FM1 §3.1 Hooke's law and equilibrium problems

Further mechanics 1

Table of contents

1. Hooke's law
2. Units
3. Modulus of elasticity ($\lambda$)
4. Examples
5. Exercise 3A

1. Hooke's law

When an elastic string or spring is stretched, the tension, $T$, produced is proportional to the extension, $x$. $$\begin{align} T \propto x \\ \\ \Rightarrow \qquad T = kx \end{align}$$ where $k$ is a constant.

 

The constant $k$ depends on the unstretched length of the string or spring, $l$, and the modulus of elasticity of the string or spring, $\lambda$.

$$\begin{align} T = \frac{ \lambda x }{ l } \end{align}$$ This relationship is called Hooke's law.

 

Note 1: An elastic spring can also be compressed. Instead of a tension, this will produce a thrust (or compression) force. Hooke's law still works for compressed elastic springs. (For an elastic spring, the constant $k$ is often called the 'spring constant'.)

 

Note 2: In this chapter, all elastic strings and springs are modelled as being light. This means they have negligible mass and do not stretch under their own weight.

2. Units

• $T$ is a force measured in newtons.
• $x$ and $l$ are both lengths.
• So the units of $\lambda$ are also newtons.

3. Modulus of elasticity ($\lambda$)

The value of $\lambda$ depends on the material from which the elastic string or spring is made, and is a measure of the 'stretchiness' of the string or spring. In this chapter, we may assume that Hooke's law applies for the values given in a question. In reality, Hooke's law only applies for values of $x$ up to a maximum value, known as the elastic limit or yield point of the string or spring. (c.f. elastic and plastic deformations.)

 

 

The modulus of elasticity, $\lambda$, tells us the amount of force required to extend an elastic string or spring by its natural length, $l$. For example, if $\lambda = 10$N and $l = 1$m, then the extension, $x$, and the required tension, $T$, read as follows.

• $x = 50$ cm, $T = 5$ N.
• $x = 1$ m, $T = 10$ N.
• $x = 1.5$ m, $T = 15$ N.
• $x = 2$ m, $T = 20$ N.

4. Examples

Example 1. An elastic string of natural length 2 m and modulus of elasticity 29.4 N has one end fixed. A particle of mass 4 kg is attached to the other end and hangs at rest. Find the extension of the string.

 

Solution.

The string is stretched such that the tension and the gravitational force due to mass balance. $$\begin{align} && T &= mg \\ &\Rightarrow& \frac{ \lambda x }{ l } &= mg \\ &\Rightarrow& x &= \frac{ mgl }{ \lambda } = \frac{ 4 \, \textrm{kg} \times 9.8 \, \textrm{m/s}^2 \times 2 \, \textrm{m} }{ 29.4 \, \textrm{N}} = \frac{8}{3} \textrm{m} \qquad \checkmark \end{align}$$

This result can be illustrated as follows:

 

Aside. This is also known as the equilibrium position, $e$, around which any vertical (simple harmonic) oscillations take place. $$\begin{align} \boxed{ e = \frac{ mgl }{ \lambda } = \frac{ mg }{ k } } \end{align}$$

Simple harmonic oscillations: Suppose that the total extension is $e+x$. Then, the Newton's second law gives $$\begin{align} && F = m \frac{ \textrm d^2 }{ \textrm dt^2 } (e + x) &= - \frac{ \lambda }{ l } (e + x) + mg \\ & \Rightarrow & m \ddot x = - \frac{ \lambda }{ l } x \\ & \Rightarrow & \ddot x = - \frac{ \lambda }{ ml } x \\ \\ & \Rightarrow & x(t) = A \cos \omega t + B \sin \omega t = A \cos \omega ( t - t_0 ) \end{align}$$ where $x(t)$ is the displacement from the equilibrium position and where $$\begin{align} \omega = \sqrt{ \frac{ \lambda }{ ml } } = \sqrt{ \frac{ k }{ l } } \qquad \Leftrightarrow \qquad f &= \frac1{ 2\pi } \sqrt{ \frac{ \lambda }{ ml } } = \frac1{ 2\pi } \sqrt{ \frac{ k }{ l } } \\ T &= 2\pi \sqrt{ \frac{ ml }{ \lambda } } = 2\pi \sqrt{ \frac{ l }{ k } } \\ \end{align}$$

 

Example 2. An elastic spring of natural length 1.5 m has one end attached to a fixed point. A horizontal force of magnitude 6 N is applied to the other end and compresses the spring to a length of 1 m. Find the modulus of elasticity of the spring.

 

Solution.

The thrust due to the compression balances the horizontal force. Since the spring is compressed to a length of 1 m, the compression is $x = 0.5$ m. $$\begin{align} && T = \frac{ \lambda x }{ l } = \frac{ \lambda \times 0.5 }{ 1.5 } &= 6 \\ &\Rightarrow & \lambda &= 18 \, \textrm{N} \qquad \checkmark \end{align}$$

 

Example 3. The elastic springs $PQ$ and $QR$ are joined together at $Q$ to form one long spring. The spring $PQ$ has natural length 1.6 m and modulus of elasticity 20 N. The spring $QR$ has natural length 1.4 m and modulus of elasticity 28 N. The ends, $P$ and $R$, of the long spring are attached to two fixed points which are 4 m apart, as shown in the diagram.

 

Find the tension in the combined spring.

 

Solution.

The knot $Q$ is located such that the tensions in the two springs balance.

Let the extension of $PQ$ be $x$. The tension in $PQ$ reads $$\begin{align} T_1 = \frac{ \lambda x }{ l } = \frac{ 20 x }{ 1.6 } = \frac{25}{2} x \end{align}$$

The extension of $QR$ is given by $1-x$ since the total length $PR=4$ m. The tension in $QR$ reads $$\begin{align} T_2 = \frac{ 28 (1-x) }{ 1.4 } = 20 (1-x) \end{align}$$

Requiring $T_1=T_2$ gives $$\begin{align} && \frac{25}{2} x &= 20 (1-x) \\ &\Rightarrow& \left( \frac{25}{2} + 20 \right) x &= 20 \\ &\Rightarrow& x&= \frac{20}{ \frac{65}{2} } = \frac{ 8 }{ 13 } \end{align}$$ Thus, for the tension, we find: $$\begin{align} T_1 = T_2 = \frac{ 25 }{ 2 } \times \frac{ 8 }{ 13 } = \frac{100}{13}\,\textrm{N} \simeq 7.69 \, \textrm N \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align}$$

 

Example 4. An elastic string of natural length $2l$ and modulus of elasticity $4mg$ is stretched between two points $A$ and $B$. The points $A$ and $B$ are on the same horizontal level and $AB = 2l$. A particle $P$ is attached to the midpoint of the string and hangs in equilibrium with both parts of the string making an angle of $30^\circ$ with the line $AB$. Find, in terms of $m$, the mass of the particle.

 

Solution.

By considering the triangle $ACP$, we find $$\begin{align} AP = \frac{ AC }{ \cos 30^\circ } = \frac{ l }{ \frac{ \sqrt{3} }{ 2 } } = \frac{ 2l }{ \sqrt{3} } \end{align}$$ Then the extension of $AP$ is $$\begin{align} x = \frac{ 2l }{ \sqrt{3} } - l = \left( \frac{ 2 }{ \sqrt{3} } - 1 \right) l \end{align}$$ The tension in $AP$ then reads $$\begin{align} T &= \frac{ \lambda x }{ l } \\ &= \frac{ 4mg }{ l } \left( \frac{ 2 }{ \sqrt{3} } - 1 \right) l \\ &= 4mg \left( \frac{ 2 }{ \sqrt{3} } - 1 \right) \end{align}$$ By symmetry, the tension in $BP$ has the same value. The vertical component of the sum of the two tensions balances the weight of the mass, i.e. $$\begin{align} && 2T \cos 60^\circ &= Mg \\ &\Rightarrow& 2\times 4mg \left( \frac{ 2 }{ \sqrt{3} } - 1 \right) \times \frac12 &= Mg \\ &\Rightarrow& M &= 4m \left( \frac{ 2 }{ \sqrt{3} } - 1 \right) \\ &&&= 4m \left( \frac{ 2 - \sqrt{3} }{ \sqrt{3} } \right) \\ &&&= \frac43 \left( 2 \sqrt{3} - 3 \right) m \\ &&&\simeq 0.619 m \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align}$$

 

Example 5. An elastic string has natural length 2 m and modulus of elasticity 98 N. One end of the string is attached to a fixed point $O$ and the other end is attached to a particle $P$ of mass 4 kg. The particle is held in equilibrium by a horizontal force of magnitude 28 N, with $OP$ making an angle $\theta$ with the vertical, as shown. Find:

(a) the value of $\theta$,

(a) the length $OP$.

 

Solution.

The tension has vertical and horizontal components. $$\begin{align} \textrm{Horizontally :}&& T \sin \theta &= 28 \\ \textrm{Vertically :}&& T \cos \theta &= 4g \end{align}$$

 

(a) We find $$\begin{align} \frac{ T \sin \theta }{ T \cos \theta } = \tan \theta = \frac{ 28 }{ 4g } = \frac{7}{g} \\ \Rightarrow \qquad \theta &= \arctan\left( \frac{7}{g} \right) = \arctan\left( \frac{7}{9.8} \right) \simeq 35.5^\circ \qquad \checkmark \end{align}$$

 

(b) For $T$ and $x$, $$\begin{align} && T &= \sqrt{ T^2 \sin^2\theta + T^2 \cos^2\theta } \\ &&&= \sqrt{ 28^2 + (4g)^2 } \\ &&&= \sqrt{ \frac{ 58016 }{ 25 } } \\ &&&= \frac{28}{5} \sqrt{ 74 } \\ \\ & \Rightarrow & \frac{ \lambda x }{ l } &= \frac{28}{5} \sqrt{ 74 } \\ & \Rightarrow & x &= \frac{ l }{ \lambda } \times \frac{28}{5} \sqrt{ 74 } \\ &&&= \frac{ 2 }{ 98 } \times \frac{28}{5} \sqrt{ 74 } \\ &&&= \frac{ 4 \sqrt{ 74 } }{ 35 } \\ &&&\simeq 0.983 \, \textrm{m} \quad \textrm{(3 s.f.)} \end{align}$$ Finally, the length $OP$ reads $$\begin{align} OP = l + x = 2 + \frac{ 4 \sqrt{ 74 } }{ 35 } \simeq 2.98 \, \textrm{m} \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align}$$

 

Example 6. Two identical elastic springs $PQ$ and $QR$ each have natural length $l$ and modulus of elasticity $2mg$. The springs are joined together at $Q$. Their other ends, $P$ and $R$, are attached to fixed points, with $P$ being $4l$ vertically above $R$. A particle of mass $m$ is attached to $Q$ and hangs at rest in equilibrium. Find the distance of the particle below $P$.

 

Solution.

• If we join the two springs at $Q$, $Q$ would be placed at the mid-point of $PR$ since they have the same modulus of elasticity and natural length, i.e. they produce an equal tension with an equal extension of $l$.
• Now, as we attach a mass $m$ to $Q$, due to gravity, $Q$ would lower its position such that the tension in $PQ$ balances the tension in $QR$ plus the weight of $m$.
• Let the extension in $PQ$ be $l+x$ and the extension in $QR$ be $l-x$.

$$\begin{align} && T_1 &= T_2 + mg \\ \\ &\Rightarrow& \frac{ \lambda ( l + x ) }{ l } &= \frac{ \lambda ( l - x ) }{ l } + mg \\ &\Rightarrow& \frac{ 2 \lambda x }{ l } &= mg \\ &\Rightarrow& \frac{ 2 \times 2mg x }{ l } &= mg \\ &\Rightarrow& x &= \frac{ l }{ 4 } \end{align}$$ Thus, the distance of the partlce below $P$ is $$\begin{align} PQ = 2l + x = 2l + \frac{ l }{ 4 } = \frac{ 9l }{ 4 } \qquad \checkmark \end{align}$$

 

Example 7. One end, $A$, of a light elastic string $AB$, of natural length 0.6 m and modulus of elasticity 10 N, is fixed to a point on a fixed rough plane inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac45$. A ball of mass 3 kg is attached to the end, $B$, of the string. The coefficient of friction, $\mu$, between the ball and the plane is $\frac13$. The ball rests in limiting equilibrium, on the point of sliding down the plane, with $AB$ along the line of greatest slope.

(a) Find:

(i) the tension in the string,

(ii) the length of the string.

 

(b) If $\mu > \frac13$, without doing any further calculation, state how your answer to part (a)(ii) would change.

 

Solution.

(a) We note that trigonometry gives: $$\begin{align} \sin \theta = \frac45 \qquad \Rightarrow \qquad \cos \theta = \frac35 \qquad \textrm{and} \qquad \tan \theta = \frac43 \end{align}$$

• Resolve the forces perpendicular to the plane: $$\begin{align} R = mg \cos \theta \end{align}$$
• Resolve the forces parallel to the plane: (the friction rads $f = \mu R$) $$\begin{align} mg \sin \theta &= T + \mu R \end{align}$$

Solving simultaneous equations gives (substituting an expresion into $R$) $$\begin{align} &\Rightarrow& mg \sin \theta &= T + \mu mg \cos \theta \\ \\ &\Rightarrow& T &= mg ( \sin \theta - \mu \cos \theta) \\ &&&= mg \left( \frac45 - \frac13 \times \frac35 \right) \\ &&&= \frac{ 3mg }{ 5 } \\ &&&= \frac{ 3 \times 3 \times 9.8 }{ 5 } \\ &&&= \frac{ 441 }{ 25 } \\ &&&= 17.64 \, \textrm{N} \qquad \checkmark \end{align}$$ For the extension, we find $$\begin{align} T &= \frac{ \lambda x }{ l } \\ \\ \Rightarrow \qquad x &= \frac{ l }{ \lambda } T = \frac{ 3mgl }{ 5 \lambda } \\ &= \frac{ 0.6 }{ 10 } \times 17.64 \\ &= \frac{1323}{1250} \\ &= 1.0584 \, \textrm{m} \end{align}$$ The total length of the string is $$\begin{align} L = l + x = 0.6 + 1.0584 = 1.6584 \, \textrm m \qquad \checkmark \end{align}$$

 

(b) As $\mu$ increases, we would need less tension $T$, and hence smaller value for $x$, i.e. $$\begin{align} && T &= mg ( \sin \theta - \mu \cos \theta) = \frac{ mg }{ 5 } ( 4 - 3 \mu ) = \frac{ \lambda x }{ l } \\ \\ &\Rightarrow & x&= \frac{ mgl }{ \lambda } ( \sin \theta - \mu \cos \theta) = \frac{ mgl }{ 5 \lambda } (4 - 3 \mu) \end{align}$$ Aside. For $\mu \ge \tan \theta = \frac43$, the friction is so high that no tension is required.

5. Exercise 3A

Question 1. One end of a light elastic string is attached to a fixed point. A force of 4 N is applied to the other end of the string so as to stretch it. The natural length of the string is 3 m and the modulus of elasticity is $\lambda$ N. Find the total length of the string when:

(a) $\lambda = 30$.

(b) $\lambda = 12$.

(c) $\lambda = 16$.

 

Solution.

As a force is applied, the string is extended such that the tension generated by it balances the external force, i.e. $$\begin{align} && T = \frac{ \lambda x }{ l } &= F \\ \\ &\Rightarrow & x &= \frac{ F l }{ \lambda } \end{align}$$ The total length of the string is $L = l + x$: $$\begin{align} \textbf{(a)}&& \lambda &= 30 && \Rightarrow & x &= \frac{ F l }{ \lambda } = \frac{ 4 \times 3 }{ 30 } = \frac25 = 0.4 && \Rightarrow & L &= 3 + 0.4 = 3.4 \, \textrm{m} \qquad \checkmark \\ \\ \textbf{(b)}&& \lambda &= 12 && \Rightarrow & x &= \frac{ F l }{ \lambda } = \frac{ 4 \times 3 }{ 12 } = 1 && \Rightarrow & L &= 3 + 1 = 4 \, \textrm{m} \qquad \checkmark \\ \\ \textbf{(c)}&& \lambda &= 16 && \Rightarrow & x &= \frac{ F l }{ \lambda } = \frac{ 4 \times 3 }{ 16 } = \frac34 = 0.75 && \Rightarrow & L &= 3 + 0.75 = 3.75 \, \textrm{m} \qquad \checkmark \end{align}$$ Comment. The modulus of elasticity, $\lambda$, tells us the amount of force required to extend an elastic string or spring by its natural length, $l$.

In other words, the higher the $\lambda$, the stronger the force required to reach the same extension. Here, the external force is fixed at 4 N, thus the higher the $\lambda$, the smaller the extension.

 

Question 2. The length of an elastic spring is reduced to 0.8 m when a force of 20 N compresses it. Given that the modulus of elasticity of the spring is 25 N, find its natural length.

 

Solution.

Let the compression by $x$. Then, we have $l - x = 0.8$, i.e. $x = l - 0.8$ and find: $$\begin{align} && T = \frac{ \lambda x }{ l } = \frac{ 25 ( l - 0.8 ) }{ l } &= 20 \\ &\Rightarrow& 5 (l - 0.8) &= 4 l \\ &\Rightarrow& ( 5 - 4 ) l &= 5 \times 0.8 \\ &\Rightarrow& l &= 4 \, \textrm{m} \qquad \checkmark \end{align}$$

 

Question 3. (P) An elastic spring of modulus of elasticity 20 N has one end fixed. When a particle of mass 1 kg is attached to the other end and hangs at rest, the total length of the spring is 1.4 m. The particle of mass 1 kg is removed and replaced by a particle of mass 0.8 kg. Find the new length of the spring.

 

Solution.

With a particle of mass 1 kg hanging, the tension balances its weight. The total length can be writte as $l + x = 1.4$ which gives $x = 1.4 - l$. $$\begin{align} && T = \frac{ \lambda x }{ l } &= mg \\ &\Rightarrow& \frac{ 20 (1.4 - l) }{ l } &= g \\ &\Rightarrow& 28 - 20l &= gl \\ &\Rightarrow& ( 20 + g ) l &= 28 \\ &\Rightarrow& l&= \frac{ 28 }{ 20 + g } = \frac{ 28 }{ 20 + 9.8 } = \frac{140}{149} \simeq 0.940 \end{align}$$ When it is replaced by a particle of mass 0.8 kg, we find $$\begin{align} && T = \frac{ \lambda x }{ l } &= mg \\ &\Rightarrow& x &= \frac{ mgl }{ \lambda } = \frac{ 0.8 \times 9.8 \times \frac{140}{149} }{ 20 } = \frac{1372}{3725} \simeq 0.368 \, \textrm{m} \quad \textrm{(3 s.f.)} \\ \\ &\Rightarrow& L &= l + x = \frac{140}{149} + \frac{1372}{3725} = \frac{4872}{3725} \simeq 1.31 \, \textrm m \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align}$$

 

Question 4. (P) A light elastic spring, of natural length $a$ and modulus of elasticity $\lambda$, has one end fixed. A scale pan of mass $M$ is attached to its other end and hangs in equilibrium. A mass $m$ is gently placed in the scale pan. Find the distance of the new equilibrium position below the old one.

 

Solution.

To start with, the elastic spring is extended such that its tension balances the weight of the scale pan of mass $M$. Let the extension be $e$ and we find: $$\begin{align} T = \frac{ \lambda e }{ a } = Mg \qquad \Rightarrow \qquad e = \frac{ Mga }{ \lambda } \end{align}$$ As a mass $m$ is gently placed, a further extension is required to produce additional to balance the additional weight due to $m$, i.e. $$\begin{align} && T = \frac{ \lambda ( e + x ) }{ a } &= (M + m) g \\ &\Leftrightarrow & \frac{ \lambda x }{ a } &= mg \\ \\ &\Rightarrow & x &= \frac{ mga }{ \lambda } \qquad \checkmark \end{align}$$

 

Question 5. (P) An elastic string has length $a_1$ when supporting a mass $m_1$, and length $a_2$ when supporting a mass $m_2$. Find the natural length and modulus of elasticity of the string.

 

Solution.

(i) When supporting a mass $m_1$, $$\begin{align} T_1 = \frac{ \lambda x_1 }{ l } = \frac{ \lambda ( a_1 - l ) }{ l } = m_1 g \qquad \Rightarrow \qquad \lambda a_1 = ( \lambda + m_1 g ) l \end{align}$$ (ii) When supporting a mass $m_2$ $$\begin{align} T_2 = \frac{ \lambda x_2 }{ l } = \frac{ \lambda ( a_2 - l ) }{ l } = m_2 g \qquad \Rightarrow \qquad \lambda a_2 = ( \lambda + m_1 g ) l \end{align}$$ Solving these simultaneous equations gives $$\begin{align} && l = \frac{ \lambda a_1 }{ \lambda + m_1 g } &= \frac{ \lambda a_2 }{ \lambda + m_2 g } \\ &\Rightarrow & a_1 ( \lambda + m_2 g ) &= a_2 ( \lambda + m_1 g ) \\ &\Rightarrow& ( a_1 - a_2 ) \lambda &= ( m_1 a_2 - m_2 a_1 ) g \\ \\ &\Rightarrow& \lambda &= \frac{ ( m_1 a_2 - m_2 a_1 ) g }{ a_1 - a_2 } \qquad \checkmark \\ \\ &\Rightarrow& l &= \frac{ \lambda a_1 }{ \lambda + m_1 g } \\ &&&= \frac{ \frac{ ( m_1 a_2 - m_2 a_1 ) g }{ a_1 - a_2 } a_1 }{ \frac{ ( m_1 a_2 - m_2 a_1 ) g }{ a_1 - a_2 } + m_1 g } \\ &&&= \frac{ ( m_1 a_2 - m_2 a_1 ) a_1 g }{ \big[ ( m_1 a_2 - m_2 a_1 ) + m_1 ( a_1 - a_2 ) \big] g } \\ &&&= \frac{ ( m_1 a_2 - m_2 a_1 ) a_1 }{ ( m_1 - m_2 ) a_1 } \\ &&&= \frac{ m_1 a_2 - m_2 a_1 }{ m_1 - m_2 } \qquad \checkmark \end{align}$$

 

Question 6. (P) When a weight, $W$ N, is attached to a light elastic string of natural length $l$ m, the extension of the string is 10 cm. When $W$ is increased by 50 N, the extension of the string is increased by 15 cm. Find $W$.

 

Solution.

(i) $x_1=10$ cm for a weight $W$ N: $$\begin{align} T_1 = \frac{ \lambda x_1 }{ l } = \frac{ 0.1 \lambda }{ l } = W \qquad \Rightarrow \qquad \frac{ \lambda }{ l } = 10W \end{align}$$ (ii) $x_2=25$ cm for a weight $W+50$ N: $$\begin{align} T_2 = \frac{ \lambda x_2 }{ l } = \frac{ 0.25 \lambda }{ l } = W + 50 \end{align}$$ Substituting $\frac{ \lambda }{ l } = 10W$ into the $T_2$-equation gives $$\begin{align} &\Rightarrow& \frac14 \times 10W &= W + 50 \\ &\Rightarrow& \frac32 W &= 50 \\ &\Rightarrow& W &= \frac23 \times 50 = \frac{100}{3} \, \textrm{N} \qquad \checkmark \end{align}$$

 

Question 7. (E/P) An elastic spring has natural length $2a$ and modulus of elasticity $2mg$. A particle of mass $m$ is attached to the midpoint of the spring. One end of the spring, $A$, is attached to the floor of a room of height $5a$ and the other end is attached to the ceiling of the room at a point $B$ vertically above $A$. The spring is modelled as light.

(a) Find the distance of the particle below the ceiling when it is in equilibrium. [8 marks]

(b) In reality, the spring may not be light. What effect will the model have had on the calculation of the distance of the particle below the ceiling? [1 mark]

 

Solution.

(a) (i) Since the particle of mass $m$ is attached to the midpoint $M$ of the spring as shown in the diagram, we may regard the spring as being composed of two elastic springs each of which has and natural length $a$ and modulus of elasticity $2mg$.

 

(ii) As the spring is stretched from the floor to the ceiling, each half experiences an extension of $1.5a$.

 

(iii) Now, let the particle fall under gravity and the extensions will adjust themselves such that the overall tension balances the weight, i.e. $$\begin{align} && T_1 &= T_2 + mg \\ \\ &\Rightarrow& \frac{ \lambda x_1 }{ l } &= \frac{ \lambda x_2 }{ l } + mg \\ &\Rightarrow& \frac{ 2mg (1.5a + x) }{ a } &= \frac{ 2mg (1.5a - x) }{ a } + mg \\ &\Rightarrow& 2 ( 1.5a + x ) &= 2 ( 1.5a - x ) + a \\ &\Rightarrow& 4x &= a \\ &\Rightarrow& x &= \frac{ a }{ 4 } \\ \\ &\Rightarrow& BM &= 2.5a + x = \frac52a+\frac14a = \frac{11}{4}a \qquad \checkmark \end{align}$$

 

Aside. Suppose that the initial extension of each half is $L$, then the equation gives $$\begin{align} && T_1 &= T_2 + mg \\ \\ &\Rightarrow& \frac{ \lambda ( L + x ) }{ l } &= \frac{ \lambda ( L - x ) }{ l } + mg \\ &\Rightarrow& \lambda ( L + x ) &= \lambda ( L - x ) + mgl \\ &\Rightarrow& 2x &= mgl \\ \\ &\Rightarrow& x &= \frac12 \left( \frac{ mgl }{ \lambda } \right) \end{align}$$ which is one half of the extension required for an elastic spring with natural length $l$ and modulus of elasticity $\lambda$ to support a mass $m$. We can understand it as one half of the weight supported by the additional tension generated in the upper spring while the other half by the thrust generated in the lower spring.

 

(b) If the spring is not light, we can model the spring as a light spring with a mass $M$ at its centre of mass. As a result, the total mass of the system would increase to $m+M$ which lowers the position of the particle. Hence, the distance $BM$ increases.

 

Aside. Length of a spring under its own weight:

(i) Step 1. We may regard the spring as a uniformly distributed massive body. Let $\rho$ be the mass density, i.e. mass per unit length, which reads $$\begin{align} \rho = \frac{ m }{ L } \end{align}$$

 

(ii) Step 2. We compare the spring extended under its own weight to its natural state, i.e. unextended, and come to the following picture.

 

(iii) Step 3. There are two formulae to describe Hooke's law $$\begin{align} T = kx = \frac{ \lambda }{ l }x \end{align}$$ The latter, which is the standard formula used in Further mechanics, is more convenient in many circumstances although it looks slightly more complicated. This is because the spring constant $k$ varies as we vary the length of the spring. For example, suppose you cut an elastic spring with the spring constant $k$ into two halves. Now, each half becomes an elastic spring with the spring constant $2k$. If we use the latter formula, this is obvious as the natural length of the spring is halved. Since we are going to consider small incremental sections with various lengths, we will stick to the standard formula, namely, $T = \frac{ \lambda }{ l }x$.

 

(iv) Step 4. We may think of a small increment as having the length $\delta y = \delta l + \delta x$ where $\delta l$ denotes the original length of the section $\delta x$ denotes the extension of the section due to the weight acting on it. The sum of the original lengths is $l$ and let the sum of the extension be $X$, i.e. $$\begin{align} \int \textrm dl = L \qquad \textrm{and} \qquad \int \textrm dx = X \end{align}$$

 

(v) Step 5. This small increment of the spring is extended by $\delta x$ in order to sustain the weight below, i.e. $$\begin{align} T = \frac{ \lambda }{ \delta l } \delta x = \rho g l \qquad \Rightarrow \qquad \lambda \delta x = \rho g l \delta l \end{align}$$ Note: One may think that the length of the increment itself should also be included on the right-hand side - with a possible ambiguity of whether it should be $\delta l$, $\delta l + \delta x$ or $\delta l + \frac12 \delta x$, but we can see that the inclusion of such terms makes no difference as it gives rise to a second-order term, i.e. $$\begin{align} \lambda \delta x = \rho g l \delta l = \rho g ( l + \delta l ) \delta l = \rho g ( l + \delta l + \alpha \delta x ) \delta l \end{align}$$

 

(vi) Step 6. Finally, integration gives $$\begin{align} && \int_0^X \lambda \, \textrm d x &= \int_0^L \rho g l \, \textrm d l \\ &\Rightarrow& \lambda X &= \left[ \frac12 \rho g l^2 \right]_0^L = \frac12 \rho g L^2 = \frac12 mgL \\ &\Rightarrow& X &= \frac{ mgL }{ 2 \lambda } \end{align}$$ Comment. The total extension is as if a particle of mass $m$ sits at the centre of mass, i.e. $$\begin{align} && T = \frac{ \lambda }{ \frac{L}{2} } X &= mg \\ &\Rightarrow & X&= \frac{ mgL }{ 2 \lambda } \qquad \checkmark \end{align}$$

 

As an side, a discrete version gives: $$\begin{align} \frac{ \lambda }{ l_k } x_k &= \rho g \sum_{i=1}^{k-1} l_i = \rho g ( l_1 + l_2 + \cdots + l_{k-1} ) \\ \frac{ \lambda }{ l_{k+1} } x_{k+1} &= \rho g \sum_{i=1}^{k} l_i = \rho g ( l_1 + l_2 + \cdots + l_{k-1} + l_k ) \\ \\ \Rightarrow \qquad \lambda \left( \frac{ x_{k+1} }{ l_{k+1} } - \frac{ x_k }{ l_k } \right) &= \rho g l_{k} \end{align}$$

 

Question 8. (E/P) A uniform rod $PQ$, of mass 5 kg and length 3 m, has one end, $P$, smoothly hinged to a fixed point. The other end, $Q$, is attached to one end of a light elastic string of modulus of elasticity 30 N. The other end of the string is attached to a fixed point $R$ which is on the same horizontal level as $P$ with $RP = 5$ m. The system is in equilibrium and $\angle PQR = 90^\circ$. Find:

(a) the tension in the string [5 marks]

(b) the natural length of the string. [3 marks]

 

[Hint: Take moments about $P$. See M2 §1. Moments.]

 

Solution.

(a) As shown in the diagram, the string is extended such that the tension supports the weight $5g$. Note that: $$\sin \theta = \frac35$$ Taking moments about $P$ gives $$\begin{align} && mg \times 1.5 \sin \theta &= T \times 3 \\ &\Rightarrow& T &= \frac{ \frac32 mg \sin\theta }{ 3 } = \frac{ 5g }{ 2 } \times \frac35 = \frac32g = 14.7 \, \textrm N \end{align}$$

 

(b) Hooke's law gives, noting that the total length of the string is $l+x=4$, $$\begin{align} && T = \frac{ \lambda x }{ l } = \frac{ 30 ( 4 - l ) }{ l } &= 14.7 \\ &\Rightarrow& 30 ( 4 - l ) &= 14.7 l \\ &\Rightarrow& ( 30 + 14.7 ) l &= 120 \\ &\Rightarrow& l &= \frac{120}{44.7} = \frac{400}{149} \simeq 2.68 \, \textrm m \quad \textrm{(3 s.f.)} \end{align}$$

 

Aside. We may answer the question by resolving forces in horizontal and vertical directions but should remember that, since it is hinged at $P$, there are horizontal and vertical forces acting at $P$. Then, We end up with three variables $H,V$ and $T$ and could obtain three equations - two from forces and one from moments. So we realise the necessity to use the moment equation anyway. Using $T = \frac32g$, $$\begin{align} \textrm{Horizontally :} && T \cos \theta &= H &&\Rightarrow& H &= \frac32g \times \frac45 = \frac65g = 11.76 \, \textrm N \\ \textrm{Vertically :} && V + T \sin \theta &= mg &&\Rightarrow& V &= mg - T \sin \theta = 5g - \frac32g \times \frac35 = \frac{41}{10}g = 40.18 \, \textrm N \end{align}$$ The force at $P$ then reads $$\begin{align} \Rightarrow \qquad F &= \sqrt{ H^2 + V^2 } = \sqrt{ \left( \frac65g \right)^2 + \left( \frac{41}{10}g \right)^2 } = \frac{ \sqrt{73} }{ 2 } g = 41.87 \, \textrm N \end{align}$$

 

Question 9. (E/P) A light elastic string $AB$ has natural length $l$ and modulus of elasticity $2mg$. Another light elastic string $CD$ has natural length $l$ and modulus of elasticity $4mg$. The strings are joined at their ends $B$ and $C$ and the end $A$ is attached to a fixed point. A particle of mass $m$ is hung from the end $D$ and is at rest in equilibrium. Find the length $AD$. [7 marks]

 

Solution.

Since the system is in equilibrium, the forces must balance at $B$ and $D$. $$\begin{align} \textrm{At $D$ :} && T_2 &= mg && \Rightarrow & \frac{ \lambda_2 x_2 }{ l_2 } &= mg && \Rightarrow & x_2 = \frac{ mg l_2 }{ \lambda_2 } = \frac{ mgl }{ 4mg } = \frac{l}{4} \\ \textrm{At $B$ :} && T_1 &= T_2 = mg && \Rightarrow & \frac{ \lambda_1 x_1 }{ l_1 } &= mg && \Rightarrow & x_1 = \frac{ mg l_1 }{ \lambda_1 } = \frac{ mgl }{ 2mg } = \frac{l}{2} \end{align}$$ The length $AD$ is $$\begin{align} AD &= l_1 + l_2 + x_1 + x_2 \\ &= l_1 + l_2 + mg \left( \frac{ l_1 }{ \lambda_1 } + \frac{ l_2 }{ \lambda_2 } \right) \\ &= \left( 1 + \frac{ mg }{ \lambda_1 } \right) l_1 + \left( 1 + \frac{ mg }{ \lambda_2 } \right) l_2 \qquad \checkmark \end{align}$$ For $\lambda_1 = 2mg$ and $\lambda_2 = 4 mg$, $$\begin{align} \Rightarrow \qquad AD &= \left( 1 + \frac{ mg }{ 2mg } \right) l + \left( 1 + \frac{ mg }{ 4mg } \right) l \\ &= \frac32l + \frac54l \\ &= \frac{11}{4}l \qquad \checkmark \end{align}$$

 

Question 10. (E/P) An elastic string $PA$ has natural length 0.5 m and modulus of elasticity 9.8 N. The string $PB$ is inextensible. The end $A$ of the elastic string and the end $B$ of the inextensible string are attached to two fixed points which are on the same horizontal level. The end $P$ of each string is attached to a 2 kg particle. The particle hangs in equilibrium below $AB$, with $PA$ making an angle of $30^\circ$ with $AB$ and $PA$ perpendicular to $PB$. Find:

(a) the length $PA$ [7 marks]

(b) the length $PB$ [2 marks]

(c) the tension in $PB$. [2 marks]

 

Solution.

(a)

• Resolving horizontally: $$\begin{align} T_1 \cos \theta = T_2 \sin \theta \qquad \Rightarrow \qquad T_2 = T_1 \cot \theta \qquad (*) \end{align}$$
• Resolving verticallly: $$\begin{align} && T_1 \sin \theta + T_2 \cos \theta &= mg \\ \\ (*) \qquad &\Rightarrow & T_1 \sin \theta + T_1 \cot \theta \cos \theta &= mg \\ &\Rightarrow & T_1 \left( \sin \theta + \frac{ \cos^2 \theta }{ \sin \theta } \right) &= mg \\ &\Rightarrow & \frac{ T_1 }{ \sin \theta } &= mg \\ &\Rightarrow & T_1 &= mg \sin \theta = \frac12 mg \end{align}$$

Hooke's law gives $$\begin{align} && T_1 = \frac{ \lambda x }{ l } &= mg \sin \theta \\ \\ &\Rightarrow& x &= \frac{ mgl \sin \theta }{ \lambda } \\ &&&= \frac{ 2 \times 9.8 \times 0.5 \times \frac12 }{ 9.8 } \\ &&&= 0.5 \, \textrm m \end{align}$$ and hence $$\begin{align} PA &= l + x = \left( 1 + \frac{ mg \sin\theta }{ \lambda } \right) l \\ &= 0.5 + 0.5 = 1 \, \textrm m \qquad \checkmark \end{align}$$

 

(b) By trigonometry, $$\begin{align} && \tan \theta &= \frac{ PB }{ PA } \\ &\Rightarrow& PB &= PA \cdot \tan \theta = \tan 30^\circ = \frac1{\sqrt{3}} \simeq 0.577 \, \textrm m \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align}$$

 

(c) $$\begin{align} T_2 &= T_1 \cot \theta \\ &= (mg \sin \theta) \cot \theta \\ &= mg \cos \theta \\ &= 2 \times 9.8 \times \cos 30^\circ \\ &= 2 \times \frac{49}{5} \times \frac{ \sqrt{3} }{ 2 } \\ &= \frac{ 49 \sqrt{3} }{ 5 } \\ &\simeq \, 17.0 \textrm N \quad \textrm{(3 s.f.)} \qquad \checkmark \end{align}$$

 

Question 11. (E/P) A particle of mass 2 kg is attached to one end $P$ of a light elastic string $PQ$ of modulus of elasticity 20 N and natural length 0.8 m. The end $Q$ of the string is attached to a point on a rough plane which is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac34$. The coefficient of friction between the particle and the plane is $\frac12$. The particle rests in limiting equilibrium, on the point of sliding down the plane, with $PQ$ along a line of greatest slope. Find:

(a) the tension in the string [5 marks]

(b) the length of the string. [2 marks]

 

Solution.

(a)

• Resolving forces perpendicular to the plane $$\begin{align} R = mg \cos \alpha \end{align}$$
• Resolving forces parallel to the plane $$\begin{align} mg \sin \alpha &= T + f \\ &= T + \mu R \\ &= T + \mu mg \cos \alpha \end{align}$$

It gives $$\begin{align} \Rightarrow \qquad T &= mg \sin \alpha - \mu mg \cos \alpha \\ &= mg \cos \alpha ( \tan \alpha - \mu ) \\ &= 2 \times 9.8 \times \frac45 \left( \frac34 - \frac12 \right) \\ &= \frac{ 98 }{ 25 } \\ &= 3.92 \, \textrm N \qquad \checkmark \end{align}$$

 

(b) Hooke's law gives $$\begin{align} && T &= \frac{ \lambda x }{ l } = \frac{ 20 x }{ 0.8 } = \frac{ 98 }{ 25 } \quad \Big[ = mg \cos \alpha ( \tan \alpha - \mu ) \Big] \\ &\Rightarrow& x &= \frac{ T l }{ \lambda } = \frac{ 98 }{ 25 } \times \frac{ 0.8 }{ 20 } = \frac{ 98 }{ 625 } = 0.1568 \, \textrm m \quad \left[ = \frac{ mgl }{ \lambda } \cos \alpha ( \tan \alpha - \mu ) \right] \\ \\ &\Rightarrow& L &= l + x = 0.8 + \frac{ 98 }{ 625 } = \frac{598}{625} = 0.9568 \, \textrm m \qquad \checkmark \end{align}$$

 

Aside. In summary, $$\begin{align} T &= mg \cos \alpha ( \tan \alpha - \mu ) = \frac{ \lambda x }{ l } \\ x &= \frac{ mgl }{ \lambda } \cos \alpha ( \tan \alpha - \mu ) \end{align}$$ If the particle rests in limiting equilibrium, on the point of sliding up the plane, $$\begin{align} T &= mg \cos \alpha ( \tan \alpha + \mu ) = \frac{ \lambda x }{ l } \\ x &= \frac{ mgl }{ \lambda } \cos \alpha ( \tan \alpha + \mu ) \end{align}$$