# Pascal’s principle – problems and solutions

1. __Known :__

The area of A_{1 }= 10 cm^{2}

The area of A_{2} = 100 cm^{2}

Force 2 (F_{2}) = 100 Newton

__Wanted :__ Force 1 (F_{1})

__Solution :__

P = F / A

*P = **pressure**, F = **force**, A = **area*

P_{1 }= F_{1} / A_{1 }

P_{2} = F_{2} / A_{2}

P_{1} = P_{2}

F_{1} / A_{1 }= F_{2 }/ A_{2}

F_{1} / 10 cm^{2 }_{ }= 100 N / 100 cm^{2}

F_{1} / 10_{ }= 1 N

F_{1} = (10)(1 N)

F_{1} = 10 Newton

2. If the area of A_{1} = 0.001 m^{2} and the area of A_{2} = 0.1 m^{2} , external input force F_{1} = 100 N, then the external output force F_{2} ?

__Known :__

The area of A_{1} = 0.001 m^{2}

The area of A_{2} = 0.1 m^{2}

External input force F_{1} = 100 Newton

__Wanted ____:__ External output force (F_{2})

__Solution :__

P_{1} = P_{2}

F_{1} / A_{1 }= F_{2 }/ A_{2}

100 N / 0.001 m^{2} = F_{2} / 0.1 m^{2}

100 N / 0.001 = F_{2} / 0.1

100,000 N = F_{2} / 0.1

F_{2} = (0.1)(100,000 N)

F_{2} = 10,000 N

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 (A_{B}) = 4000 cm^{2} = 4000 / 10,000 m^{2} = 4 / 10 m^{2} = 0.4 m^{2 }

Area of A (A_{A}) = 50 cm^{2} = 50 / 10,000 m^{2} = 0.005 m^{2 }

__Wanted :__ Force F

__Solution :__

F / A_{A} = w / A_{B}

F / 0.005 m^{2 }= 16,000 N / 0.4 m^{2 }

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 cm^{2} and area of B is 4,200 cm^{2}, determine the external input force of F.

__Known :__

Area of A (A_{A}) = 60 cm^{2}

Area of B (A_{B}) = 4200 cm^{2}

Weight w (w) = 3500 Newton

__Wanted :__ F_{1}

__Solution :__

Force of F calculated using the equation of Pascal’s principle :

F_{1} / A_{1} = F_{2} / A_{2}

F_{1} / 60 cm^{2 }= 3500 N / 4200 cm^{2}

F_{1} / 60 = 35 N / 42

F_{1} = (60)(35) / 42

F_{1} = 2100 / 42

F_{1} = 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 (A_{1}) = A

Large cross-sectional area (A_{2}) = 20A

Input force (F_{1}) = 25 N

__Wanted :__ Output force (F_{2})

__Solution :__