A 12,000-N car is raised using a hydraulic lift,which consists of a U-tube with arms of unequal areas,filled with oil and capped at both ends with tight-fitting pistons.The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm.The car rests on the piston on the wider arm of the U-tube.The pistons are initially at the same level.What is the initial force that oil must exert on the smaller piston in order to start lifting the car? For purposes of this problem,neglect the mass of the pistons.
A)727 N
B)926 N
C)2.90 kN
D)3.33 kN
E)1.20 kN

Question

Quizplus · Accepted Answer

The pressure in the wider arm of the U-tube is equal to the pressure in the narrower arm, since the oil is incompressible and therefore the pressure must be constant throughout the fluid. We can use the equation for hydraulic pressure:

$$F_1/A_1 = F_2/A_2$$

where $F_1$ is the force on the wider piston, $A_1$ is its cross-sectional area, $F_2$ is the force on the narrower piston (which we want to find), and $A_2$ is its cross-sectional area. Rearranging this equation, we get:

$$F_2 = \frac{A_2}{A_1} F_1$$

Plugging in the values given, we get:

$$F_2 = \frac{\pi (0.050\ 	ext{m})^2}{\pi (0.180\ 	ext{m})^2} (12,\!000\ 	ext{N}) = 925.9\ 	ext{N} \approx \boxed{	extbf{(B) } 926\ 	ext{N}}$$

Note that we convert the radii from centimeters to meters in the calculation.

A 12,000-N Car Is Raised Using a Hydraulic Lift,which Consists