Chapter 7, Problem 7.1P

To determine

The force required to lift to the mass.

The distance $x_1$.

The work done by the force $f_1$.

Answer to Problem 7.1P

The force required to lift to the mass is $20\text{lb}$.

The distance $x_1$ is $18\text{in}$.

The work done by the force $f_1$ is $360\text{lb}\cdot \text{in}$.

Explanation of Solution

Given information:

The area of the left side is $10{\text{in}}^2$, the area of right side is $30{\text{in}}^2$, the weight is $60\text{lb}$ the distance $x_2$ is $6\text{in}$.

Write the expression for the force required to lift the mass.

$\begin{array}{l}\frac{f_1}{A_1}=\frac{f_2}{A_2}\\ f_1=\frac{A_1}{A_2}{f}_2\end{array}$   ....... (1)

Here, force required to lift the mass is $f_1$, the area of the left side is $A_1$, the area of the right side is $A_2$ and the force on the right side is $f_2$.

Write the expression for the distance in the left side.

$\begin{array}{l}{A}_1{x}_1=A_2{x}_2\\ x_1=\frac{A_2}{A_1}{x}_2\end{array}$   ....... (2)

Here, distance in the left side is $x_1$ and the distance moved by right piston is $x_2$.

Write the expression for the work done by force $f_1$.

$W=f_1{x}_1$   ....... (3)

Here, work done is $W$.

Calculation:

Substitute $60\text{lb}$ for $f_2$, $10{\text{in}}^2$ for $A_1$ and $30{\text{in}}^2$ for $A_2$ in Equation (1).

$\begin{array}{c}{f}_1=\frac{10{\text{in}}^2}{30{\text{in}}^2}\left(60\text{lb}\right)\\ =\frac{1}{3}\left(60\text{lb}\right)\\ =20\text{lb}\end{array}$

Substitute, $10{\text{in}}^2$ for $A_1$, $6\text{in}$ for $x_2$ and $30{\text{in}}^2$ for $A_2$ in Equation (2).

$\begin{array}{c}{x}_1=\frac{30{\text{in}}^2}{10{\text{in}}^2}\left(6\text{in}\right)\\ =3\left(6\text{in}\right)\\ =18\text{in}\end{array}$

Substitute $18\text{in}$ for $x_1$ and $20\text{lb}$ for $f_1$ in equation (3).

$\begin{array}{c}W=\left(20\text{lb}\right)\left(18\text{in}\right)\\ =360\text{lb}\cdot \text{in}\end{array}$.

Conclusion:

The force required to lift to the mass is $20\text{lb}$.

The distance $x_1$ is $18\text{in}$.

The work done by the force $f_1$ is $360\text{lb}\cdot \text{in}$.