Pressure problem: the pressure of two solid cubes on the desktop is equal. If the same volume, thickness, mass and bottom area are cut off, what is the residual pressure?

This lecture is about educational information.

First, the teaching content:

Chapter 10 Pressure of Pressure Liquid (1)

Two. Important and difficult

1. Understand the concept of pressure, know the direction of pressure and know the difference between pressure and gravity;

2. Grasp the concept of pressure and the physical meaning of its definition p=F/S;

3. Understand the unit Pascal of pressure and its physical meaning;

4. Master the methods of increasing or decreasing stress.

3. Knowledge point analysis

(1) pressure

1. Definition: The force acting vertically on the surface of an object is called pressure.

2. Direction: pointing to the stressed object perpendicular to the supporting surface; Action point: on the surface of the stressed object.

3. When an object is under pressure from other objects, this object also exerts pressure on that object, and the effects of pressure between objects are mutual.

4. Pressure may or may not be caused by gravity; The magnitude and direction of pressure are not necessarily equal to that of gravity.

(2) pressure

1. Definition: The pressure per unit area of an object is called pressure.

2. Pressure is not pressure, but a physical quantity representing the action of pressure. Pressure is determined by both pressure and stress area.

3. Formula: pressure = pressure/stressed area, that is, p = f/s.

P= F/ S can be used not only for the calculation of solid pressure, but also for the calculation of liquid and gas pressure.

4. Unit: International System of Units: Pascal (Pa)

1 Pascal = 1 Newton/square meter, which means that the pressure per square meter of surface area is 1 Newton.

Pressure units include: hectopascals (HP), kilopascals (KP), megapascals (MP), etc.

5. Ways to increase and decrease stress:

(1) pressurization method: when the pressure is constant, reduce the stress area.

(2) decompression method: when the pressure is constant, increase the stress area.

6. The magnitude and direction of the transfer pressure between solids remain unchanged under the condition that the force on the object is balanced; However, the generated pressure is determined by the size of the stressed area and cannot be transmitted through solids.

Typical example

[Example 1] The height, width and thickness of a rectangular granite monument are 10m, 3m and 2m respectively, the density is 2.5× 103Kg/m3, and the maximum earth pressure is 1.2× 105Pa. In order to make the ground bear the load, the monument is now erected on the cornerstone of a cuboid, with a height of 1m and a density of 2× 103kg/m3, taking g = 10n/kg. (1) What is the mass of the monument? (2) If the monument stands directly on the ground, what is its pressure on the ground? (3) What is the minimum floor area of the cornerstone?

Analysis: The first two questions of this question are relatively simple, so it is enough to calculate with basic formulas. Knowing the length, width and height of the monument, we can get its volume, and then multiply it by its density to get its mass. Then we can get the gravity of the monument through the gravity formula, that is, its pressure on the ground, and then divide it by the bottom area to get the pressure. In order to find the minimum bottom area of the last problem, we can use the critical condition of the maximum ground pressure at this time to find the minimum bottom area by equation.

Answer:

(1) The quality of the monument is:

m =ρV = 2.5× 103kg/m3× 10m×3m×2m = 1.5× 105kg

(2) When the monument stands directly on the ground, the pressure on the ground is:

p = F/S = mg/S = 1.5× 103Kg× 10N/Kg/(3m×2m)= 2.5× 105 pa

(3) If the bottom area of the foundation stone is S stone m2, the mass of the foundation stone is:

M stone = rho stone v stone =2× 103Kg /m3× 1m×S stone m2=2× 103S stone Kg.

The pressure of the cornerstone on the ground is:

P stone =F/S stone =(m+m stone) g/ S stone

1.2×105pa = (1.5×105kg+2×103s stone Kg)× 10N/Kg/S stone.

S stone = 15m2

Description: This question examines the ability to use stress knowledge to solve practical problems. To calculate pressure, we must master the definition of p = f/s. When we know the calculation of pressure P, we must first make clear the stress area s corresponding to pressure F.. In this problem, the pressure formula p=F/S is used twice. Although the formula is the same, the corresponding pressure and stress area are different. It can be seen that when using p=F/ S, we should not only pay attention to the corresponding relationship between p, f and s, but also pay attention to the unification of units into the international system of units.

[Example 2] The length-width-height ratio of a brick is 4:2: 1, and three identical bricks are stacked on a horizontal desktop as shown in Figure A,

The pressure of bricks on the desktop is 1.5× 103N/ m2. Now stacked as shown in Figure B, what is the pressure of bricks on the desktop?

Analysis: Both of them are placed on a horizontal desktop, so the pressure on the desktop is equal to the gravity of the brick, and the gravity of the brick remains unchanged, so the pressure on the desktop remains unchanged. The two placement methods only have different stress areas, so the pressure is inversely proportional to the stress area.

Answer: FA=FB=G

pASA= pBSB

pB = pASA/SB = pa×4×2/2× 1 = 4pA = 6× 103 pa

Note: To solve the three-dimensional problem, we should follow the order of "pressure first, then pressure", that is, we must first determine the pressure and then calculate it.

[Example 3] Two cubic blocks, A and B, are placed on a horizontal desktop, and they exert equal pressure on the desktop. The pressure ratio is 9: 4, so the density ratio of A and B is ().

A.2:3 B. 3:2 C. 1: 1 D. 4:9

Analysis: The pressure of the cube on the horizontal desktop is p=F/ S=G/S=ρgV/S=ρgSh/S=ρgh. From this formula, we can know that the pressure of a cube with constant cross-sectional area on the horizontal support surface has nothing to do with the volume, stress area and gravity of the cube, but only with the height and density of the object. Using this derived formula and the definition of pressure p=F/S, it can be solved smoothly.

Answer: Let the side length of A be A and the side length of B be A and B. ..

P A = ρ A ga A，P B = ρ B ga B，P A =P B。

ρ A ga A = ρ B ga B，ρ A/ρ B =a B/a A A

P A = F A/S A = G A A A A A 2，P B =F B /S B =G B /a B 2，P A =p B。

G A /a A A 2 =G B /a B 2

G A/G B = A A A a2/A B2 = 9/4

A A /a B =3/2

ρ A/ρ B =a B /a A A =2/3

The correct option for this problem is a.

Note: The applicable condition of p=ρgh derived from the definition of pressure p=F/S must be a solid with vertical side wall, constant cross-sectional area and uniform mass distribution. Therefore, the derived formula is not only applicable to cuboids, cubes and cylinders, but also to other cylinders with the same cross-sectional area. Mastering this feature is helpful to improve the speed of solving problems, especially when analyzing multiple-choice questions and fill-in-the-blank questions.

[Example 4] As shown in the figure, a frustum weighs 10N, has an upper surface area of S 1= 1dm2, bears a horizontal pressure of p 1= 102Pa, and the pressure acting on the wall is p2= 104Pa. If the frustum of a cone is placed vertically on a horizontal desktop, and the vertical downward pressure of 12N is applied to the upper surface, how much pressure will the desktop bear?

Analysis: Solid can transmit pressure, and the pressure acting on the frustum is equal to the pressure acting on the wall, so a small area can be obtained. When the frustum is placed vertically on a horizontal table, the total pressure of the frustum on the table is equal to the gravity of the frustum plus the second pressure.

Answer: Because F 1=F2,

So p 1S 1=p2S2, S2 = p1s1/p2 =102×1104 = 0.0/kloc-0.

When the frustum of a cone is vertically placed on a horizontal desktop:

P 1'= F total/S2 = (g+f)/S2 = (10+12)/10-4 = 2.2×105 (pa).

A: The desktop will be under the pressure of 2.2× 105Pa.

Note: When calculating solid problems, we often use the conclusion that "in the case of force balance, the magnitude and direction of solid energy remain unchanged", but students should pay special attention to the fact that solids cannot transmit pressure. For example, "the pressure on the upper and lower surfaces of a cone is different" is a typical example.

[Example 5] A cylindrical copper column with a bottom area of S 1 and a height of h 1 was forged into a hexagon nut with the original height of 2/5, and placed on a horizontal table, as shown in the figure. What is the pressure after that?

Analysis: the volume of the same metal workpiece is unchanged before and after forging into different shapes, so the ratio of cylinder bottom area before and after forging can be obtained by high change; Then, according to the conditions of constant mass, constant gravity and constant horizontal pressure before and after forging, the horizontal pressure before and after forging remains unchanged. From the proportional relationship, it is not difficult to find out the ratio of pressure on the horizontal plane before and after forging.

Answer: Because the volume is constant, V 1=V2, so: S 1h 1=S2h2, S 1/S2=h2/h 1=2/5.

Because G 1=G2, p2/p1= (F2/S2)/(f1/s1) = g2s1/g1S2 = 2/.

A: Later, the pressure was 2/5 times of the original pressure.

Note: To solve the problem of constant volume after shape transformation, we should grasp two points: first, from the condition of constant volume, find out the relationship between stress area before and after change; Second, because the volume, mass and gravity of the same substance remain unchanged, the pressure on the horizontal plane remains unchanged.

[Example 6] As shown in the figure, the area of surface A of a cuboid is 5m2 more than that of surface C, and the area of surface B is 10m2 more than that of surface C, and the pressure is 500Pa. Find the range of the maximum and minimum pressure values generated when the B surface is grounded.

Analysis: To find the pressure value of B when it landed, we first need to solve the gravity expression of the object, and then express and discuss the formula of the pressure value of B when it landed.

Answer: G=F= pASA=500×(5+SC)

pB = F/SB = 500×(5+SC)/( 10+SC)

=500SC +2500 /( 10+SC)

When SC=0, pB=250Pa.

When SC→∞, pB'=500Pa.

When a:B touches the ground, the pressure range is less than 500Pa and more than 250Pa.

Explanation: This problem is a comprehensive problem combining mathematical knowledge. The key to solve this problem is whether the mathematical knowledge can be correctly applied to find the extreme value, and the lateral deformation involved in this problem is often used in our physical problems. Students should pay attention to mastering this method.

Simulated test questions (answer time: 30 minutes)

I. Fill in the blanks:

1. Explain with physical knowledge: When sewing clothes, the needle is made very sharp for _ _ _ _ _ _ _ _, and the thimble is worn on the finger for _ _ _ _ _ _ _ _.

2. A brick weighing 30 newtons is pressed on a vertical wall with a force of 20 newtons, which is perpendicular to the wall, so the pressure acting on the wall is _ _ _ _ _ _ _ _ Newton.

3. Pressure is a physical quantity reflecting _ _ _ _ _ effect. In physics, the pressure on an object is called pressure.

4. The horizontal desktop has an area of 1 m2, and an object with a mass of 5 kg is placed on it. If the contact area between the object and the desktop is 0.65,438+0 square meters, the pressure of the object on the desktop is _ _ _ _ _ _ _ PA.

5. As shown in the figure, a wooden block weighing 30 Newton is pressed on a vertical blackboard with a horizontal force of 50 Newton, and the contact area between the wooden block and the blackboard is 0.0 1 m2, so the pressure acting on the blackboard is Newton and the pressure is Pa.

2. Multiple choice questions:

1. A brick is laid flat on the horizontal ground, and half of it is cut off along the horizontal direction, and the remaining half of the brick ().

A. the density is reduced by half. B. Double the density

C, reduce the ground pressure; Keep the ground pressure constant.

2. There are uneven patterns on the bottom of sports shoes for the sake of ()

A. Reduce pressure B. Increase friction

C. Reduce friction and increase pressure

When two objects touch each other, which of the following statements will definitely increase the pressure? ( )

A. increase the pressure and reduce the stress area.

B. reduce the pressure and increase the stress area.

C. At the same time, increase the pressure and stress area between them.

D. simultaneously reduce the pressure and stress area between them.

4. The pressure of a tractor on the ground is 3 104 Pa, which means that the exposed pressure per square meter of the ground is ().

A. note: the pressure is 3 104 pa.

C. the pressure is 3 104 pa. Gravity is 3 104 pa.

5. When a chopstick stands on a horizontal surface, the pressure acting on the desktop is F and P. When a pair of identical chopsticks are tied together and stand on a horizontal desktop, the pressure and pressure acting on the desktop are () respectively.

A.2P·2F·2F，P . c . F/

6. About stress, the following statement is true ()

A. If pressure is generated by gravity, then pressure must be equal to gravity.

B. If the pressure is generated by gravity, the pressure is not necessarily equal to gravity.

C. If the pressure is generated by gravity, the direction of pressure must be consistent with the direction of gravity.

D if the pressure is generated by gravity, the direction of the pressure may be different from that of gravity.

7. As shown in the figure, two cuboids A and B are stacked on the horizontal ground. A weight 10N, B weight 30N, and the ratio of the pressure of A to B to the pressure of B to the ground is 3:2, so the ratio of the bottom areas of A and B is ().

A.3:2 B. 2:3 C. 6: 1 D. 1:6

3. Calculation problems:

Two cuboids, A and B, are made of different materials, and their base areas are S A =40cm2 and S B =30cm2, respectively, and their heights are relatively high.

The ratio of H A: H B is 3: 2, and the density ratio ρ A: ρ B is 3:1. As shown in the picture, put A on the horizontal table and B on A..

In fact, the pressure of a horizontal desktop is 7000Pa. When B is taken out and placed on a horizontal desktop, what is the supporting force of the desktop for B?

Test answer

One.

1. Increase pressure and decrease pressure

2.20

3. Pressure, unit area and pressure

4.49、490

5.50、5000

Two.

1.D 2。 B 3。 A 4。 A 5。 B 6。 BD 7。 D

Three.

G A+G B = F = pS A = 7000×4× 10-3 = 28(N)

G A/G B =ρA gS A H A/ρB gS B H B =(3/ 1)×(40/30)×(3/2)= 6/ 1