Question

A bicycle pump is used to inflate a tire. The initial tire (gauge) pressure is 210 kPa (30 psi). At the end of the pumping process, the final pressure is 310 kPa (45 psi). If the diameter of the plunger in the cylinder of the pump is 2.5 cm, what is the range of the force that needs to be applied to the pump handle from beginning to end?

Short answer

The range of the force will be \(102.9\;{\rm{N}} \le F \ge 151.9\;{\rm{N}}\).

Step-by-step solution

Given Data

The initial tire pressure is \({P_1} = 210\,{\rm{kPa}}\).

The final tire pressure is \({P_2} = 310\,{\rm{kPa}}\).

The diameter of the plunger is \(d = 2.5\;{\rm{cm}}\).

Understanding the relation of force and pressure

In this problem, the range of force will be evaluated by using the product of pressure and area of the cylinder in initial and final process.

Estimating the ranges of force

The relation offorceis given by,

\(\begin{array}{l}{F_1} = {P_1}A\\{F_1} = {P_1}\left( {\frac{{\pi {d^2}}}{4}} \right)\end{array}\)

On plugging the values in the above relation.

\(\begin{array}{l}{F_1} = \left( {210\,{\rm{kPa}} \times \frac{{{{10}^3}\,{\rm{N/}}{{\rm{m}}^2}}}{{1\;{\rm{kPa}}}}} \right)\left( {\frac{{\pi {{\left( {2.5\;{\rm{cm}} \times \frac{{1\,{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}}}{4}} \right)\\{F_1} = 102.9\;{\rm{N}}\end{array}\)

Estimating the ranges of force

The relation offorceis given by,

\(\begin{array}{l}{F_2} = {P_2}A\\{F_2} = {P_2}\left( {\frac{{\pi {d^2}}}{4}} \right)\end{array}\)

On plugging the values in the above relation.

\(\begin{array}{l}{F_2} = \left( {310\,{\rm{kPa}} \times \frac{{{{10}^3}\,{\rm{N/}}{{\rm{m}}^2}}}{{1\;{\rm{kPa}}}}} \right)\left( {\frac{{\pi {{\left( {2.5\;{\rm{cm}} \times \frac{{1\,{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}}}{4}} \right)\\{F_2} = 151.9\;{\rm{N}}\end{array}\)

Thus, the range of the force will be \(102.9\;{\rm{N}} \le F \ge 151.9\;{\rm{N}}\).