Hooke's lawEdexcel International A Level Further Maths: Revision notes

An elastic string exerts a tension only when it is stretched beyond its natural length $l$ (the length when slack, with no force). If its length is less than $l$ it is slack and the tension is zero. A spring can be stretched or compressed: when stretched it exerts a tension, and when compressed it exerts a thrust (a push) on whatever it presses against. The extension (or compression) is $x=$ length $-$ natural length, and the string or spring is called light when its mass is negligible, so the tension is the same throughout its length. All strings in this topic are elastic and obey Hooke's law.

Hooke's law says that the tension (or thrust) is proportional to the extension (or compression): $T=\frac{\lambda x}{l},$ where $\lambda$ is the modulus of elasticity (in newtons), $x$ the extension and $l$ the natural length. The ratio $\frac{\lambda}{l}$ is the stiffness: the force per metre of extension. Rearranged: $x=\frac{Tl}{\lambda}$ and $\lambda=\frac{Tl}{x}$. Example: $l=0.5$ m, $\lambda=20$ N and extension $0.2$ m gives $T=\frac{20\times0.2}{0.5}=8$ N. Doubling the extension doubles the tension. Doubling the natural length for the same $\lambda$ halves the tension for the same extension.

Combine Hooke's law with resolving forces. In equilibrium, the resultant force is zero.

• Hanging vertically: $T=mg$, then $x=\frac{Tl}{\lambda}$ gives the extension; the total length is $l+x$.
• On a smooth inclined plane: resolve parallel to the plane: $T=mg\sin\alpha$.
• On a rough plane: add friction $F\le\mu R$, with $R=mg\cos\alpha$. If the particle is on the point of slipping, $F=\mu R$, and friction acts opposite to the direction of slipping.
• Spring standing on the ground: the thrust equals the weight; the length is $l-x$. Example: $\lambda=58.8$, $l=1.2$ and a $3$ kg particle on a smooth $30^\circ$ plane: $T=14.7$ N, $x=0.3$ m, so $OP=1.5$ m.

With two strings on a smooth horizontal surface the tensions are found separately, then compared. Example: strings $AP$ ($l=0.6$, $\lambda=24$) and $BP$ ($l=0.5$, $\lambda=30$), with $AB=2$ m and $AP=x$. The extensions are $x-0.6$ and $2-x-0.5$, so $T_{AP}=40(x-0.6)$ and $T_{BP}=60(1.5-x)$. If $P$ is in equilibrium with no other force, the tensions are equal, which gives $x=1.14$ m. If another horizontal force acts, resolve along $AB$. Check that both strings are taut: each length must exceed its natural length.

A string with $l=0.5$ m, $\lambda=20$ N carries a $0.6$ kg particle: $T=0.6\times9.8=5.88$ N and $x=\frac{5.88\times0.5}{20}=0.147$ m, so the string is $0.647$ m long. Method: (1) list $l$, $\lambda$ and the extension or compression; (2) draw the forces including tension (or thrust), weight and friction; (3) resolve in equilibrium; (4) apply Hooke's law $T=\frac{\lambda x}{l}$; (5) answer what was asked, which may be the total length rather than the extension. Give final answers to three significant figures when $g=9.8$ is used.