Question Video: Strain Induced in a Solid by Compressive Force | Nagwa Question Video: Strain Induced in a Solid by Compressive Force | Nagwa

Question Video: Strain Induced in a Solid by Compressive Force

The “lead” in pencils is a graphite composition with a Young’s modulus of 1.0 × 10⁹ N/m². 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 N. The lead is 0.50 mm in diameter and 60 mm long.

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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.

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