The area of cross-section of the pump plunger and the press plunger of hydraulic press are $0.03{{m}^{2}}$ and $9{{m}^{2}}$ respectively. How much is the force acting on the pump plunger of the hydraulic pres overcomes a load of 900kgf
a) 2kgf
b) 3kgf
c) 4kgf
d) 5kgf
Answer
555.9k+ views
Hint: The press plunger(surface where object to be lifted is kept) and the pump plunger (surface where the force is applied) are inserted in a closed vessel attached together and filled with liquid. In the question we are basically asked to determine the force in terms of kgf required to lift a load of 900kgf. Hence using Pascal’s law of transmission of fluid pressure we can determine the force required to be applied on the other surface so as to lift the body.
Formula used:
$\dfrac{f}{a}=\dfrac{F}{A}$
Complete answer:
The hydraulic lift consists of two cylinders connected to each other with a pipe. The cylinders are fitted with water tight frictionless pistons of different cross sectional areas. The cylinders and the pipe contain liquid. Suppose a force ‘f’ is applied to a smaller piston(pump plunger) of cross sectional area ‘a’. Since pressure is defined as force per unit area, the pressure ‘P’ acting downwards is given by,
$P=\dfrac{f}{a}$
According to Pascal's law the same pressure is transmitted to the larger piston (press plunger). As the liquid moves downwards the piston on the other side will get pushed upwards. Let us say the area of the cross section of the second piston be ‘A’ and the upward force be equal to ‘F’. Hence using Pascal’s law we get,
$\dfrac{f}{a}=\dfrac{F}{A}$
The area of cross-section of the pump plunger and the press plunger of hydraulic press are $0.03{{m}^{2}}$ and $9{{m}^{2}}$ . The load that is lifted is 900kgf. Hence using the above equation the force that is applied on the pump plunger is,
$\begin{align}
& \dfrac{f}{a}=\dfrac{F}{A} \\
& \Rightarrow \dfrac{f}{0.03{{m}^{2}}}=\dfrac{900kgf}{9{{m}^{2}}} \\
& \Rightarrow f=\dfrac{900kgf\times 0.03}{9}=100\times 0.03kgf \\
& \therefore f=3kgf \\
\end{align}$
Therefore the correct answer of the above question is option b.
Note:
It is to be noted that a hydraulic lift is basically a mechanical advantage. If we observe carefully the above equation even to lift an object of huge mass a small application of force is sufficient. This is due to the fact that the area of the press plunger is greater than that of the pump plunger.
Formula used:
$\dfrac{f}{a}=\dfrac{F}{A}$
Complete answer:
The hydraulic lift consists of two cylinders connected to each other with a pipe. The cylinders are fitted with water tight frictionless pistons of different cross sectional areas. The cylinders and the pipe contain liquid. Suppose a force ‘f’ is applied to a smaller piston(pump plunger) of cross sectional area ‘a’. Since pressure is defined as force per unit area, the pressure ‘P’ acting downwards is given by,
$P=\dfrac{f}{a}$
According to Pascal's law the same pressure is transmitted to the larger piston (press plunger). As the liquid moves downwards the piston on the other side will get pushed upwards. Let us say the area of the cross section of the second piston be ‘A’ and the upward force be equal to ‘F’. Hence using Pascal’s law we get,
$\dfrac{f}{a}=\dfrac{F}{A}$
The area of cross-section of the pump plunger and the press plunger of hydraulic press are $0.03{{m}^{2}}$ and $9{{m}^{2}}$ . The load that is lifted is 900kgf. Hence using the above equation the force that is applied on the pump plunger is,
$\begin{align}
& \dfrac{f}{a}=\dfrac{F}{A} \\
& \Rightarrow \dfrac{f}{0.03{{m}^{2}}}=\dfrac{900kgf}{9{{m}^{2}}} \\
& \Rightarrow f=\dfrac{900kgf\times 0.03}{9}=100\times 0.03kgf \\
& \therefore f=3kgf \\
\end{align}$
Therefore the correct answer of the above question is option b.
Note:
It is to be noted that a hydraulic lift is basically a mechanical advantage. If we observe carefully the above equation even to lift an object of huge mass a small application of force is sufficient. This is due to the fact that the area of the press plunger is greater than that of the pump plunger.
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