(a) Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.65 m tall. (b) Consider a cylindrical segment of a blood vessel 2.00 cm long and 1.50 mm in diameter. What additionaloutward force would such a vessel need to withstand in the person’s feet compared to a similar vessel in her head?

• Height of person = 1.65 m
• Length of cylindrical segment = 1.61 N
• Density of blood$\rho =1.06\times {10}^{3}kg/m^3$

Pressure is defined as the load applied perpendicularly of an item per unit area across which this force is dispersed.

Pressure is measured in pounds per square inch. It is the pressure measured in relation to the surrounding atmosphere that is referred to as gauge pressure. Pressure difference can be given by,

$p-p_0=\rho g\left(y_1-y_2\right)\dots \text{(i)}$

Pressure difference can be evaluated using equation (i),

$\begin{array}{r}{p}_1-p_2=\left(1.06\times {10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(1.65\mathrm{m}\right)\left(\frac{\mathrm{Pa}}{\mathrm{kg}/{\mathrm{ms}}^2}\right)\\ p_1-p_2=1.71\times {10}^4\mathrm{Pa}\end{array}$

As outward force is perpendicular to surface area,

$F_{out}=pA...\left(ii\right)$… (ii)

Substitute value in equation (ii),

Substitute value in above equation we get,

$\begin{array}{r}{F}_{\text{out}}=\left(1.71\times {10}^4\mathrm{Pa}\right)\times \left[\pi \left(1.50\times {10}^{-3}\mathrm{m}\right)\left(2.00\times {10}^{-2}\mathrm{m}\right)\right]\left(\frac{\mathrm{N}}{{\mathrm{Pam}}^2}\right)\\ F_{\text{out}}=1.61\mathrm{N}\end{array}$

Thus, outward force of vessel is 1.61 N.

The blood pressure difference between person’s feet and top is$1.71\times {10}^4\mathrm{Pa}$