegraphics{figure_2} A uniform rigid rod \(AB\) of length \(1.2\text{ m}\) and weight \(8\text{ N}\) has a particle of weight \(2\text{ N}\) attached at the end \(B\). The end \(A\) of the rod is freely hinged to a fixed point. One end of a light elastic string of natural length \(0.8\text{ m}\) and modulus of elasticity \(20\text{ N}\) is attached to the hinge. The string passes over a small smooth pulley \(P\) fixed \(0.8\text{ m}\) vertically above the hinge. The other end of the string is attached to a small light smooth ring \(R\) which can slide on the rod. The system is in equilibrium with the rod inclined at an angle \(\theta°\) to the vertical (see diagram).

1. Show that the tension in the string is \(20\sin\theta\text{ N}\). [1]
2. Explain why the part of the string attached to the ring is perpendicular to the rod. [1]
3. Find \(\theta\). [3]

\includegraphics{figure_2} A uniform rigid rod \(AB\) of length \(1.2\,\text{m}\) and weight \(8\,\text{N}\) has a particle of weight \(2\,\text{N}\) attached at the end \(B\). The end \(A\) of the rod is freely hinged to a fixed point. One end of a light elastic string of natural length \(0.8\,\text{m}\) and modulus of elasticity \(20\,\text{N}\) is attached to the hinge. The string passes over a small smooth pulley \(P\) fixed \(0.8\,\text{m}\) vertically above the hinge. The other end of the string is attached to a small light smooth ring \(R\) which can slide on the rod. The system is in equilibrium with the rod inclined at an angle \(\theta°\) to the vertical (see diagram).

1. Show that the tension in the string is \(20\sin\theta\,\text{N}\). [1]
2. Explain why the part of the string attached to the ring is perpendicular to the rod. [1]
3. Find \(\theta\). [3]

\includegraphics{figure_1} A rod \(AB\), of mass \(2m\) and length \(2a\), is suspended from a fixed point \(C\) by two light strings \(AC\) and \(BC\). The rod rests horizontally in equilibrium with \(AC\) making an angle \(\alpha\) with the rod, where \(\tan \alpha = \frac{3}{4}\), and with \(AC\) perpendicular to \(BC\), as shown in Fig. 1.

1. Give a reason why the rod cannot be uniform. [1]
2. Show that the tension in \(BC\) is \(\frac{4}{5}mg\) and find the tension in \(AC\). [5]

The string \(BC\) is elastic, with natural length \(a\) and modulus of elasticity \(kmg\), where \(k\) is constant.

1. Find the value of \(k\). [4]

\includegraphics{figure_13} A step-ladder has two sides AB and AC, each of length \(4a\). Side AB has weight \(W\) and its centre of mass is at the half-way point; side AC is light. The step-ladder is smoothly hinged at A and the two parts of the step-ladder, AB and AC, are connected by a light taut rope DE, where D is on AB, E is on AC and AD = AE = \(a\). A man of weight \(4W\) stands at a point F on AB, where BF = \(x\). The system is in equilibrium with B and C on a smooth horizontal floor and the sides AB and AC are each at an angle \(\theta\) to the vertical, as shown in Fig. 13.

1. By taking moments about A for side AB of the step-ladder and then for side AC of the step-ladder show that the tension in the rope is $$W\left(1 + \frac{2x}{a}\right)\tan\theta.$$ [7]

The rope is elastic with natural length \(\frac{1}{2}a\) and modulus of elasticity \(W\).

1. Show that the condition for equilibrium is that $$x = \frac{1}{2}a(8\cos\theta - \cot\theta - 1).$$ [5]

In this question you must show detailed reasoning.

1. Hence determine, in terms of \(a\), the maximum value of \(x\) for which equilibrium is possible. [5]

END OF QUESTION PAPER