Problem Statement
An 8 cm diameter piston compresses manometer oil into an inclined 7 mm diameter tube. When a weight \(W\) is added to the top of the piston, the oil rises an additional distance of 10 cm up the tube. Calculate the weight \(W\) in Newtons.
Given data:
- Diameter of piston (\(D\)) = 8 cm = 0.08 m
- Diameter of tube (\(d\)) = 7 mm = 0.007 m
- Height of oil rise (\(h\sin(15^\circ)\)) = 0.10 \(\times \sin(15^\circ)\)
Solution
1. Calculate Oil Rise in Tube
Determine the vertical component of the oil rise:
\(h = 0.10 \times \sin(15^\circ) = 0.0258 \, \text{m}\)
2. Volume Conservation
The volume of oil displaced by the piston equals the volume of oil rising in the tube:
\( \frac{\pi}{4} \times (0.08)^2 \times \Delta h = \frac{\pi}{4} \times (0.007)^2 \times 0.10 \)
\( \Delta h = 0.000766 \, \text{m}\)
3. Pressure Balance
Equate pressures at the new equilibrium level:
\( \frac{W}{\text{Area of piston}} + P_X + \gamma_{\text{air}} \Delta h = \gamma_{\text{oil}} \times (h + \Delta h) + \gamma_{\text{oil}} \times L_1 \sin(15^\circ) \)
Neglecting small terms involving \(\Delta h\):
\( \frac{W}{\pi/4 \times (0.08)^2} = \gamma_{\text{oil}} \times h \)
Substituting values:
\( W = \frac{\pi}{4} \times (0.08)^2 \times 0.827 \times 9810 \times 0.0258 = 1.05 \, \text{N} \)
Weight of the piston:
\( W = 1.05 \, \text{N} \)
Explanation
- Volume Conservation: The displacement caused by the piston is exactly balanced by the volume rise in the tube.
- Pressure Balance: Pressure at the new equilibrium level includes contributions from the weight of the piston and the rise in oil height.
- Simplifications: Small terms involving the height increment are neglected to simplify calculations.
Physical Meaning
This numerical illustrates the principles of volume conservation and pressure equilibrium in a practical scenario. Such analyses are fundamental in designing hydraulic systems and understanding fluid behavior under compression.
