# C3-1: Determining the Young’s Modulus of Wood Along the Grain Using a Cantilever¶

## Apparatus¶

Wooden metre rule; $$100\text{g}$$ mass; elastic band; G-clamp; block of wood; vernier calipers; stopwatch; graph paper

## Procedure¶

1. Clamp the loaded metre rule firmly to the end of a bench with a definite length, $$L$$, projecting from the edge of the bench.
2. Start the metre rule vibrating vertically and find the periodic time, $$T$$, for one complete oscillation. Do this by timing $$20$$ oscillations and dividing by $$20$$. Find $$T$$ for the following lengths $$L: 0.5, 0.6, 0.7, 0.75, 0.8, \text{ and } 0.9\text{m}$$. Tabulate your readings of $$L$$ and $$T$$.
3. Using the callipers, measure the dimensions $$b$$ and $$d$$ of the metre rule. Take six readings for each dimension at different positions along the rule. Record the readings, then calculate the mean values of $$b$$ and $$d$$.

## Observations¶

$$M =$$ mass at end of the metre rule = ________ kg

6 readings for $$b$$: ______, ______, ______, ______, ______, ______

Average $$b$$ : ________ m

6 readings for $$d$$: ______, ______, ______, ______, ______, ______

Average $$d$$ : ________ m

Tabulate:

## Theory¶

Bending theory gives $$s = \frac{4 F L^3}{b d^3 E}$$ where $$F$$ is a force applied to the end of the metre rule and $$E$$ is known as the Young’s modulus(reference Scholarship Physics by Nelkon, fifth edition, p44). Thus if the rule is depressed a distance, $$s$$, from equilibrium the restoring force is:

$F=-\frac{b d^3 E s}{4 L^3}=-ks \quad \text{ where } \quad k=\frac{b d^3 E}{4 L^3}$

This force acts on the mass at the end of the rule. Ignoring the mass of the metre rule itself, the following is derived:

$F = M a = - k s \quad \text{and therefore} \quad a = - \frac{k}{M} s$

The solution to this equation comes from the theory of simple harmonic motion. The equation describes an oscillation with $$\omega ^2 = \frac{k}{M}$$. In terms of the period this is:

$T=\frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{M}{k}}$

Therefore:

$T^2 =\frac{4 \pi ^2 M}{k}=\frac{16 \pi ^2 M}{b d^3 E}L^3$

## Analysis¶

1. Plot a graph of $$T^2$$ against $$L^3$$ and find the gradient.
2. From the equation $$T^2 =\frac{16 \pi ^2 M}{b d^3 E}L^3$$ and the gradient of your graph determine $$E$$, the Young’s modulus of the wood along the grain.
3. The Young’s modulus across the grain is about $$0.5\text{GPa}$$. Compare this with your value of $$E$$ from (2.) and give a reason for the difference.
4. Calculate the longitudinal tension that would stretch the metre rule by $$0.1\text{mm}$$. Use the dimensions of the rule, your calculated value for $$E$$, and the relation: $$E = \frac{\text{stress}}{\text{strain}}$$