Chapter 13.3, Problem 33SSC

To determine

To Calculate: The additional force transmitted to $\text{4}{\text{.0×10}}^{\text{-4}}{\text{m}}^{\text{2}}$ launch tube.

Answer to Problem 33SSC

  $\text{24 N}$

Explanation of Solution

Given data:

Applied force, $F_1=150\text{N}$

Area of launch tube, $A_2=\text{4}\text{.0}\times {\text{10}}^{-4}{\text{m}}^{\text{2}}$

Area of piston, $A_1=\text{2}\text{.5}\times {\text{10}}^{-3}{\text{m}}^{\text{2}}$

Formula used:

According to Pascal's law, when pressure is applied at the one end then the pressure gets distributed in all directions equally. A force  $F_1$  is applied to a small piston of area  $A_1$ . The pressure is transmitted through a liquid to a larger piston of area  $A_2$ . Since the pressure is the same on both sides, so that

  $P\text{ =}\frac{F_1}{A_1}= \frac{F_2}{A_2}$

Therefore, the force  $F_2$  is larger than  $F_1$  by multiplying factor  $\frac{A_2}{A_1}$ .

Force exerted by hydraulic lift, $F_2\text{ =}\frac{F_1{A}_2}{A_1}$

Calculation:

Now, by using the formula, $F_2\text{ =}\frac{F_1{A}_2}{A_1}$ and putting all the values in this formula,

  $\begin{array}{c}{F}_2\text{=}\frac{\left(\text{150}\text{N}\right)\left(\text{4}{\text{.0×10}}^{\text{-4}}{\text{m}}^{\text{2}}\right)}{\text{2}{\text{.5×10}}^{\text{-3}}}\\ \text{=}\text{24}\text{N}\end{array}$

Conclusion:

The additional force transmitted to $\text{4}{\text{.0×10}}^{\text{-4}}{\text{m}}^{\text{2}}$ launch tubeis $\text{24 N}$ .