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

A light elastic string has a natural length $l$, the length when unstretched. When stretched to a length greater than $l$, the string has an extension $x$ and a tension $T$. Hooke's law states that the tension is proportional to the extension: $T=\frac{\lambda x}{l},$ where $\lambda$ is the modulus of elasticity (in newtons, N) of the string. A string can only pull: if its length is less than $l$ it is slack and the tension is zero. For $\lambda=40$ N and $l=0.8$ m, an extension of $0.392$ m gives $T=\frac{40(0.392)}{0.8}=19.6$ N.

A spring obeys the same law. It can be stretched, giving a tension, or compressed, giving a thrust (a pushing force) of $T=\frac{\lambda x}{l}$ with $x$ the compression. The stiffness or spring constant is $k=\frac{\lambda}{l}$, so $T=kx$. A spring with $\lambda=30$ N and $l=0.6$ m has $k=50$ N m$^{-1}$, and compressed to $0.45$ m the thrust is $50(0.15)=7.5$ N.

For equilibrium, resolve forces and set the resultant to zero. A particle hanging from a single string has $T=mg$. A $3$ kg particle on the string above ($\lambda=40$, $l=0.8$): $T=29.4$ N, $x=\frac{29.4(0.8)}{40}=0.588$ m, and the length is $1.388$ m. On a slope, resolve along and perpendicular to the plane. If a particle on a rough surface is on the point of moving, friction is $\mu R$. Always find the extension first, then the tension, then use the equilibrium equation.

When a particle is held by two strings, give one extension a symbol and write the other in terms of it using the total distance. For $A$ above $B$ with $AB=3$, $l_{AP}=1$, $l_{BP}=0.8$: if $AP$ has extension $x$, then $BP=3-(1+x)=2-x$, so its extension is $1.2-x$. The tensions are $40x$ and $30(1.2-x)$, and equilibrium of a $2$ kg particle gives $40x=30(1.2-x)+19.6$, so $x=0.794$ m. Check that both extensions are positive, so that both strings are taut.

When the particle is not in equilibrium, use Newton's second law: resultant force $=ma$. The tension is found from Hooke's law at the instant concerned. Pulling the $0.5$ kg particle from the earlier string ($\lambda=24$, $l=1.2$) down to a length of $1.8$ m and releasing gives $T=\frac{24(0.6)}{1.2}=12$ N, so $12-4.9=0.5a$ and $a=14.2$ m s$^{-2}$ upwards. If a string is cut, a remaining string's length (and tension) is unchanged at that instant, but the cut string's force vanishes, so the resultant changes at once.