Pascal's Principle — Question 4

Question 4

A spring load makes hydraulic work grow during the stroke

An ideal horizontal hydraulic device has area ratio $A_2/A_1=10$. The output piston compresses a spring of constant $k=20000\,\mathrm{N/m}$ from its natural length. Both piston backs have equal pressure; ignore gravity, friction and liquid compression.

1. Express input force as a function of input displacement $x_1$. Find the final input force when $x_1=0.20\,\mathrm{m}$.

2. Find the input work for that stroke and compare it with stored spring energy. Why is final force times total stroke incorrect?

Question 4 - Solution

A spring load makes hydraulic work grow during the stroke

(a) Derivation and calculation

$$A_1x_1=A_2x_2\Rightarrow x_2=x_1/10.$$$$p_gA_2=kx_2,\quad F_1=p_gA_1\Rightarrow\boxed{F_1=kx_1/100}.$$$$\boxed{x_2=.0200\,\mathrm m},\qquad\boxed{F_{1,f}=20000(.200)/100=40.0\,\mathrm N}.$$

(b) Result and interpretation

$$\boxed{W_1=\int_0^{.200}\frac{kx_1}{100}\,dx_1=\frac{k(.200)^2}{200}=4.00\,\mathrm J}.$$$$\boxed{U_s=\tfrac12kx_2^2=\tfrac12(20000)(.0200)^2=4.00\,\mathrm J}.$$$$F_{1,f}x_{1,f}=8.00\,\mathrm J=2W_1.$$The force rises linearly; its mean is half its final value.

Physical check

The effective stiffness seen at the input is $k/100$, the squared inverse area ratio.