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 with a density of 800 kg/m³ 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 force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? For purposes of this problem,neglect the weight of the pistons.

Question

Quizplus · Accepted Answer

The pressure at any point in the U-tube is given by the equation:

$P = \frac{F}{A}$

where P is the pressure, F is the force applied, and A is the area of the piston.

The pressure is constant throughout the U-tube, so we can set the pressure at the level of the pistons to be equal:

$P_1 = P_2$

where the subscripts 1 and 2 refer to the wider and narrower arms of the U-tube, respectively.

We can use the equation for pressure to relate the forces and areas of the two pistons:

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

We can solve for $F_2$ in terms of $F_1$:

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

We can plug in the values for the areas:

$F_2 = \frac{\pi r_2^2}{\pi r_1^2}F_1 = \frac{r_2^2}{r_1^2}F_1$

Now we need to find $F_1$. We know the weight of the car is 12,000 N, and we can use the density of the oil to find the weight of the oil displaced by the car:

$w_{oil} = V_{displaced} 	imes ho_{oil} = (\pi r_1^2 h) 	imes (ho_{oil})$

where $V_{displaced}$ is the volume of oil displaced by the car and $h$ is the height of the oil column in the wider arm of the U-tube. The height of the oil column in the narrower arm of the U-tube is $1.2	ext{ m} - h$.

Setting the weight of the car equal to the weight of the displaced oil, we can solve for $h$:

$12,000	ext{ N} = (\pi r_1^2 h) 	imes (ho_{oil})$

$h = \frac{12,000	ext{ N}}{\pi r_1^2 ho_{oil}}$

We can plug in the values and solve for $h$:

$h = \frac{12,000	ext{ N}}{\pi (0.18	ext{ m})^2 (800	ext{ kg/m}^3)} \approx 0.274	ext{ m}$

Now we can plug in the values for $r_1$, $r_2$, $h$, and $F_1$ to solve for $F_2$:

$F_2 = \frac{(0.05	ext{ m})^2}{(0.18	ext{ m})^2}F_1 \approx 0.287F_1$

Setting $P_1 = P_2$, we can use the equation for pressure to relate $F_1$ to the weight of the displaced oil:

$\frac{F_1}{A_1} = P = ho_{oil}gh$

$F_1 = (ho_{oil}gh)(A_1) = (ho_{oil}g)(\frac{12,000	ext{ N}}{\pi r_1^2 ho_{oil}})(\pi r_1^2) = 12,000	ext{ N}$

Now we can plug in $F_1$ to find $F_2$:

$F_2 = 0.287(12,000	ext{ N})\approx 3,444	ext{ N} \approx 1.96	ext{ kN}$

Therefore, the answer is (C) 1.96 kN.

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