1. Known :
The area of A1 = 10 cm2
The area of A2 = 100 cm2
Force 2 (F2) = 100 Newton
Wanted : Force 1 (F1)
Solution :
P = F / A
P = pressure, F = force, A = area
P1 = F1 / A1
P2 = F2 / A2
P1 = P2
F1 / A1 = F2 / A2
F1 / 10 cm2 = 100 N / 100 cm2
F1 / 10 = 1 N
F1 = (10)(1 N)
F1 = 10 Newton
[irp]
2. If the area of A1 = 0.001 m2 and the area of A2 = 0.1 m2 , external input force F1 = 100 N, then the external output force F2 ?
Known :
The area of A1 = 0.001 m2
The area of A2 = 0.1 m2
External input force F1 = 100 Newton
Wanted : External output force (F2)
Solution :
P1 = P2
F1 / A1 = F2 / A2
100 N / 0.001 m2 = F2 / 0.1 m2
100 N / 0.001 = F2 / 0.1
100,000 N = F2 / 0.1
F2 = (0.1)(100,000 N)
F2 = 10,000 N
[irp]
3. Car’s weight = 16,000 N. What is the external input force F…
Known :
Car’s weight (w) = 16,000 N
Area of B (AB) = 4000 cm2 = 4000 / 10,000 m2 = 4 / 10 m2 = 0.4 m2
Area of A (AA) = 50 cm2 = 50 / 10,000 m2 = 0.005 m2
Wanted : Force F
Solution :
F / AA = w / AB
F / 0.005 m2 = 16,000 N / 0.4 m2
F / 0.005 = 16,000 N / 0.4
F / 0.005 = 40,000 N
F = (40,000 N)(0.005)
F = 200 Newton
4.
Area of A is 60 cm2 and area of B is 4,200 cm2, determine the external input force of F.
Known :
Area of A (AA) = 60 cm2
Area of B (AB) = 4200 cm2
Weight w (w) = 3500 Newton
Wanted : F1
Solution :
Force of F calculated using the equation of Pascal’s principle :
F1 / A1 = F2 / A2
F1 / 60 cm2 = 3500 N / 4200 cm2
F1 / 60 = 35 N / 42
F1 = (60)(35) / 42
F1 = 2100 / 42
F1 = 50 Newton
5. The hydraulic lift has a large cross section and a small cross section. Large cross-sectional area is 20 times the small cross-sectional area. If on the small cross section is given an input force of 25 N, then determine the output force.
Known :
Small cross section area (A1) = A
Large cross-sectional area (A2) = 20A
Input force (F1) = 25 N
Wanted : Output force (F2)
Solution :
