Chapter 13, Problem 105A

(a)

To determine

To calculate: the force exerted on the piston.

Answer to Problem 105A

The force is $2.7\times {10}^3\text{N}$ .

Explanation of Solution

Given:

Diameter of piston is $22\text{m}$ and $6.3\text{mm}$ .

Force of $3$ tones is $3.0\times {10}^4\text{N}$ .

Formula used:

Pressure on the piston is equal to the force applied on the piston divided by its area. That is,

  $P=\frac{F}{A}$.......(1)

Here, $F$ and $A$ are the force applied on the piston and surface area.

Surface area of the piston is given by

  $A=\pi r^2$.......(2)

Here, $r$ is the radius of the piston.

Calculation:

Substitute the value of $\pi r^2$ for $A$ in equation (1),

  $P=\frac{F}{\pi r^2}$.......(3)

Radius of the piston is equal to haft of its diameter, so it can be written as,

  $r=\frac{d}{2}$.......(4)

Substitute the value of $\frac{d}{2}$ for $r$ in equation (3),

  $\begin{array}{c}P=\frac{F}{\pi {\left(\frac{d}{2}\right)}^2}\\ =\frac{4F}{\pi r^2}\end{array}$

This equation can be used to get the pressure on the small piston as,

  $P_1=\frac{4F_1}{\pi d_1^2}$

Here, $F_1$ and $d_1$ are the force on small piston and diameter.

Pressure on the larger piston can be written as,

  $P_2=\frac{4F_2}{\pi d_2^2}$

Here, $F_2$ is the force on the large piston and $d_2$ is its diameter.

From Pascal law, it can be written that,

  $P_1=P_2$

Substitute the values of $\frac{4F_1}{\pi d_1^2}$ for $P_1$and $\frac{4F_2}{\pi d_2^2}$ for $P_2$ in the above equation,

  $\begin{array}{l}\frac{4F_1}{\pi d_1^2}=\frac{4F_2}{\pi d_2^2}\\ F_1=\frac{F_2}{d_2^2}\left(d_1^2\right)\\ F_1=F_2\left(\frac{d_1^2}{d_2^2}\right)\end{array}$

Substitute $3.0\times {10}^4\text{N}$ for $F_2$ and $22\text{mm}$ for $d_2$ and $6.33\text{mm}$ for $d_1$ ,

  $\begin{array}{l}{F}_1=3.0\times {10}^4\left(\frac{6.3\text{mm}}{22\text{mm}}\right)\\ F_1=2.7\times {10}^3\text{N}\end{array}$

Conclusion:

Hence, the force that has to be applied on the small piston is $2.7\times {10}^3\text{N}$ .

(b)

To determine

To calculate: length of the effort arm.

Answer to Problem 105A

The length of the effort arm should be $81\text{cm}$ .

Explanation of Solution

Given:

Diameter of piston is $22\text{m}$ and $6.3\text{mm}$ .

Force of $3$ tones is $3.0\times {10}^4\text{N}$ .

Formula used:

Force multiplied by the perpendicular distance is equal to the torque applied on the lever. That is,

  $\tau =Fl$.......(1)

Here, $F$ is the force applied and $l$ is the perpendicular distance respectively.

Calculation:

Principle involved in using the lever is the torque applied on effort arm is equal to the torque applied on the resistance arm.

  ${\tau }_e={\tau }_R$.......(2)

Substitute $Fl$ for $\tau$ in the equation (2),

  $F_e{l}_e=F_R{I}_R$.......(3)

Here, $F_e$ is the force applied on the effort arm, $l_e$ is the length of the effort arm, $F_R$ is the resistance and $I_R$ is the length of the resistance arm.

Rearrange the equation (3),

  $I_e=\frac{F_R{I}_R}{F_e}$.......(4)

Substitute $2.7\times {10}^3\text{N}$ for $F_R$ , $3.0\text{cm}$ for $I_R$ and $100.0\text{N}$ for $F_e$ in equation (4),

  $\begin{array}{l}{l}_e=\frac{2.7\times {10}^3\text{N}\times \text{3}\text{.0}\text{m}}{100.0\text{N}}\\ l_e=81\text{cm}\end{array}$

Conclusion:

Hence, the length of the effort arm should be $81\text{cm}$ .