Interactive Solution 11.35 at www.wiley.com/college/cutnell...

Interactive Solution $11.35$ at www.wiley.com/college/cutnell is a model for this problem. Multiple-Concept Example 8 also presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of $8.30 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}$. The weight of the input piston is negligible. The radii of the input piston and output plunger are $7.70 \times 10^{-3} \mathrm{~m}$ and $0.125 \mathrm{~m}$, respectively. What input force $F$ is needed to support the $24500-\mathrm{N}$ combined weight of a car and the output plunger, when $\quad$ (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is $1.30 \mathrm{~m}$ above that of the input piston?

Key Concepts

Force and Pressure Relationship

The force and pressure relationship is governed by the equation Pressure = Force / Area. In hydraulic systems, this relationship is used to determine how a small input force, applied over a small area, can generate a larger output force when transmitted through the fluid to a larger area, underlining the mechanism of force multiplication.

Circular Area Calculation

Calculating the area of circular pistons is fundamental in hydraulic systems since the relationship between force and pressure depends on the area of the piston surfaces. The area is determined using the formula Area = ?r², where r is the radius of the circle, making this a key step in analyzing and designing hydraulic mechanisms.

Pascal's Principle

Pascal's Principle states that any change in pressure applied to an enclosed incompressible fluid is transmitted undiminished throughout the fluid. This principle forms the basis for hydraulic systems, where a small force applied at one point is magnified to generate a larger force at another point, making it essential for understanding the operation of devices such as hydraulic lifts.

Hydrostatic Pressure

Hydrostatic pressure refers to the pressure exerted by a fluid at equilibrium due to the force of gravity, which increases with depth. When different parts of a hydraulic system are at different heights, this pressure difference must be taken into account, as it affects the overall pressure experienced by the system and the force required to balance loads.

Transcript

0:00 Hi there.

00:01 So for this problem, we have an hydraulic oil in a car leaf that has a density of 8 .30 times 10 to 2 kilograms per meter cube.

00:12 This value in here.

00:16 And the weight of the input piston is negligible.

00:22 So we note that the rady of the impact piston and output plunger of the following, 7 .70 times 10 to the minus 3 meters.

00:32 For the radii of the input piston.

00:37 And for the plunger, we have, for the output piston, we have a rady 2 that has a value of 0 .125 meters.

00:50 And for this problem, we need to calculate the force that we need to apply in the input piston so that it supports the weight of a car, whose weight is to 24 ,500 mutons.

01:10 So the first part of this problem tell us to obtain that force that we need to apply in the input piston, when the piston and the plunger are at the same level.

01:23 So the piston, which is this, the input piston, and the plunger are at the same height.

01:32 So with that information, we need to calculate the force f.

01:43 So for this we use the definition that in this case, the pressure in the piston should be equal to the pressure in the plunker.

01:54 So we know that the definition of pressure is the force applied over the area of application.

02:03 So we have this.

02:08 We are going to call f1, the force that we need to apply f.

02:13 And f2 is the force that in the, this mechanism needs to lift that or to support what is the weight of the cart.

02:26 And the area two also corresponds to the area of the plunger and area one corresponds to the area of the input piston.

02:35 So with that information, we can solve for f1, that is also equal to f.

02:41 So in this case, we will have that this is the ratio between a1, the area of the piston over the area of the plunger, times the force f2, which is the weight of the car.

02:55 So we note that the area in this case is the area of a circle, and we know that the area of circle is p times its radius a square.

03:04 So we have the radius of 1 square and the radius of the plunger also square.

03:11 We can console these pi terms.

03:14 And this times the weight of the card that needs to be support.

03:18 So that is 24 ,500 meters.

03:24 So we substitute the values for the rati in here.

03:28 So i'm going to put that explicitly.

03:31 So we have the radi 1, which is 7...