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Chapter 14: Problem 8

The small piston of a hydraulic lift has a cross-scrional area of \(3.00 \mathrm{cm}^{2},\) and its large piston has a cross-sectional area of \(\left.200 \mathrm{cm}^{2} \text { (Figure } 14.4\right) .\) What force must be applied to the small piston for the lift to raise a load of \(15.0 \mathrm{kN} ?\) (In service stations, this force is usually exerted by compressed air.)

Short Answer

Expert verified
The force that must be applied to the small piston for the lift to raise a load of 15.0 kN is 2.25 N.

Step by step solution

01

Understanding Pascal's Principle

To start with, recall Pascal's principle: the pressure is equal everywhere in an enclosed fluid. In mathematical terms, this means \( F1/A1 = F2/A2 \), where F is force and A is area, and the subscripts 1 and 2 refer to the small and large pistons, respectively.
02

Substitute Known Values

We can solve this equation for \( F1 \), the force on the small piston: \( F1 = F2 * A1 / A2 \). We know the values of \( F2 = 15.0 \, kN = 15000 \, N \), \( A1 = 3.00 \, cm^2 = 3.00*10^{-4} \, m^2 \) (because 1 cm^2 equals \( 10^{-4} m^2 \)), and \( A2 = 200 \, cm^2 = 2 \, m^2 \). So, \( F1 = 15000 \, N * 3.00*10^{-4} \, m^2 / 2 \, m^2 \)
03

Calculate the Result

Compute the value of \( F1 \) from the above expression to determine the force required to be applied on the small piston. This results in: \( F1 = 2.25 \, N \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydraulic Lift
A hydraulic lift works by using liquid to transfer force from one point to another, taking advantage of Pascal's Principle. This device comprises two connected pistons of different sizes. The smaller piston, where force is applied, is connected via fluid to a larger piston, which does the heavy lifting. By applying force to the smaller piston, you can lift heavy objects with relative ease. The magic of the hydraulic lift lies in the fluid's incompressibility, ensuring consistent pressure throughout the system, allowing the force applied on the small piston to be distributed over the large piston. This principle allows even a small force applied on the smaller area to create a larger force at the larger area, enabling the lifting of substantial weights with precision and efficiency.
Pressure in Fluids
Understanding pressure in fluids is key to comprehending how hydraulic systems work. Pressure is defined as the force exerted per unit area and is expressed as: \[ P = \frac{F}{A} \]In the context of hydraulic systems, pressure inside the fluid must remain constant throughout. This means the pressure exerted on the smaller piston must equal the pressure exerted on the larger piston. Pascal's Principle underlines this, stating that changes in pressure applied to a confined fluid are transmitted undiminished throughout the fluid. Therefore, in a hydraulic lift, the force applied to the small piston can raise a much larger load. This is possible because as area increases, the same pressure allows for a greater force, thus enhancing our ability to perform tasks that require heavy lifting.
Force Calculation
Force calculation in a hydraulic system involves leveraging the relationship between pressure, force, and area. To find the force necessary to raise a load using a hydraulic lift, we use the formula derived from Pascal's principle:\[ F_1 = \frac{F_2 \times A_1}{A_2} \]Here, \( F_1 \) is the force applied on the smaller piston, \( F_2 \) is the force exerted by the load, \( A_1 \) is the area of the small piston, and \( A_2 \) is the area of the large piston. By maintaining the same pressure across different areas, we find that a smaller force on a smaller piston can move a larger piston and lift larger loads, as seen in hydraulic lifts. This useful property allows industries and service stations to efficiently move and lift heavy objects with minimal effort.

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Most popular questions from this chapter

Normal atmospheric pressure is \(1.013 \times 10^{5} \mathrm{Pa}\). The approach of a storm causes the height of a mercury barometer to drop by \(20.0 \mathrm{mm}\) from the normal height. What is the atmospheric pressure? (The density of mercury is \(\left.13.59 \mathrm{g} / \mathrm{cm}^{3} .\right)\)

For the cellar of a new house, a hole is dug in the ground, with vertical sides going down \(2.40 \mathrm{m} .\) A concrete foundation wall is built all the way across the \(9.60-\mathrm{m}\) width of the excavation. This foundation wall is \(0.183 \mathrm{m}\) away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\).

Water falls over a dam of height \(h\) with a mass flow rate of \(R,\) in units of \(\mathrm{kg} / \mathrm{s} .\) (a) Show that the power available from the water is $$\mathscr{P}=R g h$$ where \(g\) is the frec-fall acceleration. (b) Each hydroclectric unit at the Grand Coulce Dam takes in watcr at a rate of \(8.50 \times 10^{5} \mathrm{kg} / \mathrm{s}\) from a height of \(87.0 \mathrm{m} .\) The power developed by the falling water is converted to clectric power with an efficiency of \(85.0 \% .\) How much electric power is produced by each hydroelectric unit?

A swimming pool has dimensions \(30.0 \mathrm{m} \times 10.0 \mathrm{m}\) and a flat bottom. When the pool is filled to a depth of \(2.00 \mathrm{m}\) with fresh water, what is the force caused by the water on the bottom? On each end? On each side?

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. (a) Suppose that a horizontal wind blows in streamline flow with a speed of \(11.2 \mathrm{m} / \mathrm{s}\) outside a large pane of plate glass with dimensions \(4.00 \mathrm{m} \times 1.50 \mathrm{m}\) Assume the density of the air to be uniform at \(1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? (b) What If? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is \(22.4 \mathrm{m} / \mathrm{s}\), twice as high.
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