## Derivation

Before paperclips were hung at the cotton loop, the metre rule of mass M was in the state of equilibrium where the pivot point was located at the midpoint of the metre rule (about 50.0 cm from the edge): After the paperclips were hung at the cotton loop, The pivot has now moved to a new position (denoted as P'). The paperclips acts a force which is equal to the product of the (number of paperclip $\times$ mass of a paperclip ) and the gravitational acceleration, g. There is a net moment in the direction of the force equals to $\tau_c=nmg \sin \theta \times$ (distance from the new pivot). The centre of gravity of the metre rule exerts a force which results in the anticlockwise moment: \begin{equation} \tau_{ac}=Mg(\sin \theta )l \end{equation} Where $M=$mass of the metre rule According to the principle of moment, for a system to be in the state of rotational equilibrium, its clockwise moment must be equal in magnitude to the anticlockwise moment. \begin{aligned} \tau_{ac}&=\tau_{c}\\ Mgl \sin \theta &= nmg \sin \theta (x-l)\\ Ml &= nm(0.15-l)\\ xnm&= (M+nm)l\\ \frac{1}{l}&=\frac{M+nm}{xnm} ---(1) \end{aligned} where $m$ is the mass of a paperclip and n is the number of paperclip. We consider that the length $l$ (distance between P' and P) is also equal to the magnitude of the arc $s$ of the circle (beaker). Therefore: \begin{equation} l=s=r\theta ---(2) \end{equation} r is the radius of the beaker.Now, let's turn our attention to the tilted metre rule, as shown in this diagram: We apply the small angle approximation: \begin{equation} \frac{h_1-h_2}{L}=\theta ---(3) \end{equation} Substitute the equation (3) into equation (2): \begin{equation} l=\frac{(h_1-h_2)r}{L} \end{equation} Substitute (4) into (1): \begin{aligned} \frac{1}{h_1-h_2}&=\frac{(M+nm)r}{xnmL}\\ &=\frac{Mr}{xnmL}+\frac{r}{xL}\\ &=\frac{1}{n}a+b \end{aligned} where $a$=gradient. $b=\frac{r}{xL}$