If a 240 N force F1 is applied to the piston with the 4 cm diameter, what is the magnitude of the force F2 that can be resisted by the piston with the 10 cm diameter

38.4 N.

Explanation:

From pascal's principle,

Pressure applied to the first piston = pressure applied to the second piston

F1/A1 = F2/A2 Equation 1

Where F1 = Force applied to the first piston, A1 = Area of the first piston, F2 = force applied to the second piston, A2 = Area of the second piston.

But,

A1 = πd₁²/4 Equation 2

A2 = πd₂²/4 Equation 3

Where d₁ and d₂ are the diameter of the first and second piston respectively.

Substitute equation 2 and equation 3 into equation 1

F1/πd₁²/4 = F2/πd₂²/4

F1/d₁² = F2/d₂² Equation 4

make F2 the subject of the equation,

F2 = (F1×d₂)/d₁² Equation 5

Given: F1 = 240 N, d₁ = 4 cm, d₂ = 10 cm.

Substitute into equation 5

F2 = (240×4²)/10²

F2 = 38.4 N.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west