The third year of mechanics urgently seeks advice, and the third year of mechanics problems

A right. The most important feature of the balance of forces is that it acts on the same object.

So A is right, B is wrong, C is wrong.

Looking at d again, it should be the balance between Xiaohong's pressure on the scale, the gravity of the scale, and the support force on the scale.

a, the balance of two forces.

b, Relative force. The force bearer is different, so it is not a two-force balance.

CD, the same as B.

Choose a two forces balance remember:

Equal size, directional, collinear, homogeneous.

a Yes, because the forces are balanced, the two forces are balanced.

b is false, because this is the interaction force between two objects, although the magnitude is equal and the direction is opposite, it cannot be said to be balanced.

c is false, because it acts on two objects, and the size is not necessarily equal.

D is false, because it acts on two objects, and the size must not be equal (the scale also has its own weight).

a Because the so-called equilibrium force refers to two or two sets of forces acting on the same object, causing the object to be in a state of rest or uniform motion.

The pressure in b is not on the same object as the support force.

The force in c is also not on the same object.

In d, the resultant force of the gravity of the pressure and the scale is the supporting force that is symmetrical with the ground as balanced with each other.

So, the answer is A

The answer is b....In this question....The pressure and symmetry of Xiaohong are a pair of action and reaction forces....So they're a pair of equilibrium....You can also use the method of elimination....The other options other than b are problematic....a…Red is under gravity but not on the scale...Nothing to do with it....c…The gravity force and the symmetrical support force on the ground are a pair of balancing forces....What does it have to do with the support for Xiaohong...? d Xiaohong's symmetrical pressure and the support received by the scale...? It is said that the support received is given by the ground....Is this related...?

The balance of the forces should be chosen for the same object.

B is two objects, and it cannot be said that they are balanced. C is absolutely wrong, needless to say. d is weighed by the support force and the gravity of the scale, and the symmetrical pressure of the little red is balanced.

I'm really embarrassed to choose B, I don't think there's anything to talk about this question.

Hope to give points, thank you.

Kindness. First of all, it is necessary to understand the relationship between the force applied and the force being forced.

So choose B

A is the balance force.

b is the action force, the reaction force.

The equilibrium force acts on a point, and you can draw a diagram for a simple force analysis. (The schematic diagram of the force analysis of mechanics is very important, very important).

A force should be chosen to act on Xiaohong.

For pulley blocks:

When rope weight and friction are not considered: when different objects are lifted to the same height, the extra work is certain. The mechanical efficiency increases as the weight of the object increases; Because w forehead g moves h

When considering the weight of the rope and friction: there is no specific rule, only that if the object is heavy, the friction becomes greater and the extra work increases, but how the mechanical efficiency changes is unknown.

The extra work can only be calculated indirectly by: the formula w has w total.

Absolutely.

If you think about it, the actual force will be greater.

"Regardless of rope weight and friction" is an idealized pulley block condition, with which the influence of rope weight and friction on the mechanical efficiency of the pulley block can not be considered, and the extra work is mainly the work done to overcome the gravity of the moving pulley, so there is w = g moving h. And because W total = W has + W amount, there is FS=G object H+G motion H, that is: NF= G object + G motion.

Friction and rope weight are not considered, just for the convenience of calculation, plus the friction is very small and the rope weight is very light, it has no effect on the pulley, and the calculation can be ignored. For example, the motion of objects ignores the resistance of air, and the motion of some particles with dots ignores gravity.

When friction and rope weight are considered, there will be an impact on the mechanical efficiency, and the extra work will also be increased, and the form of the formula will remain the same, only the additional force (friction and rope weight) or additional work (the work done by friction and rope weight will be useless);

Regardless of friction and rope weight, the pulley block efficiency is 100%.

When friction and rope weight are not considered in a pulley block, you are asked to consider the weight of the moving pulley. The formula is f=1 n(g+g). This is not the case when considering friction and rope weight. Because the extra work has increased.

If you don't take it into account, the first is the power saving, that is, the mechanical efficiency is high. In the case of the same pulley set.

Excluding rope weight and friction f=[g object g motion] divided by n

Regardless of rope weight, friction, pulley weight f = g divided by n

n refers to the number of rope segments connected to the movable pulley.

If this is a question raised by a comrade who does not know how to learn or learn as a junior high school student, I often encounter such a question when teaching, and there will be only one answer"Take a serious look at the book", but you are a second-year junior high school student, you can think of these problems, you have a good attitude, I hope to continue to work hard and carry forward, to the next level!

This is the definition of a connector that you don't really understand: a container that is open at the top and connected at the bottom is called a connector. The glass tube used in Torricelli's experiment is not a communicator because it has only one end open and is at the lower end and is sealed with mercury, so it is not a communicator and cannot be explained by the communicator principle (i.e., the liquid in the communicator is always the same height in each vessel when it does not flow).

The mercury in Torricelli's experiment is held up by atmospheric pressure. If it is replaced by water, it can be calculated by p=pgh, p is the atmospheric pressure at that time, p is the density of water, and h is the height to be calculated. In the case of standard atmospheric pressure and pure water, the water height is about meters.

The communicator must be open at both ends.

The one in the Torricelli experiment is a straight pipe, one section is open, one section is closed, not a communicator If the water is used for the experiment, the length of the straight pipe is at least needed, so it should not be replaced by water.

The upper end of the support tube is closed, and the communicator is open.

No, the feed-through is an upper opening, while the mercury tube in Torricelli's experiment is vacuum, so it's different.

Hope you understand.

Uh: You misunderstood. How can the inside and outside of the tube be a communicator? The inside and outside of the tube you are talking about here actually means the inside of the balloon and the outside of the balloon, I don't know if you can understand it.

Atmospheric pressure refers to the difference between the outside air pressure and the air pressure in the container or utensil, so the pressure is generated! Do you understand?

The communicator is a U-shaped tube, and the two ends of the U-shaped tube are mouths, and the one that Tolly disassembles is equivalent to a test tube that does not pass through.

Not. The second element of the communicator is the opening.

In the Torricelli experiment, the tube is filled with water, that is to say, the vacated part is vacuum, there is no air pressure, and the liquid surface outside the tube is atmospheric pressure, so the pressure generated by the height difference between the liquid level inside and outside the tube is the vacuum and the pressure difference outside the tube, that is, the atmospheric pressure.

But in Torricelli's experiment, isn't the inside and outside of the tube also equivalent to a communicator?

What do you mean?

Torricelli's test tube is not open, the upper end is closed, and if it is open, the mercury column cannot be lifted by atmospheric pressure.

Let the center of gravity be d from one end and d from the other end'。It is known from the principle of leverage

200(d+d')=gd

300(d+d')=gd'

The above two formulas add up to:

500(d+d')=g(d+d')

Thus there is: g=500n

The answer is 500n. The formal method is the center of gravity analysis method, assuming a center of gravity point, someone above has analyzed it well.

Quick Method: Imagine that the end resting on the ground is also pulled up with a spring scale, and the effect is the same. That is to say, both ends of the rod are pulled vertically with spring scales, one end is high, the other end is low, and the total weight is of course equal to the sum of the readings of the two spring scales.

There is also a special method, imagine a 400N weight uniform rod, each end must be 200N, and then add a concentrated mass of 100N at one end of the rod, then this end is 300N, a total weight of 500N.

You don't understand friction, it is divided into dynamic friction and static friction.

When the box moves, the box is subjected to dynamic friction, and the kinetic friction is a fixed value, which is what you call 10n, but when your force is 10n, the box is subjected to static friction, which is equal to the magnitude of the force you push, and the force in the opposite direction.

Static friction is a range value, 0< = static friction< = f, where f is the maximum static friction, and f is a little bit greater than the kinetic friction.

When considering that the static friction force is a range value, then your problem is easy to understand, the force of 5N is not enough to push the box to move, then the box is subjected to static friction, and the range of static friction is 0 to 10N (let's think it is 10, but it is actually a little bit larger than 10). So the box doesn't move.

10N is the dynamic friction and 5N is the static friction. 5n did not reach the level that it was a box that moved. And when 5N is used, the friction of the box also becomes 5N.

The magnitude of sliding friction is related to the roughness of the contact surface and the magnitude of the pressure.

There are three forces horizontally on object A: the frictional force experienced by object A, F2, and the thrust force on object B, and the thrust force on object B is subjected to a thrust force and a frictional force, the thrust and friction force are balanced forces, and the thrust of A on B is an interaction force, and the magnitude is equal, so F2 is greater than the thrust of A on B.

The scene in the diagram is derived.

1, F1=F2 pressure 9Ga+GB) and the contact surface are unchanged, the friction force F is unchanged, and the uniform linear line F=F

2. In Figure B: The thrust of A against B< f2 on the grounds that f2 has to overcome the friction of a, and the rest is added to b.

Let's first consider a and b as a whole, and since they move in a uniform linear line, f2=f

The total frictional force equal to the thrust f2 is equal to the sum of the frictional forces experienced by a, b, and it is also equivalent to f2 being divided into two branches, so the thrust of b is less than f2...

You can't look at how many moving pulleys you have, you have to count the number of segments of the rope, and in this case, the rope is two sections, so the pulling force is half of the total weight of the object.

Tips, don't ignore the essence of dismantling the sail because the teacher said that the number of pulleys is the essence of the number of rope segments.

When calculating the tensile force f, it is to look at how many sections of rope "bear" the weight of the world, not how many wheel cover sellers.

There are only two ropes here, and there are only two ropes to search for the elevator and the person. Hence w*1 2.

The elevator is a whole, as a moving pulley, the total gravity is 2 * 10 to the fourth power, and half of the force of the moving pulley is 1 * 10 to the fourth power.

The elevator is a whole. 2 movable pulleys are equivalent to one and they are a segment, so it is 1 2 instead of 1 4

Look at the elevator car as a whole!