### Video Transcript

The “lead” in pencils is a graphite
composition with a Young’s modulus of 1.0 times 10 to the ninth newtons per meter
squared. Calculate the change in length of
the lead in an automatic pencil when it is tapped straight into a pencil with a
force of 4.0 newtons. The lead is 0.50 millimeters in
diameter and 60 millimeters long.

We want to calculate the change in
the length of the lead, which we can call Δ𝐿. And we’re told the Young’s modulus
of this material, which we’ll label 𝑌 sub 𝑚. The force with which the lead is
tapped into the pencil of 4.0 newtons we’ll call 𝐹. And we’re told the lead’s diameter
as well as its original length. We’ll label these 𝐷 and 𝐿,
respectively. We can start on our solution by
recalling that the Young’s modulus, 𝑌 sub 𝑚, is equal to stress divided by strain,
where stress is equal to a force divided by an area and the strain 𝜀 is equal to
the change in length Δ𝐿 divided by the original length.

Combining all three of these
relationships, we can say that the Young’s modulus, 𝑌 sub 𝑚, is equal to force
divided by area, all divided by Δ𝐿, the change in length, over 𝐿, the original
length. We want to solve of course for
Δ𝐿. So we rearrange this equation. Δ𝐿 is equal to 𝐹 times 𝐿 divided
by 𝑌 sub 𝑚 times 𝐴. The area 𝐴 is the cross section of
our circular shaft of graphite and therefore equals 𝜋 over four times 𝐷
squared. Altogether then, we have an
expression now for Δ𝐿 in terms of variables we’ve been given in the problem
statement.

Plugging in all these values, being
sure to convert our length to units of meters as well as our diameter 𝐷, we find
that Δ𝐿, to two significant figures, is 1.2 millimeters. That’s the change in the length of
this shaft of graphite in response to the applied stress.