Basic Physics

Pascal’s principle – problems and solutions

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
P
2 = F2 / A2

P1 = P2
F
1 / A1 = F2 / A2

F1 / 10 cm2  = 100 N / 100 cm2

F1 / 10 = 1 N

F1 = (10)(1 N)

F1 = 10 Newton

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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
F
1 / A= 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

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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 :

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