Solution Found!
Refer to Figure 7.1.4a, and suppose that p1 p2 = 10
Chapter , Problem 7.5(choose chapter or problem)
Refer to Figure 7.1.4a, and suppose that \(p_{1}-p_{2}=10 \ \mathrm{lb} / \mathrm{in}^{2}, A=3 \text { in. }^{2}\), and mg = 600 lb. If the mass starts from rest at x(0) = 0, how far will it move in 0.5 sec, and how much hydraulic fluid will be displaced?
Questions & Answers
QUESTION:
Refer to Figure 7.1.4a, and suppose that \(p_{1}-p_{2}=10 \ \mathrm{lb} / \mathrm{in}^{2}, A=3 \text { in. }^{2}\), and mg = 600 lb. If the mass starts from rest at x(0) = 0, how far will it move in 0.5 sec, and how much hydraulic fluid will be displaced?
ANSWER:Step 1 of 3
Consider the following diagram:
Figure 1
Consider the following data:
The difference between two pressures, \({p_1} - {p_2} = 10\;{\rm{lb/i}}{{\rm{n}}^2}\), the area of the piston, \(A = 3\;{\rm{i}}{{\rm{n}}^2}\), the time t = 0.5 sec, the weight of the load is \(mg = 600\;{\rm{lb}}\).
Convert the area of the piston into square meters.
\(A = 3\;{\left( {0.0254\;{\rm{m}}} \right)^2}\)
\(A = 1.935 \times {10^{ - 3}}\;{{\rm{m}}^2}\)
Convert the value of pressure difference into \({\rm{N/}}{{\rm{m}}^2}\).
\({p_1} - {p_2} = 10\left( {\frac{{4.4482}}{{0.0254}}} \right)\;{\rm{N/}}{{\rm{m}}^2}\)
\({p_1} - {p_2} = 68947.23\;{\rm{N/}}{{\rm{m}}^2}\)