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Q82GP

Expert-verifiedFound in: Page 260

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**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?**

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

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 \(d = 2.5\;{\rm{cm}}\).

**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. **

The relation of force is 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.

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

The relation of force is 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 \(102.9\;{\rm{N}} \le F \ge 151.9\;{\rm{N}}\).

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