Factors influencing the deformation of plastic objects when they fall freely are added

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After reading your question, I feel that first of all, this is a problem of the conversion of mechanical energy, so I think the first two conclusions must be correct, because the expression of gravitational potential energy is mgh, and you consider the two factors that increase the gravitational potential energy, which will inevitably lead to a greater velocity before grounding, resulting in greater deformation, so these two factors must be correct.

And then the third aspect you raised, if I'm not mistaken, should be to focus on the size of the air resistance during the fall, the resistance is large, the initial velocity when it falls to the ground will be small, so it is not good for causing deformation, and I understand that this is what it means. For this question, I think you should also keep the overall thinking when you put forward the first two factors, because the expression of air resistance is f=1 2 (p*c*a*u 2), p is the density of the air, a is the positive windward area you said, c is the air resistance coefficient, and u can be regarded as the falling speed here. It can be seen that this resistance is not only a function of the windward area, but also related to the air resistance coefficient, so you can't separate the connection of these factors, the two windward areas are the same, but one is an iron ball, and the other is the speed before grounding caused by the iron plate falling from a high altitude.

So I think we also need to consider the influence of the whole air resistance, and the windward area is only one of the factors.

The contact area mentioned in the information is the contact area between the ground and the ground after falling on the ground, and you refer to the contact area with the air during the fall process, which is the windward area mentioned above.

No, when the free fall motion is introduced, it has nothing to do with the gravity of its falling height and its own gravity, this question should be from the perspective of energy, the problem is that the mgh gravitational potential energy at different heights is different So the kinetic energy converted before landing is different, in the analysis of the formula of impulse, the greater the velocity, of course, the greater the deformation, and the same mass also affects the potential energy.

The 3rd positive windward area has no effect when the air resistance is ignored, if there is air resistance, the larger the windward area The greater the air resistance, the greater the heat generated by overcoming the air resistance work, and the more mechanical energy consumed, which affects the speed before landing and affects the size of the deformation.

It is related to the area of the face perpendicular to the air. When considering the free fall motion of an object, it has nothing to do with its falling height and the gravitational force it is subjected to, because for the free fall motion of any object, its velocity is certain, so the only factor affecting the deformation of the object itself is related to its surface area in contact with the ground.

I read the answers of the heroes above.

I hope my answer is the simplest and most understandable.

Consider your question for now. I think the main reason for what you call the size of the shape is the maximum pressure that the object is subjected to in the process.

So your three observations are easy to explain.

1.The greater the height of the object when other conditions remain unchanged (including the mass of the object and the size of the contact area), the greater the velocity of the object when it lands, and the greater the impulse. The greater the force, the greater the pressure, and the more powerful the shape.

ft=mvp=f/s

V is big, F is big, P is big. The size becomes larger.

2.All other things being equal, the greater the mass (including the height of the object and the size of the contact area), the greater the impulse of the object when it lands. The greater the force, the greater the pressure, and the greater the deformation.

ft=mvp=f/s

M is large, F is large, P is large, and the type becomes larger.

3.When other conditions remain unchanged (including the mass of the object, including the height of the object), the impulse does not change, and the force does not change. The area becomes smaller, the pressure becomes larger, and the deformation is more severe.

ft=mvp=f/s

m does not change, v does not change, f does not change, but s becomes smaller. p becomes larger. The type becomes larger.

The larger the contact area, the smaller the impulse of the object when it hits the ground, and the less force is generated, so the deformation is smaller.

I think the contact area should be changed to the frontal contact area, because when the raindrops fall, the contact area of the lower end is the largest, and the shape variable is not large. In addition, the falling process of the object is a process of reaching equilibrium, and if the equilibrium is reached, even if the height is increased, the deformation will not change greatly. So it has a lot to do with the height.

Also relevant is the magnitude of air resistance (related to the proportion of air components).

In general, it is because of external forces. Sometimes it's an intermolecular force.

The effect of force has two aspects: one is to change the state of motion of the object; The second is to deform the object (contact force).

When an object is subjected to a balanced force, the state of motion does not change, but the object may be deformed by one or both of these contact forces. For example, an object placed on a level ground is subjected to gravity and a supporting force that causes the object to deform.

Yes. The equilibrium force cannot change the state of motion of an object, but it can change the shape of an object.

Imagine that you are standing on a railroad and there are trains on both sides at the same time squeezing you with the same force, will you deform?

No, the equilibrium force can only make the object move in a straight line at a stationary or uniform speed.

Again, no, pressing the balloon by hand is not a balancing force, the force received by the balloon is the pressure of the hand on the balloon, and the elastic force of the balloonist, and does not act on the same object.

So it's not a balance force.

Of course. Press the balloon with your hands.

It's balance.

Isn't the balloon deformed?

Conditions under which pressure is generated: objects are in contact with each other and deformed;

In-depth understanding: 1) the two conditions for generating pressure (contact, deformation) must exist at the same time, and one is indispensable;

2) Pressure is generated by mutual extrusion (deformation) between two objects in contact with each other, and its existence is closely related to deformation;

3) If the objects do not touch each other, there will be no pressure; However, if the objects are in contact with each other, if they do not deform, there is no pressure. As shown in the figure, in Figure A, ball A is placed on the horizontal plane, and it is impossible to generate pressure on the contact surface Mn (perpendicular to the horizontal plane); In Figure B, A hangs vertically and will not squeeze each other with the contact surface Mn, and only A in Figure C will exert pressure on the inclined plane.

I hope it helps you, and if you have any questions, you can ask them

For example, two wooden blocks are in contact with each other, and the upper wooden blocks are suspended by ropes, although they are in contact with the wooden blocks below, they are not squeezed from each other, that is, they happen to be in contact, and there is no deformation, at this time, there is no pressure between the two wooden blocks. That is, the lower wooden block is not subjected to the pressure of the upper wooden block. This is only contact, no deformation, all without pressure.

Another example is two magnets, the same sex repels each other, and there is no need to touch, there is repulsion. This is deformed, but not in contact, all without pressure.

Only when two objects are in contact and deformed can there be pressure.

The magnitude of elastic force is related to the size of the deformation of the object, in addition, the elastic force is also related to the type of object that produces elastic force (for example, the elastic force of rubber bands and springs is different in the case of the same deformation), and it also depends on whether the object exceeds the elastic deformation, and the elastic force is proportional to the size of the deformation when it does not exceed the elastic deformation (Hooke's law).

Depending on the material of the object, the greater the degree of deformation of the same object, the greater the elastic force (within the elastic limit of the object), and vice versa.

Thank you for adopting

The main factor affecting the deformation of parts when machining - the rigidity of the process system (composed of machine tools, tools, fixtures, machined parts, etc.); the magnitude of the cutting force; Processing temperature, temperature difference, etc.

FYI.

I think we should start with four aspects: 1: material.

The processing stress of different materials is different, such as 45 steel and SUS303 of the same workpiece of different materials, the processing difficulty is much worse. 2: Structural characteristics of the workpiece, 3:

Selection of machining process (passing, cutting tool,). 4: The cutting stress generated by the machine tool is different depending on the machine tool selected for processing.

The total distortion rate is generally denoted as THR (Total Distortion Ratio).

Refers to the ratio of the root mean square of the total distortion content to the root mean value of the fundamental wave.

The total distortion factor is generally denoted as TDF (Total Distortion Factor), which is the ratio of the root mean square value of the total distortion content to the root mean square value of the voltage or current.

The numerator is the same, but the difference is that the root mean square of the voltage or current is greater than or equal to the root mean square of its fundamental, so the total distortion rate is greater than or equal to the total distortion factor.

The condition that ensures the integrity and continuity of the object after deformation is called the deformation coordination condition.

The coordination condition is the condition that ensures that the integrity and continuity of the continuous solid remain after deformation. These conditions can be used by six strain components at points within the solid (i.e., in Cartesian Cartesian coordinate systems.

three stretching strains and three shear strains) are expressed in six second-order partial differential equations.

Introduction to coordinationAfter the object is deformed, it must still maintain its integrity and continuity, i.e. the coordination of the deformation. From a mathematical point of view, the displacement function is required to be in its defined domain.

The inner is a single-valued continuous function.

Phenomena that disrupt the integrity and continuity of objects: the phenomenon of "tearing", the phenomenon of "nesting".

The compatibility condition is the condition that ensures that a continuous solid remains a continuum after deformation. These conditions can be expressed by six second-order partial differential equations satisfied by the six strain components at each point in the solid (i.e., three telescopic strains and three shear strains in Cartesian Cartesian coordinate systems). Coordination conditions can be used to test where there is cracking or embedding when the rock is deformed.

Geomechanics regards the coordination condition as one of the factors controlling the stress mode in the block or rock block, and it is necessary to consider the coordination condition when studying the stress activity mode in the process of various tectonic systems and inferring the boundary conditions of the area where the tectonic system is spreading. [1]

1. The shape of the foundation of the building.

2. The depth of embedding of the foundation.

3. The moisture content of the foundation soil.

4. The inner friction angle of the foundation soil.

5 types of foundation soil.