Question

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg. The area of the cross section of the piston carrying the load is \[425\,{\text{c}}{{\text{m}}^2}\] . What maximum pressure would the smaller piston have to bear?

Answer 1

Hint:First of all, we will find the weight force of the automobile, which is simply the function of mass and acceleration due to gravity. Then we will find the pressure, which is termed as force per unit area. We will substitute the required values and obtain the answer.

Complete step by step solution:
In the given question, we are supplied the following data:
There is a hydraulic automobile lift which is designed to lift cars. The maximum capacity that the lift can hold with getting collapsed is \[3000\,{\text{kg}}\]. The area of the cross section of the piston that is carrying the load is \[425\,{\text{c}}{{\text{m}}^2}\]. We are asked to find the maximum pressure that the smaller have to bear.

To begin with, as we can see that the maximum capacity of the lift is given, in terms of mass. We can proceed to solve the numerical after we get the weight of the car with maximum mass. Let us proceed to solve the numerical. We know, weight force which is a function of mass and acceleration due to gravity, is given by the formula, which is shown below:
\[F = m \times g\] …… (1)
Where,
\[F\] indicates the weight force.
\[m\] indicates the mass of the car.
\[g\] indicates acceleration due to gravity.

Now, we substitute the required values in the equation (1) and we get:
$F = m \times g \\
\Rightarrow F = 3000 \times 9.8 \\
\Rightarrow F = 29400\,{\text{N}} \\$
Therefore, weight force is found to be \[29400\,{\text{N}}\] .

Again, we know the pressure is related to the force by a relation, as given below:
\[P = \dfrac{F}{A}\] …… (2)
Where,
\[P\] indicates the pressure.
\[A\] indicates the area of the cross section of the piston.

Now, we substitute the required values in the equation (2), we get:
$\Rightarrow P = \dfrac{{29400\,{\text{N}}}}{{425\,{\text{c}}{{\text{m}}^2}}} \\
\Rightarrow P = \dfrac{{29400\,{\text{N}}}}{{425 \times {{\left( {{{10}^{ - 2}}} \right)}^2}\,{{\text{m}}^2}}} \\
\Rightarrow P = \dfrac{{29400\,{\text{N}}}}{{425 \times {{10}^{ - 4}}\,{{\text{m}}^2}}} \\
\therefore P = 6.92 \times {10^6}\,{\text{Pa}} \\$
Hence, the maximum pressure that the smaller have to bear is \[6.92 \times {10^6}\,{\text{Pa}}\].

Note:While solving this problem, most of the students forget to convert the unit of the area of the cross section, which is square centimetres to square metres. Students get wrong and irrelevant answers, due to this reason. It is important to remember that smaller is the cross-sectional area of the object, higher is the pressure. Due to this reason a needle can easily go inside another object due to its small cross-sectional area.