Chapter 10, Problem 68GP

(a)

To determine

The area of the input piston.

Answer to Problem 68GP

The area of the input piston is $6.69\times {10}^{-4}{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given:

Diameter at the output of the piston, $D_o=0.18\text{m}$

Input force, $F_i=250\text{N}$

Formula used:

From Pascal’s Law,

  $\frac{F_i}{A_i}=\frac{F_o}{A_o}$

Here, $F_i$ is the input force,

  $F_o$ is the output force.

  $A_o$ is the output area.

  $A_i$ is the input area.

Calculation:

  $d=0.18\text{m}$ , $r=0.09\text{m}$

  $\frac{F_i}{A_i}=\frac{F_o}{A_o}$

  $\begin{array}{c} A_i=\frac{A_o{F}_i}{F_o}\\ =\frac{A_o{F}_i}{m_{o}g}\\ =\frac{\pi {\left(0.09\right)}^2\times 250}{970\times 9.8}\\ =6.69\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Conclusion The area of the input piston is $6.69\times {10}^{-4}{\text{m}}^{\text{2}}$ .

(b)

To determine

The work done in lifting the car $12\text{cm}$ .

Answer to Problem 68GP

The work done in lifting the car is found to be $1140.72\text{J}$ .

Explanation of Solution

Given:

Height to which car raised= $12\text{cm}$

Mass of car = $970\text{kg}$

Formula Used:

Work done to lift the car:

  $W=m\times g\times h$

Here $m$ is the mass of the car.

  $g$ is the acceleration due to gravity.

  $h$ is the height to which car is raised.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}W=m\times g\times h\\ =970\times 9.8\times 0.12\\ =1140.72\text{J} \end{array}$

Conclusion:

The work done in lifting the car is found to be $1140.72\text{J}$ .

(c)

To determine

The height that car moves up for each stroke.

Answer to Problem 68GP

The car moves up $0.342\text{cm}$ high for each stroke.

Explanation of Solution

Given:

Height $\left(h_i\right)$ of the input piston $13\text{cm}$ for each stroke.

Formula Used:

Write the expression for the volume for input and output.

  $A_i\times h_i=A_o\times h_o$

This can be written as:

  $D_i{}^2\times h_i=D_o{}^2\times h_o$

Here, $D_i$ is the diameter at the input,

  $D_o$ is the diameter at the output,

  $h_i$ is the height at the input,

  $h_o$ is the height at the input.

Calculation:

Diameter at the input can be calculated as follows:

  $\begin{array}{l}{A}_i=6.69\times {10}^{-4}{\text{m}}^2\\ \pi {\left(\frac{d_i}{2}\right)}^2=6.69\times {10}^{-4}{\text{m}}^2\\ d_i=\sqrt{\frac{4\left(6.69\times {10}^{-4}{\text{m}}^2\right)}{3.14}}\\ d_i=2.92\times {10}^{-2}\text{m}\end{array}$

Substitute the values in the above expression.

  $h_o={\left(\frac{D_i}{D_o}\right)}^2\times h_i$

  $={\left(2.92/18\right)}^2\times 13=0.342\text{cm}$

Conclusion:

The car moves up $0.342\text{cm}$ high for each stroke.

(d)

To determine

The number of strokes required to jack the car up $12\text{cm}$ high.

Answer to Problem 68GP

The number of strokes required to jack the car up $12\text{cm}$ high is $35.09$ .

Explanation of Solution

Given:

Required height $h_i=12\text{cm}$

Formula Used:

Write the expression for the number of strokes required to jack the car up.

  $N=\frac{h_i}{h_o}$

Here $h_i$ is the input height,

  $h_o$ is the output height.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{l}N =\frac{12}{0.342}\hfill \\ N=35.09\hfill \end{array}$

Conclusion:

The number of strokes required to jack the car up $12\text{cm}$ high is $35.09$ .

(e)

To determine

To find: The energy is conserved.

Answer to Problem 68GP

The energy is conserved with work done as $32.5\text{J}$ .

Explanation of Solution

Given:

Diameter at the output of the piston, $D_o=0.18\text{m}$

Input force, $F_i=250\text{N}$

Distance moved by the piston, $s=0.13\text{m}$

Height to which car raised= $12\text{cm}$

Mass of car = $970\text{kg}$

Formula Used:

Write the expression for the work done at the output.

  $W_{\text{output}}=m\times g\times h$

Here $m$ is the mass of the car,

  $g$ is the acceleration due to gravity,

  $h$ is the height to which car is raised.

Write the expression for the work done at input piston:

  $W_{\text{Input}}=F_i\times s$

As per the law of conservation of energy,

  $W_{\text{output}}=W_{\text{Input}}$

Calculation:

On input piston:

  $W_{\text{input}}=F_i\times s=\left(250\times 0.13\right)=32.5\text{J}$

On output piston:

  $\begin{array}{c}{W}_{\text{output}}=mgh\\ =970\times 9.8\times 0.00342=32.5\text{J}\end{array}$

Here $W_{\text{output}}=W_{\text{Input}}$ .

Conclusion:

The energy is conserved with work done $32.5\text{J}$ .