Chapter 4, Problem 42Q

(Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without this designation typically require integrating or extending the concepts presented thus far.)
42. Pistons are fitted to two cylindrical chambers connected through a horizontal tune to form a hydraulic system. The piston chambers and the connecting tube are Filled with an incompressible fluid. The cross-sectional areas of piston 1 and piston 2 are A₁and A₂, respectively. A force F_fis excited on piston 1. Rank the resultant force F₂on piston 2 that results from the combinations of F₁ A ₁ and A₂ given from greatest to smallest. If any of the combinations yield the same force, give them the same ranking.
(a) F₁= 2.0 N; A₁ = 0.5 m²; and A₂= 1.0 m²
(b) F_l= 1.0 N; A₁= 0.5 m²; and A₂ = 0.25 m².
(c) F₁= 1.0 N; A₁= 1.0 m²; and A₂= 2.0 m².
(d) F₁ = 2.0 N; A₁= 0.25 m²; and A₂ = 1.0 m².
(e) F₁ = 2.0 N; A₁= 0.25 m²; and A₂ = 0.5 m²
(f) F₁ = 1.0 N; A₁ = 1.0 m²; and A₂= 0.5 m².

To determine

Rank the resultant force F₂ on piston 2 that results from the combinations of F₁, A₁, and A₂ given from greatest to smallest. If any of the combinations yield the same force, give them the same ranking.

Answer to Problem 42Q

The order of force $F_2$ from greatest to lowest is as follows:

$d>(a=e)>c>(b=f)$.

Explanation of Solution

Given:

Pistons are fitted to two cylindrical chambers connected through a horizontal tube to form a hydraulic system. The piston chambers and the connecting tube are filled with an incompressible fluid. The cross-sectional areas of piston 1 and piston 2 are A₁ and A₂, respectively. A force F₁ is exerted on piston 1.

(a)F₁ = 2.0 N; A₁ = 0.5 m² ; and A₂ = 1.0 m²

(b)F₁ = 1.0 N; A₁ = 0.5 m² ; and A₂ = 0.25 m²

(c)F₁ = 1.0 N; A₁ = 1.0 m² ; and A₂ = 2.0 m²

(d)F₁ = 2.0 N; A₁ = 0.25 m² ; and A₂ = 1.0 m²

(e)F₁ = 2.0 N; A₁ = 0.25 m² ; and A₂ = 0.5 m²

(f)F₁ = 1.0 N; A₁ = 1.0 m² ; and A₂ = 0.5 m².

Formula used:

The expression for pressure is given as

$\begin{array}{l}P=\frac{F}{A}\\ \text{Where}\\ P\text{ = Pressure, }F\text{ = Force, }A\text{ = Area}\end{array}$

Since the pressure is the same

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

Using the above expression, we can find the resultant force $F_2$ on the piston 2.

Calculation:

(a) $F_1=2.0N,A_1=0.5m^2,A_2=1.0m^2$

$\begin{array}{l}{F}_2=\left(\frac{A_2}{A_1}\right)F_1\\ \Rightarrow F_2=\left(\frac{1}{0.5}\right)\times 2\Rightarrow F_2=4.0N\end{array}$

(b) $F_1=1.0N,A_1=0.5m^2,A_2=0.25m^2$

$\begin{array}{l}{F}_2=\left(\frac{A_2}{A_1}\right)F_1\\ \Rightarrow F_2=\left(\frac{0.25}{0.5}\right)\times 1\Rightarrow F_2=0.5N\end{array}$

(c) $F_1=1.0N,A_1=1.0m^2,A_2=2.0m^2$

$\begin{array}{l}{F}_2=\left(\frac{A_2}{A_1}\right)F_1\\ \Rightarrow F_2=\left(\frac{2}{1}\right)\times 1\Rightarrow F_2=2N\end{array}$

(d) $F_1=2.0N,A_1=0.25m^2,A_2=1.0m^2$

$\begin{array}{l}{F}_2=\left(\frac{A_2}{A_1}\right)F_1\\ \Rightarrow F_2=\left(\frac{1}{0.25}\right)\times 2\Rightarrow F_2=8N\end{array}$

(e) $F_1=2.0N,A_1=0.25m^2,A_2=0.5m^2$

$\begin{array}{l}{F}_2=\left(\frac{A_2}{A_1}\right)F_1\\ \Rightarrow F_2=\left(\frac{0.5}{0.25}\right)\times 2\Rightarrow F_2=4N\end{array}$

(f) $F_1=1.0N,A_1=1.0m^2,A_2=0.5m^2$

$\begin{array}{l}{F}_2=\left(\frac{A_2}{A_1}\right)F_1\\ \Rightarrow F_2=\left(\frac{0.5}{1}\right)\times 1\Rightarrow F_2=0.5N\end{array}$

The order of force $F_2$ from greatest to lowest is as follows:

$d>(a=e)>c>(b=f)$.

Conclusion:

The order of force $F_2$ from greatest to lowest is as follows:

$d>(a=e)>c>(b=f)$.