The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm2. what downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is fg = 15.0 kn?

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

225 N

Explanation:

From Pascal's principle,

F/A = f/a Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.