A uniform shear force of magnitude 0.01 Newton (N) is applied to the top surface of a cubical tissue sample with side of length 1 cm (marked with dashed lines in the figure below). In the deformed configuration (marked with solid lines), the angle $\theta = 5$ degrees.
The shear modulus of the tissue is kilopascals (kPa).
Assume the tissue to be a linear, isotropic and homogenous elastic solid.

Question

At time t, the cardiac dipole is oriented at - 45° (minus forty five degrees) to the horizontal axis. The magnitude of the dipole is 3 mV. Assuming Einthoven frontal plane configuration, what is the magnitude (in mV) of the electrical signal in lead II? (Round off the answer to two decimal places.)

Answer 1

Let's redo this calculation by converting everything to SI units from the very first step and cross-checking the angle with its radian value, so we can also see how close the tangent and radian values are for a small angle.

Convert the geometry to SI units first: the square has a side of $1$ cm $= 0.01$ m, so its area is $A = (0.01\,\text{m})^2 = 1\times10^{-4}\,\text{m}^2$.
Compute stress: $\tau = F/A = 0.01\,\text{N} / (1\times10^{-4}\,\text{m}^2) = 100\,\text{Pa}$.
Convert the angle to radians as a cross-check: $\theta = 5^{\circ} \times \pi/180 \approx 0.0873$ rad. Compare this to $\tan(5^{\circ}) \approx 0.0875$: the two are very close, since $5^{\circ}$ is small enough that $\tan\theta \approx \theta$ (in radians) is a good approximation.
Use the strict definition of shear strain, $\gamma = \tan\theta$, since the figure defines $\theta$ as the actual geometric tilt angle of the originally vertical face: $\gamma = \tan(5^{\circ}) = 0.0875$.
Solve for $G$ from Hooke's law in shear: $G = \tau/\gamma = 100/0.0875 \approx 1143\,\text{Pa}$.
Express the result in kilopascals: $1143\,\text{Pa} \times (1\,\text{kPa}/1000\,\text{Pa}) \approx 1.14\,\text{kPa}$.

Since $5^{\circ}$ is a small angle, using $\theta$ in radians instead of $\tan\theta$ would give $G \approx 100/0.0873 \approx 1146\,\text{Pa} \approx 1.15\,\text{kPa}$, which rounds to essentially the same value among the given choices. This confirms the result does not depend much on which small-angle form we use here.

Let's summarize:

Stress is force over the loaded area, here $100$ Pa.
Strain is the tangent of the shear angle, here about $0.0875$.
Dividing gives a shear modulus of about $1.14$ kPa.

So the shear modulus of the tissue is about $1.14$ kPa, which is option (A).

Answer 2

Let's redo this calculation by converting everything to SI units from the very first step and cross-checking the angle with its radian value, so we can also see how close the tangent and radian values are for a small angle.

Convert the geometry to SI units first: the square has a side of $1$ cm $= 0.01$ m, so its area is $A = (0.01\,\text{m})^2 = 1\times10^{-4}\,\text{m}^2$.
Compute stress: $\tau = F/A = 0.01\,\text{N} / (1\times10^{-4}\,\text{m}^2) = 100\,\text{Pa}$.
Convert the angle to radians as a cross-check: $\theta = 5^{\circ} \times \pi/180 \approx 0.0873$ rad. Compare this to $\tan(5^{\circ}) \approx 0.0875$: the two are very close, since $5^{\circ}$ is small enough that $\tan\theta \approx \theta$ (in radians) is a good approximation.
Use the strict definition of shear strain, $\gamma = \tan\theta$, since the figure defines $\theta$ as the actual geometric tilt angle of the originally vertical face: $\gamma = \tan(5^{\circ}) = 0.0875$.
Solve for $G$ from Hooke's law in shear: $G = \tau/\gamma = 100/0.0875 \approx 1143\,\text{Pa}$.
Express the result in kilopascals: $1143\,\text{Pa} \times (1\,\text{kPa}/1000\,\text{Pa}) \approx 1.14\,\text{kPa}$.

Since $5^{\circ}$ is a small angle, using $\theta$ in radians instead of $\tan\theta$ would give $G \approx 100/0.0873 \approx 1146\,\text{Pa} \approx 1.15\,\text{kPa}$, which rounds to essentially the same value among the given choices. This confirms the result does not depend much on which small-angle form we use here.

Let's summarize:

Stress is force over the loaded area, here $100$ Pa.
Strain is the tangent of the shear angle, here about $0.0875$.
Dividing gives a shear modulus of about $1.14$ kPa.

So the shear modulus of the tissue is about $1.14$ kPa, which is option (A).

Show Hint

The Correct Option is A

Solution and Explanation

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