Chapter 15, Problem 40PQ

To allow a car to slow down or stop, hydraulic brakes transmit forces from a master cylinder to the brake pads through a fluid. Imagine this system as a tube filled with an incompressible fluid and a piston on each end. A force of 95.0 N is applied to a piston 2.65 cm in diameter on one end of the tube. a. What is the magnitude of the force that is exerted on the piston 5.15 cm in diameter on the other side? b. If the 2.65-cm piston is displaced by 1.00 cm, by how much is the 5.15-cm piston displaced?

To determine

Magnitude of force that is exerted on the piston on the other side.

Answer to Problem 40PQ

The magnitude of the force exerted on the piston on the other side is $\underset{\_}{359\text{N}}$.

Explanation of Solution

Pascal’s law states that pressure applied to a point of the fluid is transmitted equally and undiminished throughout the fluid. Therefore the pressure at either ends of the piston is equal.

  $P_1=P_2$

Here, $P_1$ is the pressure at end 1 and $P_2$ is the pressure at end 2.

Write the equation to find $P_1$.

  $P_1=\frac{F_1}{A_1}$                                                                                                         (I)

Here, $F_1$ is the force at end 1 and $A_1$ is the area of cross-section of end 1.

Write the equation to find $P_2$.

  $P_2=\frac{F_2}{A_2}$                                                                                                          (II)

Here, $F_2$ is the force at end 2 and $A_2$ is the area of cross-section of end 2.

Equate equations (I) and (II) and solve for $F_2$.

  $\begin{array}{c}\frac{F_1}{A_1}=\frac{F_2}{A_2}\\ F_2=\frac{A_2}{A_1}{F}_1\end{array}$                                                                                                     (III)

Write the equation to find the area of cross-section of end 1.

  $A_1=\pi r_1{}^2$                                                                                                 (IV)

Here, $r_1$ is the radius of first piston.

Write the equation to find the cross- section of end 2.

  $A_2=\pi r_2{}^2$                                                                                                     (V)

Here, $r_2$ is the radius of the second piston.

Write the equation to find $r_1$.

  $r_1=\frac{d_1}{2}$                                                                                                        (VI)

Here, $d_1$ is the diameter of first piston.

Write the equation to find $r_2$.

  $r_2=\frac{d_2}{2}$                                                                                                     (VII)

Here, $d_2$ is the diameter of second piston.

Conclusion:

Substitute $5.15\text{cm}$ for $d_2$ in equation (VII) to get $r_2$.

  $\begin{array}{c}{r}_2=\frac{5.15\text{cm}}{2}\\ =2.57\text{cm}\end{array}$

Substitute $2.65\text{cm}$ for $d_1$ in equation (VI) to get $r_1$.

  $\begin{array}{c}{r}_1=\frac{2.65\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}{2}\\ =1.32\text{cm}\end{array}$

Substitute $0.02\text{m}$ for $r_2$, $0.01\text{m}$ for $r_1$ and $95.0\text{N}$ for $F_1$ in equation (III) to get $F_2$.

  $\begin{array}{c}{F}_2=\frac{\pi {\left(2.57\text{cm}\right)}^2}{\pi {\left(1.32\text{cm}\right)}^2}\left(95.0\text{N}\right)\\ =359\text{N}\end{array}$

Therefore, the magnitude of the force exerted on the piston on the other side is $\underset{\_}{359\text{N}}$.

To determine

Displacement of $5.15\text{cm}$ piston if the first piston is displaced by $1.00\text{cm}$.

Answer to Problem 40PQ

The $5.15\text{cm}$ piston is displaced by $\underset{\_}{0.265\text{cm}}$.

Explanation of Solution

The fluid used in the brakes of automobiles are incompressible. So the volume of fluid displaced by the first piston must be equal to the volume of fluid displaced by second piston.

  $V_1=V_2$

Here, $V_1$ is the volume of fluid displaced by first piston and $V_2$ is the volume of fluid displaced by second piston.

Write the equation to find $V_1$.

  $V_1=A_1{d}_{1p}$                                                                                                (VIII)

Here, $d_{1p}$ is the displacement of first piston.

Write the equation to find $V_2$.

  $V_2=A_2{d}_{2p}$                                                                                                   (IX)

Here, $d_{2p}$ is the displacement of second piston.

Substitute equations (VIII) and (IX) in above equation and solve for $d_2$.

  $\begin{array}{c}{A}_1{d}_{1p}=A_2{d}_{2p}\\ d_{2p}=\frac{A_1{d}_{1p}}{A_2}\end{array}$                                                                                       (X)

Conclusion:

Substitute $1.32\text{cm}$ for $r_1$ in equation (IV) to get $A_1$.

  $\begin{array}{c}{A}_1=\pi {\left(1.32\text{cm}\right)}^2\\ =5.52{\text{cm}}^2\end{array}$

Substitute $2.57\text{cm}$ for $r_2$ in equation (V) to get $A_2$.

  $\begin{array}{c}{A}_2=\pi {\left(2.57\text{cm}\right)}^2\\ =20.8{\text{cm}}^2\end{array}$

Substitute $5.52{\text{cm}}^2$ for $A_1$, $20.8\text{cm}$ for $A_2$ and $1.00\text{cm}$ for $d_{1p}$ in equation (X) to get $d_{2p}$.

$\begin{array}{c}{d}_{2p}=\frac{\left(5.25{\text{cm}}^2\right)\left(1.00\text{cm}\right)}{20.8{\text{cm}}^2}\\ =0.265\text{cm}\end{array}$

Therefore, the $5.15\text{cm}$ piston is displaced by $\underset{\_}{0.265\text{cm}}$.