The concept with which we are familiar from early childhood is mass. And yet, in a physics course, there are some difficulties associated with its study. Therefore, it is necessary to clearly define how it can be recognized? And why is it not equal to weight?
Determination of mass
The natural scientific meaning of this value is that it determines the amount of substance contained in the body. To denote it, it is customary to use the Latin letter m. The unit of measurement in the standard system is the kilogram. In tasks and Everyday life Non-systemic ones are also often used: gram and ton.
In a school physics course, the answer to the question: “What is mass?” given when studying the phenomenon of inertia. Then it is defined as the ability of a body to resist changes in the speed of its movement. Therefore, the mass is also called inert.
What is weight?
Firstly, this is force, that is, a vector. Mass is a scalar weight that is always attached to a support or suspension and is directed in the same direction as the force of gravity, that is, vertically downward.
The formula for calculating weight depends on whether the support (suspension) is moving. When the system is at rest, the following expression is used:
P = m * g, where P (in English sources the letter W is used) is body weight, g is acceleration free fall. For the earth, g is usually taken equal to 9.8 m/s 2.
From this the mass formula can be derived: m = P / g.
When moving downwards, that is, in the direction of the weight, its value decreases. Therefore the formula takes the form:
P = m (g - a). Here “a” is the acceleration of the system.
That is, if these two accelerations are equal, a state of weightlessness is observed when the weight of the body is zero.
When the body begins to move upward, we speak of weight gain. In this situation, an overload condition occurs. Because body weight increases, and its formula will look like this:
P = m (g + a).
How is mass related to density?
Solution. 800 kg/m3. In order to take advantage of already well-known formula, you need to know the volume of the spot. It is easy to calculate if you take the spot as a cylinder. Then the volume formula will be:
V = π * r 2 * h.
Moreover, r is the radius, and h is the height of the cylinder. Then the volume will be equal to 668794.88 m 3. Now you can count the mass. It will turn out like this: 535034904 kg.
Answer: the mass of oil is approximately 535036 tons.
Task No. 5. Condition: The length of the longest telephone cable is 15151 km. What is the mass of copper that went into its manufacture if the cross-section of the wires is 7.3 cm 2?
Solution. The density of copper is 8900 kg/m3. The volume is found using a formula that contains the product of the area of the base and the height (here the length of the cable) of the cylinder. But first you need to convert this area to square meters. That is, divide this number by 10,000. After calculations, it turns out that the volume of the entire cable is approximately equal to 11,000 m 3.
Now you need to multiply the density and volume values to find out what the mass is equal to. The result is the number 97900000 kg.
Answer: the mass of copper is 97900 tons.
Another problem related to mass
Task No. 6. Condition: The largest candle, weighing 89867 kg, had a diameter of 2.59 m. What was its height?
Solution. Wax density is 700 kg/m3. The height will need to be found from That is, V needs to be divided by the product of π and the square of the radius.
And the volume itself is calculated by mass and density. It turns out to be equal to 128.38 m 3. The height was 24.38 m.
Answer: the height of the candle is 24.38 m.
I regularly come across the fact that people do not understand the difference between weight and mass. This is generally understandable, since we spend our entire lives in the Earth’s incessant gravitational field, and these quantities are constantly connected for us. And this connection is also linguistically reinforced by the fact that we find out mass with the help of scales, “weigh” ourselves or, say, food in a store.
But let's still try to untangle these concepts.
In subtlety (such as a different g in different places Land and other things) we won’t go into. I note that all this is included in school course physics, so if all of the following is obvious to you, do not swear at those who did not manage to understand these things, and at the same time at those who decided to explain this for the hundredth time.) I hope that there will be people for whom this note will add to their apparatus of understanding the surrounding world.
So, let's go. The mass of a body is a measure of its inertia. That is, a measure of how difficult it is to change the speed of this body in magnitude (accelerate or decelerate) or in direction. In the SI system it is measured in kilograms (kg). Usually denoted by the letter m. It is an unchangeable parameter, whether on Earth or in space.
Gravity is measured in SI units in Newtons (N). This is the force with which the Earth attracts a body, and is equal to the product m*g. The coefficient g is 10 m/s2, called the acceleration of gravity. With this acceleration, a body begins to move relative to the earth's surface, deprived of support (in particular, if the body started from a stationary state, its speed will increase by 10 m/s every second).
Now consider a body of mass m lying motionless on a table. To be specific, let the mass be 1 kg. This body is acted vertically downward by the force of gravity mg (the vertical itself is determined precisely by the direction of the force of gravity), equal to 10 N. V technical system Units of this force are called kilogram-force (kgf).
The table does not allow our body to accelerate, acting on it with a force N directed vertically upward (it is more correct to draw this force from the table, but so that the lines do not overlap, I will also draw from the center of the body):
N is called the support reaction force, balances the force of gravity (in this case equal in magnitude to the same 10 Newtons), so that the resultant force F (the sum of all forces) is equal to zero: F = mg - N = 0.
And we see that the forces are balanced from Newton’s second law F = m*a, according to which if the acceleration of body a is zero (that is, it is either at rest, as in our case, or moves uniformly and rectilinearly), then the resultant force F is also zero.
Now we can finally say what weight is - this is the force with which the body acts on a stand or suspension. According to Newton's third law, this force is opposite to the force N and is equal to it in absolute value. That is, in this case it is the same 10 N = 1 kgf. It may seem to you that all this is unnecessarily complicated, and you should have said right away that weight and gravity are the same thing? After all, they coincide both in direction and in magnitude.
No, in fact they differ significantly. The force of gravity acts constantly. Weight changes depending on the acceleration of the body. Let's give examples.
1. You start up on a high-speed elevator (high-speed so that the acceleration phase is more impressive/more noticeable). Your mass is, say, 70 kg (you can recalculate all the numbers below for your mass). Your weight in a stationary elevator (before the start) is 700 N (or 70 kgf). At the moment of upward acceleration, the resulting force F is directed upward (it is what accelerates you), the reaction force N exceeds the force of gravity mg, and since your weight (the force with which you act on the floor of the elevator) coincides in absolute value with N, you experience the so-called overload. If the elevator accelerated with acceleration g, then you would experience a weight of 140 kgf, that is, a g-force of 2g, 2 times the resting weight. In fact, in normal operation such overloads do not occur in elevators; acceleration usually does not exceed 1 m/s2, which leads to an overload of only 1.1g. The weight in our case will be 77 kgf. When the elevator has accelerated to the required speed, the acceleration is zero, and the weight returns to the initial 70 kgf. When decelerating, the weight, on the contrary, decreases, and if the acceleration in absolute value is 1 m/s2, then the overload will be 0.9g. When moving in the opposite direction (down), the situation is reversed: when accelerating, the weight decreases, in a uniform section the weight is restored, and when decelerating, the weight increases.
2. You are running and your resting weight is still 70 kgf. At the moment of running, when you push off from the ground, your weight exceeds 70 kgf. And while you are flying (one leg has left the ground, the other has not yet touched), your weight is zero (since you are not influencing either the stand or the gimbal). This is weightlessness. True, it’s quite short. Thus, running is an alternation of overload and weightlessness.
Let me remind you that the force of gravity in all these examples did not go away, did not change, and was your “hard-earned” 70 kgf = 700 N.
Now let’s significantly lengthen the weightlessness phase: imagine that you are on the ISS (international space station). At the same time, we have not eliminated the force of gravity - it still acts on you - but since both you and the station are in the same orbital motion, you are weightless relative to the ISS. You can imagine yourself anywhere in outer space, just the ISS is a little more realistic.)
How will your interaction with objects be? Your mass is 70 kg, you take an object weighing 1 kg in your hand and throw it away from you. In accordance with the law of conservation of momentum, the main speed will be received by a 1-kg object, as it is less massive, and the throw will be approximately as “light” as on Earth. But if you try to push off from an object weighing 1000 kg, then you will actually push yourself away from it, since in this case you will receive the main speed yourself, and to accelerate your 70 kg you will have to develop more force. To roughly imagine what it’s like, you can now go up to the wall and push off from it with your hands.
Now you have left the station into outer space and want to manipulate some massive object. Let its mass be five tons.
To be honest, I would be very careful about handling a five-ton object. Yes, weightlessness and all that. But only its small speed relative to the ISS is enough to press your finger or something more serious. These five tons are difficult to move: to accelerate, to stop.
And I don’t even want to imagine, as one person suggested, between two objects weighing 100 tons. The slightest oncoming movement from them, and they will easily crush you. In complete, characteristically, weightlessness.)
And finally. If you are happily flying around the ISS and hit a wall/bulkhead, then it will hurt you exactly the same as if you were running at the same speed and hit a wall/jamb in your apartment. Because the impact reduces your speed (that is, it gives you a negative acceleration), and your mass is the same in both cases. This means that according to Newton’s second law, the force of influence will be proportionate.
I am glad that in films about space ("Gravity", "Interstellar", the TV series "The Expanse") more and more realistically (albeit not without flaws like George Clooney hopelessly flying away from Sandra Bullock) they display the basic things described in this post.
Let me summarize. Mass is "inalienable" from the object. If an object is difficult to accelerate on Earth (especially if you tried to minimize friction), then it is just as difficult to accelerate it in space. As for the scales, when you stand on them, they simply measure the force with which they are compressed, and for convenience, display this force not in Newtons, but in kgf. Without adding the letter “s”, so as not to confuse you.)
Quite a lot of mistakes and non-random slips by students are related to the strength of the weight. The phrase “power of weight” itself is not very familiar, because we (teachers, authors of textbooks and problem books, methodological manuals and reference books) are more accustomed to saying and writing “body weight”. Thus, the phrase itself takes us away from the concept that weight is force, and leads to the fact that body weight is confused with body weight (in the store we often hear people asking to weigh several kilograms of a product). The second common mistake students make is confusing the force of weight with the force of gravity. Let's try to understand the force of weight at the level of a school textbook.
First, let's look at the reference literature and try to understand the authors' point of view on this question. Yavorsky B.M., Detlaf A.A. (1) in the reference book for engineers and students, the weight of a body is the force with which this body acts due to gravity towards the Earth on the support (or suspension) that holds the body from free fall. If the body and the support are motionless relative to the Earth, then the weight of the body is equal to its gravity. Let's ask a few naive questions about the definition:
1. What reporting system are we talking about?
2. Is there one support (or suspension) or several (supports and suspensions)?
3. If a body gravitates not towards the Earth, but, for example, towards the Sun, will it have weight?
4. If the body is in spaceship, moving with acceleration, “almost” does not gravitate toward anything in observable space, will it have weight?
5. How is the support located relative to the horizon, is the suspension vertical for the case of equality of body weight and gravity?
6. If a body moves uniformly and rectilinearly along with a support relative to the Earth, then the weight of the body is equal to its gravity?
In the reference guide to physics for those entering universities and self-education by B.M. Yavorsky. and Selezneva Yu.A. (2) give an explanation of the last naive question, leaving the first ones unattended.
Koshkin N.I. and Shirkevich M.G. (3) it is proposed to consider body weight as a vector physical quantity, which can be found by the formula:
The examples below will show that this formula works in cases where no other forces act on the body.
Kuhling H. (4) does not introduce the concept of weight as such at all, identifying it practically with the force of gravity; in the drawings the weight force is applied to the body, and not to the support.
In the popular “Physics Tutor” by I.L. Kasatkina. (5) the weight of a body is defined as the force with which the body acts on a support or suspension due to attraction to the planet. In the following explanations and examples given by the author, answers are given only to the 3rd and 6th of the naive questions.
Most physics textbooks give definitions of weight that are more or less similar to the definitions of the authors (1), (2), (5). When studying physics in the 7th and 9th grades, this may be justified. In 10th specialized classes with such a definition, when solving a whole class of problems, one cannot avoid various kinds of naive questions (in general, there is no need to strive to avoid any questions at all).
Authors Kamenetsky S.E., Orekhov V.P. in (6) distinguishing and explaining the concepts of gravity and body weight, they write that body weight is a force that acts on a support or suspension. That's all. There is no need to read between the lines. True, I still want to ask, how many supports and suspensions, and can a body have both support and suspension at once?
And finally, let's look at the definition of body weight given by V.A. Kasyanov. (7) in the 10th grade physics textbook: “the weight of a body is the total elastic force of the body, acting in the presence of gravity on all connections (supports, suspensions).” If we remember that the force of gravity is equal to the resultant of two forces: the force of gravitational attraction to the planet and the centrifugal force of inertia, provided that this planet rotates around its axis, or some other inertial force associated with the accelerated movement of this planet, then One could agree with this definition. Since no one is stopping us from imagining a situation where one of the components of gravity will be negligible, for example, the case of a spaceship in deep space. And even with these reservations, it’s tempting to remove the mandatory presence of gravity from the definition, because situations are possible when there are other inertial forces not associated with the movement of the planet or Coulomb forces of interaction with other bodies, for example. Or agree with the introduction of a certain “equivalent” force of gravity in non-inertial reporting systems and give a definition of weight force for the case when there is no interaction of the body with other bodies, except for the body creating gravitational attraction, supports and suspensions.
And yet, let’s decide when the weight of a body is equal to the force of gravity in inertial reporting systems?
Let's assume we have one support or one suspension. Is it sufficient that the support or suspension is motionless relative to the Earth (we consider the Earth inertial system report), or do they move uniformly and in a straight line? Let's take a fixed support located at an angle to the horizontal. If the support is smooth, then the body slides along an inclined plane, i.e. does not rest on a support and is not in free fall. And if the support is rough enough that the body is at rest, then either the inclined plane is not a support, or the weight of the body is not equal to the force of gravity (you can, of course, go further and question that the weight of the body is not equal in magnitude and not opposite in direction ground reaction force, and then there will be nothing to talk about at all). If we consider the inclined plane to be a support, and the sentence in parentheses to be irony, then, solving the equation for Newton’s second law, which for this case will also be the condition for the equilibrium of a body on an inclined plane, written in projections onto the Y axis, we will obtain the expression for weight other than gravity:
So, in this case, it is not enough to say that the weight of a body is equal to the force of gravity when the body and the support are motionless relative to the Earth.
Let us give an example with a suspension and a body on it that are motionless relative to the Earth. A positively charged metal ball on a thread is placed in a homogeneous electric field so that the thread makes a certain angle with the vertical. Let us find the weight of the ball from the condition that the vector sum of all forces is equal to zero for a body at rest.
As we see, in the above cases, the weight of the body is not equal to the force of gravity when the condition of immobility of the support, suspension and body relative to the Earth is met. The peculiarities of the above cases are the existence of the friction force and the Coulomb force, respectively, the presence of which actually leads to the fact that the bodies are kept from moving. For vertical suspension and horizontal support, additional forces are not needed to keep the body from moving. Thus, to the condition of immobility of the support, suspension and body relative to the Earth, we could add that the support is horizontal and the suspension is vertical.
But would this addition solve our question? Indeed, in systems with vertical suspension and horizontal support, forces can act that reduce or increase the weight of the body. These could be the Archimedes force, for example, or the Coulomb force directed vertically. Let's summarize for one support or one suspension: the weight of a body is equal to the force of gravity when the body and the support (or suspension) are at rest (or uniformly and linearly moving) relative to the Earth, and only the reaction force of the support (or the elastic force of the suspension) and the force act on the body gravity. The absence of other forces, in turn, assumes that the support is horizontal and the suspension is vertical.
Let us consider cases when a body with several supports and/or suspensions is at rest (or moves uniformly and rectilinearly with them relative to the Earth) and no other forces act on it except the reaction forces of the support, the elastic forces of the suspensions, and attraction to the Earth. Using the definition of weight force by Kasyanov V.A. (7), we will find the total elastic force of the body connections in the first and second cases presented in the figures. Geometric sum of elastic forces of bonds F, in modulus equal to the weight of the body, based on the equilibrium condition, is really equal to the force of gravity and opposite to it in direction, and the angles of inclination of the planes to the horizon and the angles of deviation of the suspensions from the vertical do not affect the final result.
Let's consider an example (figure below), when in a system stationary relative to the Earth the body has a support and suspension and no other forces act in the system except the forces of elastic connections. The result is similar to the above. The weight of the body is equal to the force of gravity.
So, if a body is on several supports and (or) suspensions, and is at rest with them (or moves uniformly and rectilinearly) relative to the Earth, in the absence of other forces except gravity and the forces of elasticity of connections, its weight is equal to the force of gravity. In this case, the location of supports and hangers in space and their number do not affect the final result.
Let's consider examples of finding body weight in non-inertial reporting systems.
Example 1. Find the weight of a body of mass m moving in a spaceship with acceleration A in “empty” space (so far away from other massive bodies that their gravity can be neglected).
In this case, two forces act on the body: the inertial force and the support reaction force. If the acceleration in magnitude is equal to the acceleration of gravity on Earth, then the weight of the body will be equal to the force of gravity on Earth, and the bow of the ship will be perceived by astronauts as the ceiling, and the tail as the floor.
The artificial gravity created in this way for the astronauts inside the ship will be no different from the “real” earthly one.
IN in this example We neglect, due to its smallness, the gravitational component of gravity. Then the force of inertia on the spaceship will be equal to the force of gravity. In view of this, we can agree that the cause of body weight in this case is gravity.
Let's return to Earth.
Example 2.
Relative to the ground with acceleration A A cart is moving, on which a body is attached to a thread of mass m, deflected at an angle from the vertical. Find the weight of the body, neglect air resistance.
The problem is with one suspension, therefore, the weight is equal in magnitude to the elastic force of the thread.
Thus, you can use any formula to calculate the elastic force, and, therefore, the weight of the body (if the air resistance force is sufficiently large, then it will need to be taken into account as a term to the inertia force).
Let's work with the formula some more
Therefore, by introducing the “equivalent” force of gravity, we can state that in this case the weight of the body is equal to the “equivalent” force of gravity. And finally we can give three formulas for its calculation:
Example 3.
Find the weight of a racing driver of mass m in a moving vehicle with acceleration A car.
At high accelerations, the reaction force of the seat back support becomes significant, and we will take it into account in this example. The total elastic force of the connections will be equal to the geometric sum of both support reaction forces, which in turn is equal in magnitude and opposite in direction to the vector sum of the forces of inertia and gravity. For this problem, we find the module of the weight force using the formulas:
The effective acceleration of gravity is found as in the previous problem.
Example 4.
A ball on a string of mass m is attached to a rotating one with a constant angular velocityω platform at a distance r from its center. Find the weight of the ball.
Finding the body weight in non-inertial reporting systems in the given examples shows how well the formula for body weight proposed by the authors in (3) works. Let's complicate the situation a little in example 4. Let's assume that the ball is electrically charged, and the platform, together with its contents, is in a uniform vertical electric field. What is the weight of the ball? Depending on the direction of the Coulomb force, the weight of the body will decrease or increase:
It so happened that the question of weight naturally came down to the question of gravity. If we define the force of gravity as the resultant of the forces of gravitational attraction to a planet (or to any other massive object) and inertia, taking into account the principle of equivalence, leaving the origin of the force of inertia itself in the fog, then both components of gravity, or one of them, at least will cause body weight. If in the system, along with the force of gravitational attraction, the force of inertia and the forces of elastic connections, there are other interactions, then they can increase or decrease the weight of the body, leading to a state where the weight of the body becomes equal to zero. And these other interactions can cause weight gain in some cases. Let's charge a ball on a thin non-conducting thread in a spaceship moving uniformly and rectilinearly in distant “empty” space (we will neglect gravitational forces due to their smallness). Let's place the ball in an electric field, the thread will stretch, and weight will appear.
Summarizing what has been said, we conclude that the weight of a body is equal to the force of gravity (or the equivalent force of gravity) in any system where no other forces act on the body except the forces of gravity, inertia and elasticity of connections. Gravity or "equivalent" gravity is most often the cause of weight force. The force of weight and the force of gravity have different natures and are applied to different bodies.
Bibliography.
1. Yavorsky B.M., Detlaf A.A. Handbook of physics for engineers and university students, M., Nauka, 1974, 944 p.
2. Yavorsky B.M., Selezneva Yu.A. Physics Reference Guide for
entering universities and self-education., M., Nauka, 1984, 383 p.
3. Koshkin N.I., Shirkevich M.G. Handbook of elementary physics., M., Nauka, 1980, 208 p.
4. Kuhling H. Handbook of physics., M., Mir, 1983, 520 p.
5. Kasatkina I.L. Physics tutor. Theory. Mechanics. Molecular physics. Thermodynamics. Electromagnetism. Rostov-on-Don, Phoenix, 2003, 608 p.
6. Kamenetsky S.E., Orekhov V.P. Methods for solving problems in physics in high school., M., Education, 1987, 336 p.
7. Kasyanov V.A. Physics. 10th grade, M., Bustard, 2002, 416 p.
Exploring the differences between weight and body weight Newton did. He reasoned like this: we know very well that different substances taken in equal volumes weigh differently.
Weight
Newton called the amount of substance contained in a particular object mass.
Weight- something common that is inherent in all objects without exception - it doesn’t matter whether they are shards from the old clay pot or gold watch.
For example, a piece of gold is more than twice as heavy as an identical piece of copper. Probably, particles of gold, Newton suggested, are capable of packing more densely than particles of copper, and more substance fits in gold than in a piece of copper of the same size.
Modern scientists have established that the different densities of substances are explained not only by the fact that the particles of the substance are packed more densely. The smallest particles themselves - atoms - differ in weight from each other: gold atoms are heavier than copper atoms.
Whether any object lies motionless, or freely falls to the ground, or swings, suspended on a thread, its the mass remains unchanged under all conditions.
When we want to find out how large the mass of an object is, we weigh it on ordinary commercial or laboratory scales with cups and weights. We place an object on one pan of the scale, and weights on the other, and thus compare the mass of the object with the mass of the weights. Therefore, commercial and laboratory scales can be transported anywhere: to the pole and to the equator, to the top high mountain and into a deep mine. Everywhere and everywhere, even on other planets, these scales will show correctly, because with their help we determine not weight, but mass.
It can be measured at different points on the earth using spring scales. By attaching an object to the hook of a spring scale, we compare the force of gravity of the Earth that this object experiences with the elastic force of the spring. The force of gravity pulls down, (more details:) the force of the spring pulls up, and when both forces are balanced, the scale pointer stops at a certain division.
Spring scales are only correct at the latitude where they are made. At all other latitudes, at the pole and at the equator, they will show different weights. True, the difference is small, but it will still be revealed, because the force of gravity on Earth is not the same everywhere, and the elastic force of the spring, of course, remains constant.
On other planets this difference will be significant and noticeable. On the Moon, for example, an object that weighed 1 kilogram on Earth will weigh 161 grams on spring scales brought from Earth, on Mars - 380 grams, and on huge Jupiter - 2640 grams.
How more mass planet, the greater the force with which it attracts a body suspended on spring scales.
That is why a body weighs so much on Jupiter and so little on the Moon.
We often use phrases like: “A pack of sweets weighs 250 grams” or “I weigh 52 kilograms.” The use of such offers is automatic. But what is weight? What does it consist of and how to calculate it?
First you need to understand that it is wrong to say: “This object weighs X kilogram.” In physics there is two different concepts - mass and weight. Mass is measured in kilograms, grams, tones, etc., and body weight is calculated in newtons. So when we say, for example, that we weigh 52 kilograms, we actually mean mass, not weight.
Weight – it is a measure of the body's inertia. The more inert a body is, the longer it will take to give it speed. Roughly speaking, the higher the mass value, the harder it is to move an object. In the International System of Units, mass is measured in kilograms. But it is also measured in other units, for example;
- ounce;
- lb;
- stone;
- US ton;
- English ton;
- gram;
- milligram and so on.
When we say one, two, three kilograms, we compare the mass with a reference mass (the prototype of which is in France in the BIPM). Mass is denoted by m.
Weight – this is the force that acts on the suspension or support due to an object attracted by gravity. It is a vector quantity, which means it has a direction (like all forces), unlike mass (a scalar quantity). The direction always goes to the center of the Earth (due to gravity). For example, if we are sitting on a chair whose seat is parallel to the Earth, then the force vector is directed straight down. Weight is designated P and calculated in newtons [N].
If the body is in motion or at rest, then the force of gravity (Fgravity) acting on the body is equal to the weight. This is true if the motion occurs along a straight line relative to the Earth, and it has a constant speed. Weight acts on the support, and gravity acts on the body itself (which is located on the support). This different sizes, and regardless of the fact that they are equal in most cases, they should not be confused.
Gravity- this is the result of the body’s attraction to the ground, weight is the effect of the body on the support. Since the body bends (deforms) the support with its weight, another force arises, it is called the elastic force (Fel). Newton's third law states that bodies interact with each other with forces of the same magnitude, but different in vector. It follows from this that for the elastic force there must be an opposite force, and this is called the support reaction force and is denoted N.
Modulo |N|=|P|. But since these forces are multidirectional, then, opening the module, we get N = - P. That is why weight can be measured with a dynamometer, which consists of a spring and a scale. If you hang a load on this device, the spring will stretch to a certain mark on the scale.
How to measure body weight
Newton's second law states that acceleration is equal to force divided by mass. Thus, F=m*a. Since Ft is equal to P (if the body is at rest or moving in a straight line (relative to the Earth) with the same speed), then P of the body will be equal to the product of mass and acceleration (P=m*a).
We know how to find mass, and we know what the weight of a body is, all that remains is to figure out the acceleration. Acceleration is a physical vector quantity that denotes the change in the speed of a body per unit time. For example, an object moves for the first second at a speed of 4 m/s, and in the second second its speed increases to 8 m/s, which means its acceleration is equal to 2. According to the international system of units, acceleration is calculated in meters per second squared [m/s 2 ].
If you place a body in a special environment where there is no air resistance force - a vacuum, and remove the support, the object will begin to fly at uniform acceleration. The name of this phenomenon is acceleration of gravity, which is denoted by g and is calculated in meters per second squared [m/s 2 ].
It is interesting that acceleration does not depend on the mass of the body, which means that if we throw a piece of paper and a weight on Earth under special conditions in which there is no air (vacuum), then these objects will land at the same time. Since the leaf has large area surface and a relatively small mass, then in order to fall, it has to face great air resistance . This doesn't happen in a vacuum., and therefore a pen, a piece of paper, a weight, a cannonball and other objects will fly at the same speed and fall at the same time (provided that they start flying at the same time and their initial speed is zero ).
Since the Earth has the shape of a geoid (or otherwise an ellipsoid), and not an ideal ball, then the acceleration of free fall in different areas Lands are different. For example, at the equator it is 9.832 m/s 2 , and at the poles 9.780 m/s 2 . This happens because in some parts of the Earth the distance to the core is greater, and in others less. The closer an object is to the center, the more strongly it is attracted. The farther the object is, the less gravity there is. Usually, at school this value is rounded to 10, this is done for convenience of calculations. If it is necessary to measure more accurately (in engineering or military affairs, and so on), then specific values are taken.
Thus, the formula for calculating body weight will look like this: P=m*g.
Examples of problems for calculating body weight
First task. A load weighing 2 kilograms is placed on the table. What is the weight of the cargo?
To solve this problem we need a formula for calculating the weight P=m*g. We know the mass of the body, and the acceleration due to gravity is approximately 9.8 m/s 2 . We substitute this data into the formula and get P=2*9.8=19.6 N. Answer: 19.6 N.
Second task. A paraffin ball with a volume of 0.1 m 3 was placed on the table. What is the weight of the ball?
This problem must be solved in the following sequence;
- First, we need to remember the weight formula P=m*g. We know the acceleration - 9.8 m/s 2 . All that remains is to find the mass.
- Mass is calculated using the formula m=p*V, where p is density and V is volume. The density of paraffin can be seen in the table; we know the volume.
- It is necessary to substitute the values into the formula to find the mass. m=900*0.1=90 kg.
- Now we substitute the values into the first formula to find the weight. P=90*9.9=882 N.
Answer: 882 N.
Video
This video lesson covers the topic of gravity and body weight.
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