Pascal's Principle — Question 8

Section 14-5 · Physics pages 2146–2147

Question 8 Seal friction changes both the threshold and the energy balance A hydraulic device has input area 2.0 cm2 and output area 20 cm2. During slow upward output motion, the output seal exerts 100 N downward friction in addition to a 900 N load. Input friction is negligible. Both backs are atmospheric and height differences are neglected. F1 F2 A1 A2 Connected piston schematic; areas and elevations are specified in the text. (a) Find the required input force. For a 0.020 m output rise, find the input travel. (b) Calculate input work, useful load work and dissipated energy. Compare with the frictionless force ratio.
Show solutionHide solution
Question 8 - Solution Seal friction changes both the threshold and the energy balance F1 F2 A1 A2 (a) Derivation and calculation For slow upward output motion, pgA2 = L + Ff = 900 + 100 = 1000 N. pg = 1000/.00200 = 500000 Pa, F1 = pgA1 = 100 N . x1 = (A2/A1)x2 = 10(.0200) = .200 m . (b) Result and interpretation Win = 100(.200) = 20.0 J , Wuseful = 900(.0200) = 18.0 J . Edissipated = 100(.0200) = 2.00 J . L/F1 = 9 < 10 = A2/A1, Win = Wuseful + Edissipated. Physical check Pressure transmission remains valid while the moving seal dissipates mechanical energy.

Question 8

Seal friction changes both the threshold and the energy balance

A hydraulic device has input area 2.0cm22.0\,\mathrm{cm^2} and output area 20cm220\,\mathrm{cm^2}. During slow upward output motion, the output seal exerts 100N100\,\mathrm{N} downward friction in addition to a 900N900\,\mathrm{N} load. Input friction is negligible. Both backs are atmospheric and height differences are neglected.

Diagram shown in the original SVG worksheet.

  1. Find the required input force. For a 0.020m0.020\,\mathrm{m} output rise, find the input travel.

  2. Calculate input work, useful load work and dissipated energy. Compare with the frictionless force ratio.

Question 8 - Solution

Seal friction changes both the threshold and the energy balance

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For slow upward output motion,pgA2=L+Ff=900+100=1000N.p_gA_2=L+F_f=900+100=1000\,\mathrm N.pg=1000/.00200=500000Pa,F1=pgA1=100N.p_g=1000/.00200=500000\,\mathrm{Pa},\quad\boxed{F_1=p_gA_1=100\,\mathrm N}.x1=(A2/A1)x2=10(.0200)=.200m.\boxed{x_1=(A_2/A_1)x_2=10(.0200)=.200\,\mathrm m}.

(b) Result and interpretation

Win=100(.200)=20.0J,Wuseful=900(.0200)=18.0J.\boxed{W_{\mathrm{in}}=100(.200)=20.0\,\mathrm J},\quad\boxed{W_{\mathrm{useful}}=900(.0200)=18.0\,\mathrm J}.Edissipated=100(.0200)=2.00J.\boxed{E_{\mathrm{dissipated}}=100(.0200)=2.00\,\mathrm J}.L/F1=9<10=A2/A1,Win=Wuseful+Edissipated.L/F_1=9<10=A_2/A_1,\qquad W_{\mathrm{in}}=W_{\mathrm{useful}}+E_{\mathrm{dissipated}}.

Physical check

Pressure transmission remains valid while the moving seal dissipates mechanical energy.