A hydraulic lift is shown in the figure. The radii of the movable pistons \(P_1\) and \(P_2\) are \(2\,\text{m}\) and \(5\,\text{m}\) respectively. If a block of mass \(x\) is placed on \(P_2\), then the minimum mass that should be kept on \(P_1\) to lift the block on \(P_2\) is
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In hydraulic lift problems, use Pascal's law:
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]
and remember that piston area is proportional to the square of its radius.
Step 1: State Pascal's Law as the governing principle. Pascal's Law states that pressure applied to an enclosed fluid is transmitted equally in all directions throughout the fluid. In a hydraulic lift, this means the pressure at piston $P_1$ equals the pressure at piston $P_2$: \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \] where $F_1$ and $F_2$ are the forces on the pistons and $A_1$, $A_2$ are their respective cross-sectional areas. Step 2: Identify the forces on each piston. Let the mass placed on $P_1$ (radius $r_1 = 2$ m) be $m$. The force on $P_1$ is $F_1 = mg$. The block of mass $x$ is on $P_2$ (radius $r_2 = 5$ m), so $F_2 = xg$. For the block on $P_2$ to just be lifted, the pressure from $P_1$ must equal the pressure from $P_2$. Step 3: Compute the areas of the pistons. The pistons are circular, so their areas are: \[ A_1 = \pi r_1^2 = \pi(2)^2 = 4\pi \, \text{m}^2 \] \[ A_2 = \pi r_2^2 = \pi(5)^2 = 25\pi \, \text{m}^2 \] Step 4: Apply Pascal's Law to find the minimum mass. \[ \frac{mg}{4\pi} = \frac{xg}{25\pi} \] Cancelling $g$ and $\pi$ from both sides: \[ \frac{m}{4} = \frac{x}{25} \] \[ m = \frac{4x}{25} = 0.16x \] Step 5: Understand why a smaller mass on $P_1$ can lift a larger mass on $P_2$. This is the mechanical advantage of a hydraulic lift. Since $A_2 > A_1$, a smaller force (smaller mass) on the smaller piston $P_1$ generates the same pressure as a larger force on the larger piston $P_2$. This is the fluid equivalent of a lever: a small input force over a small area produces a large output force over a large area at equal pressures. Step 6: Verify the answer. With $m = 0.16x$ on $P_1$: pressure $= 0.16xg / 4\pi = 0.04xg/\pi$. On $P_2$: pressure $= xg / 25\pi = 0.04xg/\pi$. The pressures match, confirming the block is just lifted. \[ \boxed{m = 0.16x} \]
