Chapter 17, Problem 102A

To determine

The pressure blowed into the straw to lift the person sitting on the sealed garbage bag.

Answer to Problem 102A

The pressure blowed into the straw to lift the person sitting on the sealed garbage bag is $2.4\text{ kPa}$ .

Explanation of Solution

Given:

The force due to the person is $\Delta F_1=600\text{ N}$ .

The cross sectional area of the person sitting on the bag is $A_1=0.25{\text{ m}}^2$ .

The cross sectional area of the soda straw is $A_2=2\times {10}^{-5}{\text{ m}}^2$ .

Formula used:

The expression for the Pascal’s principle is,

  $F_2=\frac{F_1{A}_2}{A_1}$

Here, $F_1$ is the force exerted by the first piston, $A_2$ is the area of the second piston, and $A_1$ is the area of the first piston.

The expression for the pressure of the water is,

  $P=\frac{F_g}{A}$

Here, $F_g$ is the weight of the column of water above and $A$ is the cross sectional area of the column.

Calculation:

The force exerted by the straw is,

  $\begin{array}{l}\Delta F_2=\frac{\Delta F_1{A}_2}{A_1}\\ \Delta F_2=\frac{\left(600\text{ N}\right)\times \left(2\times {10}^{-5}{\text{ m}}^2\right)}{\left(0.25{\text{m}}^2\right)}\\ \Delta F_2=0.048\text{ N}\end{array}$

The pressure to blow into the straw to lift the person that is sitting on the sealed garbage bag is,

  $\begin{array}{c}P=\frac{\Delta F_2}{A_2}\\ P=\frac{0.048\text{ N}}{2\times {10}^{-5}{\text{ m}}^2}\\ P=2,400\text{N}/{\text{m}}^2\times \frac{1\text{ kPa}}{1,000\text{N}/{\text{m}}^2}\\ P=2.4\text{ kPa}\end{array}$

Conclusion:

Thus, the pressure blowed into the straw is $2.4\text{ kPa}$ .