A physical force analysis problem is solved.

a,b to do acceleration motion together,first see a,b as a whole,this whole is only supported by gravity g and the inclined plane force fn,from which we get:fn=(m+m)gcos ; There is also a force f=(m+m)gsin along the inclined plane downward; So the overall acceleration a=gsin

Therefore, the acceleration of object A is A=GSIN, that is, the magnitude of the resultant force of the force on A is Fa=mgsin, and the direction is downward along the inclined face. So.

Analysis of the force on object A alone, gravity, support force of B, friction force of B against A.

The resultant force Fa on A is decomposed into F1=MGSIN Sin in the vertical direction and F2=MGSIN Cos in the horizontal direction

So we can get: the frictional force of b on a ff = f2 = mgsin cos b support force f (ba) = mg-mgsin sin so the pressure of a on b is mg-mgsin sin, and the direction is straight down.

The friction force between A and B is mgsin cos and the direction is to the right.

If you don't understand, you can ask.

Thank you.

AAB is the overall M+M)GSIN inclination bevel angle = (M+M)A. Find a for a separate analysis of a Friction equal to Mgcos bevel pressure minus mg equals masin bevel.

Hehe, you're thinking too carefully, and it seems that you're interested in physics.

Since there is only gravity and support, there is no other external force. Then you can see whether the gravity on the center of mass of the object and its support point are on a vertical line, such as a sphere, which just coincides, so its gravity is all on the support point of the ground, just balanced. If it is subjected to the elastic force of the inclined plane, there is no other force to balance.

The gravity of the center of mass of the leaning lever is not on a perpendicular line with the support point, so it is necessary to rely on the resultant force of the support force at both ends and the frictional force given by the ground to achieve the balance with gravity. This is more complicated, the lever is subject to gravity, the elastic force of the ground, the elastic force of the inclined plane, and the frictional force given by the ground (the horizontal component used to balance the elastic force of the inclined plane).

This is just an explanation of understanding, not the basis for disintegration, just look at it yourself, the force or understand the relationship between the force balance, and then analyze more, do more and you will be able to catch it. Hehe.

If the ball is given the elastic force by the inclined plane, first of all, it can be sure that the direction of the elastic force is perpendicular to the inclined plane, and it must be at an angle with the vertical direction, which will produce a component in the horizontal direction, and the ball has no other force that produces a horizontal component to balance the component, and the ball cannot be stationary, so it will not be affected by the elastic force of the inclined plane.

If you remove the inclined plane, the long rod will definitely fall down, so it will be elastic by the inclined plane. Not to mention the ground.

The presence of elastic force requires not only the object to be contacted, but also to be elastically deformed.

Although the object in the problem is in contact with the inclined plane, there is no extrusion between the two, and there is no deformation of the inclined plane, so there is no elastic force against the ball.

First of all, is the inclined plane fixed to the ground? If it is not fixed, for example, the inclined plane is just a panel placed there, then the ball will be affected by the pressure of the inclined plane, the friction of the ground, the support force, the gravity, and the ball is stationary, so their net force is 0

If the inclined plane and the plane of the ground are fixed, then the ball is only affected by gravity and support, and the net force is 0

As for the long club, it is definitely subject to 2 support forces, and if there is no bevel support force, then the long rod will fall, but it does not fall. In addition, the direction of the support force is the same as that of the long rod.

If the bevel has elasticity on it, then it is smooth horizontally, then it must move in the opposite direction to the bottom where you say the teacher, this is an independent thinking process, in fact, it is separation, and I am the same. Suppose there's a platform, there's a ball on it, and you assume that the platform is gone, and the ball just falls down, so all the other elastic forces in it are because of that gravity, and that's what I'm used to, it's called the source of force. If you take the platform away and the ball is still there, then it has another force supporting it. So the platform didn't support it before.

Leaning against an inclined plane, if there is a supporting force, which is elasticity, then it has to move in the opposite direction, that is, on a horizontal, smooth plane, there is a vertical upward wall, and a small ball is stationary next to it. It's not going to support the ball, otherwise it's going to move in the opposite direction, and you'll have to go back and ask the teacher, and the other students, it's a conceptual problem, and you might have a hard time understanding it just by saying that.

Your teacher is right, it is enough to move the bevel away while the ball is still in balance and it is not affected by the elastic force of the bevel. If the ball is bounced by an inclined plane, how does it keep its balance? (Gravity and elasticity have been balanced).

The thin rod is different, the end of the inclined plane will press the inclined plane, on the contrary, the inclined plane will give it a support force, plus the support force given to it by the ground, and the gravity of the thin rod will reach a balance, so it will remain in a balanced state and be stationary.

The direction of gravity is vertically downward, so the vertical upward force given by the horizontal plane is just in balance with it, and the counter-proof: if the inclined plane also has a supporting force, the balance will be broken, and the ball will not be able to stand still (in fact, it is what your teacher said). And which surface the support force is given, it depends on which surface the center of gravity is projected vertically, that is, the point of action of the support force is the intersection point of the vertical line of the center of gravity and the surface.

The combined effect of gravity has been replaced by the center of gravity, and there is no need to analyze the gravity of the individual parts.

Problem Solving Guide] Pay attention to the following three points when answering this question:

1) Decompose the force f according to the actual effect.

2) Then decompose the force of the light rod according to the actual effect.

3) The tension of the thin wire is calculated from the balance of two forces.

4 points) The diagonal downward pressure f1 will produce two effects: the force f1 that presses down vertically against the slider and the force f1 that pushes the slider horizontally, therefore, decomposes f1 along the vertical and horizontal directions, as shown in Figure B, considering that the slider is not subjected to friction, the tension on the thin line is equal to the component f1 in the horizontal direction of f1, i.e.:

5 points) Answer: (3 points) Answer:

Cut the arc-shaped wire in half from the middle and take half of the wire as the object of study! The tensile force at point o is horizontal, the tensile force at point A is in the tangential direction, and the line is also subject to gravity! The intersection of the three forces is concentrated in one point!

Force analysis is sufficient. The force at point A is F sin，m=f/tanβ.

The object is acted on by four forces:

Gravity mg, downward;

thrust f, oblique upward;

wall support force n, vertical wall outward;

Friction f= n, downward.

Horizontal force balance:

n=fsinα

Vertical force balance:

fcosα=f+mg=μn+mg=μfsinα+mgf=mg/(cosα-μsinα）

If the object moves in a straight line at a uniform speed, the force is balanced.

Horizontal: fsin = n

Vertical: FCOS +F=MG

Also: f= n

The simultaneous three-formula solution yields f=mg (cos + sin).

Let's start by looking at ABCD as a whole.

a, the first gravity of the force, f1=mg, direction downward.

The static friction of the second template against a f2 = 2mg upwards in the direction of the third b against a is not known at this time and is set to f because the acceleration of a is 0 (a is at rest).

f-combined=0, so f1-f2+f=0

So f=mg in the same direction as f1 downward.

Re-analysis b first gravity f1 = mg direction downward.

The second is subject to the static friction force of a against b f2

From the action force and reaction force, it can be seen that f2=mg direction is upward.

Third, the static friction f of c against b

In the same way, f1+f2+f=0

So f=0 so there is no force between the bcs.

There is no relative movement trend of BC, and there is no friction between them.

The friction force of B against C is 1 2mg, C is subjected to three forces, the downward gravity of itself, the friction of B against C upward, and the friction of D against C upward, the three forces are balanced.

2mg Because A has an upward friction with the wall and this friction cancels out with the gravity of A, there is no friction between AB, and there is no friction between CD, so the friction between B and C can only cancel out the gravity of BC when the friction between B and C is 2mg.

Because C is at rest, C is subject to a balanced force. That is, the friction of the upward b and d to c is balanced by the gravitational force of c. Because B and D give C the same friction, it is 1 2mg.

On the first floor, B and C have a tendency to slide relatively. Explanation: Suppose you replace the contact surface of BC with ice, and C will slide down.

After the power of ball A is doubled, ball B is still subject to gravity G, rope tension T, and Coulomb force F, but the direction of the three forces no longer has a special geometric relationship.

If you use the orthogonal decomposition method, set the angles, and column equations, it is difficult to get results. At this point, we should change our thinking and compare whether there is a necessary connection between the two equilibrium states.

Therefore, the orthogonal decomposition is the synthesis of force, pay attention to the observation, it is not difficult to find: the triangle enclosed by AOB and FBT is similar, then there is: ao g=ob t.

It shows that when the system is in different equilibrium states, the tensile force t does not change.

This special state can be obtained by the fact that the power of ball A is not doubled: t=gcos30.

After the ball A is doubled and balanced, the tension of the rope is still GCOS30.

Accelerated sliding:

Acceleration: a=(mgsina-umgcosa) m=gsina-ugcosa

The horizontal component of acceleration: ax=acosa=(gsina-ugcosa)cosa

The vertical component of acceleration: ay=asina=(gsina-ugcosa)sina

So the static friction of the ground on the inclined plane is: f=m*ax=m(gsina-ugcosa)cosa (direction to the left).

The support force of the ground to the inclined plane: n=(m+m)g-m*ay=(m+m)g-m(gsina-ugcosa)sina

Deceleration slide: Acceleration: a=(umgcosa-mgsina) m=ugcosa-gsina

Horizontal component of acceleration: ax=acosa=(ugcosa-gsina)cosa

The vertical component of acceleration: ay=asina=(ugcosa-gsina)sina

So the static friction of the ground on the inclined plane is: f=m*ax=m(ugcosa-gsina)cosa (direction to the right).

The support force of the ground to the inclined plane: n=(m+m)g+m*ay=(m+m)g+m(ugcosa-gsina)sina

The end result could be appropriately simplified.

Qualitatively analyzed, no specific calculations are required.

Use the size of arctanu and a to determine whether it will accelerate and slow down... When accelerating down, the block has a horizontal acceleration, and this acceleration is due to the support force of the inclined plane. In turn, the block will exert a force opposite to the horizontal motion below, and the block is not moving because it is affected by the friction of the ground, so the direction of the static friction is the same as the horizontal direction of the block's motion.

In the same way, it can be seen that the direction of static friction is opposite to the horizontal direction of the block's motion when decelerating and sliding. If the block and the inclined plane are treated as a whole, the acceleration of the descent is equivalent to the overall weightlessness, so the support force n facing the inclined plane is less than (m+m)g when the deceleration is equivalent to the overall overweight, and the supporting force n of the ground facing the inclined plane is greater than (m+m)g

Whether accelerating or decelerating, n is equal to (m m)g. This is very easy to understand, you can look at the inclined plane and the inclined plane as a whole, and the internal force between objects will not affect the external force, so n is equal to the sum of their gravitational forces, this theorem teacher said. Give points, it's not easy to do questions during the summer vacation.

I can't draw a diagram of this problem.