MYP Physics · topic MYP-08.01
Pressure in solid contact
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A hydraulic lift below the reservoir surface
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An ideal hydraulic system contains oil of density 800 kg m⁻³. Its small piston has area 2.0 cm² and its large piston has area 40 cm². The large piston is 0.50 m below the small piston. A 60 N force is applied downward to the small piston; atmospheric pressure acts on both external faces. Use g = 10 N kg⁻¹ and neglect piston weights and friction. Supplied data (Depth below small piston / m; Additional pressure / Pa): (0, 0), (0.25, 2000), (0.5, 4000). (a) Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance. [3 marks] (b) For a short motion, the small piston moves down 0.10 m. Find the large-piston rise and explain why force amplification does not create energy. [3 marks]
Hydrostatic gauge-pressure contribution below the small piston
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Additional pressure / Pa
Horizontal axis (x): Depth below small piston / m
Read the exact data as a table
| Depth below small piston / m | Additional pressure / Pa |
|---|---|
| 0 | 0 |
| 0.25 | 2000 |
| 0.5 | 4000 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance.
Reveal the worked solution
- Convert 2.0 cm² to 2.0×10⁻⁴ m². Applied gauge pressure is F/A = 60/(2.0×10⁻⁴) = 300000 Pa.
- The lower piston has an additional ρgh = 800(10)(0.50) = 4000 Pa, giving 304000 Pa gauge.
- Its area is 40 cm² = 0.0040 m², so the balanced load is pA = 304000(0.0040) = 1216 N. Atmospheric contributions cancel across each piston.
- Incompressibility gives equal displaced volumes: As xs = AL xL, so xL = (2/40)(0.10) = 0.0050 m.
- The large force acts through a smaller displacement. For equal piston heights the ideal input and output work are equal.
- Here the initial pressure difference also involves gravitational potential energy of the oil. For a finite stroke the height difference changes; a full energy audit must include the oil and the varying pressure, rather than multiplying the initial force by any arbitrary stroke.
Answer (a) Small-piston gauge pressure = 300 kPa; large = 304 kPa; load = 1216 N. (b) Large-piston rise = 0.0050 m; input work and gravitational-energy changes must both be included.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.