Question

The drawing shows a hydraulic chamber with a spring (spring constant $=1600\, N / m$ ) attached to the input piston and a rock of mass $40.0\, kg$ resting on the output plunger. The piston and plunger are nearly at the same height, and each has a negligible mass. By how much is the spring compressed (in cm) from its unstrained position?

Answer 1

Let $F_{1}$ and $F_{2}$ are the magnitudes of the force the spring exerts on the piston and the rock exerts on the plunger respectively. Initially the piston and the plunger are at the same height, so the fluid pressure at the piston is equal to the fluid pressure at the plunger
$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$ where $A_{1}$ is the area of the piston and $A_{2}$ is the area of the plunger.
Therefore, the magnitude of the force $F_{2}$ that the rock exerts on the plunger is given by
$F_{2}=F_{1}\left(\frac{A_{2}}{A_{1}}\right)$
Here $F_{1}=k x$, where $k$ is the spring constant of the spring, and $x$ is the amount by which the spring is compressed from its unstrained position. and the magnitude $F_{2}$ of the force the rock exerts on the plunger is equal to the magnitude $W=m g$ where $m$ is the rock's mass and $g$ is the magnitude of the acceleration due to gravity.
Solving $F_{2}=F_{1}\left(\frac{A_{2}}{A_{1}}\right)$ for $F_{1}$ and substituting $F_{2}=m g$ yields
$F_{1}=F_{2}\left(\frac{A_{1}}{A_{2}}\right)=m g\left(\frac{A_{1}}{A_{2}}\right)$ ...(1)
Solving for $x$, we obtain, $x=\frac{F_{1}}{k}$ ...(2)
Substituting Eq. (1) into Eq. (2), we find that
$x=\frac{F_{1}}{k}=\frac{m g}{k}\left(\frac{A_{1}}{A_{2}}\right)$
$=\frac{(40\, kg )\left(10 m / s ^{2}\right)}{1600\, N / m }\left(\frac{15\, cm ^{2}}{75\, cm ^{2}}\right)$
$=5.0 \times 10^{-2} m$