Ask the Emperor for help. What is the law of af

In this system, only gravity does the work, the mechanical energy is conserved, m first does the variable acceleration motion with reduced acceleration, knows until it decreases to 0, v reaches the maximum value, and then does the variable deceleration motion with increased acceleration. until the speed is 0

vmax, a=0

The tension in the rope t=mg

mg = 2mgsin sin = m 2m and h = ltan

Therefore, h=ml root number (4m -m).

Note: This is the angle between the AC and BC segments and the horizontal plane when the rope is lowered.

m The object accelerates first and then decelerates, and when the resultant force is 0, the speed is maximum, and the angle formed by the rope is set at this time

It is obtained from the force analysis:

mg=2mg cosα/2

from geometric relations.

h=l cotα/2

It is obtained from the above formula.

h=lm/√（4m²-m²)

After the middle ball M starts to fall, the string begins to produce an inclination angle on the inside of the two pulleys A and B

At this time, there are three forces at the middle hanging m:

MG to the upper left, MG to the upper right, MG to the bottom of the positive

The magnitude of the combined force of the three forces is.

f=mg-2mgsin (where: m m 2m).

Gradually increasing from 0°, sin also gradually increases, the resultant force f gradually decreases, and the acceleration gradually decreases.

When f=mg-2mgsin =0, that is, sin = m (2m), the three forces are balanced, and the acceleration is zero, and when the three forces balance point m continues to fall, the direction of the resultant force is changed to upward, that is, m begins to decelerate and fall.

Therefore, when the balance point of the three forces is f=mg-2mgsin=0, m moves the fastest.

In this case, sin = m (2m).

tan = (m 2m) under the root [1-(m 2m) 2] = m under the root (4m 2-m 2).

The height of the ball fall h=ltn =lm under the root (4m 2-m 2) = lm under the root (4m 2-m 2) (4m 2-m 2).

When the speed is maximum, the force is balanced in the vertical direction.

mg=2mg cosθ

cosθ= m/2mg

again cos = h (h2+l2).

So m 2m = h (h 2 + l 2).

2mh=m√(h^2+l^2)

h^2+l^2)= 2mh/ m

h^2+l^2=4m^2h^2/m^2

m^2 h^2+ m^2 l^2=4m^2h^24m^2h^2-m^2 h^2+ = m^2 l^2(4m^2-m^2) h^2 = m^2 l^2h^2= m^2 l^2/(4m^2-m^2)h=√m^2 l^2/(4m^2-m^2)

3L>H>L under the root number

According to the law of quadrilaterals, the tensile force of AC and BC on m should be equal to m, since m > m>2mSo 60< acb<90

Since ac=bc=l, so: 3l>h>l under the root number

It should be frictional.

The second law: f=ma is ideal, and in general, if it is accompanied by friction, it should be f-f=ma

So there is a translation f on the graph

1 Coriolis Force

Also known in some places as the Goriolis force, or simply Coriolis force, it is a description of the offset of a particle in a rotating system that is moving in a straight line due to inertia relative to the linear motion produced by the rotating system. The Coriolis force is derived from the inertia of the motion of an object.

In order to describe the motion of a rotating system, an imaginary force needs to be introduced into the equation of motion, which is the Coriolis force.

The Coriolis force is practically non-existent due to the fact that the uniform linear motion that man thinks of when he is in a rotational frame is different from the uniform linear motion in an inertial frame.

2 The formula for calculating the Coriolis force is as follows:

f = 2mv'×ω

where f is the Coriolis force; m is the mass of the particle; v'is the velocity (vector) of the mass point relative to the rotating reference frame; is the angular velocity (vector) of the rotating system; The outer product sign ( v. ) representing two vectors': The size is equal to , and the direction satisfies the right-handed spiral rule).

The mathematical derivation process is described in detail.

High school physics is getting easier and easier these days.

It's wrong to ignore the quality upstairs, don't look at it.

Use the wooden board as a reference.

1: block a = -f m = -umg m = -ug = a wood = f m = 8 8 = 1m s 2

A block = a block - a wood = -3m s 2

t=(end of v-beginning of v) a=-(0-12) -3t=4s

2: s=(v2-v2) 2a=24m3: self-counting.

It's relatively simple:

It can be divided into two phases:

1) Relative to the ground, the speed of the small wooden block changes from 2m s to the left, and the time of this stage is t1, t1=2 (, and the speed of the wooden block becomes v= 10+((8-0) 8)*t1 (for the wooden plank, the mass of the small wooden block is ignored).

2) Relative to the ground, the speed of the small wooden block changes from 0 to the same as that of the plank, and the time of this stage is t2, then (2 (, t2 t1+t2 can be solved, that is, the time is known, and the remaining two questions are needless to say.

Let the two objects reach a common velocity v after t seconds, the speed of the board is v1=10m s when the block is just on the board, the initial velocity of the block is v2=2m s, the acceleration of the weight m of the block is a2, the acceleration of the weight m of the board is a1, and the friction between the block and the board is f.

1) f= mg (friction).

2) ft=mv-total+mv2 (momentum theorem of blocks) (3) f-f)t=mv-mv1 (momentum theorem of boards) (4) (m+m)v-total=mv1-mv2 (conservation of momentum) (5) v-total=v1-a1t (uniform acceleration motion formula of the board) (6) v-total=v2-a2t (uniform acceleration motion formula of blocks) (7) f-f=ma1 (ox2 of planks).

8) f=ma2 (block's ox2).

8 equations 8 unknowns, solved.

The first question is easy to ask the second question, making sure that when they are at the same speed, the blocks move to the far left of the plank. The key is to find the displacement.

The third question is done with the conservation of energy, the initial energy = the last energy + the work done by friction, or the kinetic energy theorem is used for the block.

The main problem is that the force and motion analysis are not clear.

The wooden board accelerates uniformly before the small object is put on, and the small object is still uniformly accelerated by the friction of the small object, but the acceleration is smaller than the original.

The small object begins to decelerate uniformly under the action of friction, and after the speed is reduced to zero, it accelerates uniformly in reverse, but the acceleration is greater than the acceleration of the wooden board, after a period of time, after the two velocities are equal at a certain time, after which the small object provides a uniform acceleration with the common acceleration of the wooden board by static friction.

It's very simple, draw a column, with a numerical value, if you don't know, it means that you don't have a grasp of the momentum theorem, think again, don't look at the answer!

I'm sure I can't do it, it's the quality of the small pieces. No matter how much you have to have the quality of a small object. I don't know if those people made it one by one, and the god who ignored the mass of small objects, amazing!

Since it is the second big question of physics, and it is not the finale, it is impossible that there is no mass, and there are still so many people coming....

You can't do it without quality!! It's a great god who ignores quality! Without mass, all formulas have become floating clouds!!

Think of m and m as a whole. Mass g=m+m

f=(m+m)gu friction is f

Let the combined external force n=f-f

Overall acceleration a=n (m+m).

Therefore, the acceleration of the wooden block should also be a, and it should be in the same direction as the acceleration of the wedge - horizontally to the right.

So what exactly is the force that provides the acceleration of the block?

At this point, the block is supported by gravity and wedges. Break down gravity. One is the force f1 = mgsina that is oblique downwards

One is the force f2=mgcos of the vertical inclined plane

To make the acceleration horizontal to the right, it is not enough to have only f1, so the support force is greater than f2That is, F2 canceled out some of the support.

Set the support force n1

The resultant force n2 = n1 - the effect acceleration a formed by f2 and f1 together.

So the acceleration a is equal to form an equation:

I couldn't hit some symbols for technical reasons, so I ......It can only be explained in the abstract.。。 n (m+m) = the net force m of n2 and f1

Hehe, it's really a limited technology......It might be a bit confusing. But I hope this analysis can help you, I hope you can understand it...

First of all, m and m are regarded as a whole, supported by gravity mg force n, left f to the right f, left acceleration is, f-f=ma, that is, f = (m+m)ug + (m+m)a, in isolating m, by gravity mg, support force n, cos * n=mg, n = mg cos, sin *n=ma, synthesize the above, get tan *g = a, in the substitution of the top, get f = (m + m) gu + (m + m) tan g, simplification gets, f =(m+m) (ug+tan g), sent with a mobile phone, forgive me for seeing clearly.

The whole method is used to obtain a=f (m+m)-gu by selecting a wedge frame of reference to add an inertial force ma to the object, and in the wedge system, the matter is at rest and the force is balanced, so ma=mgtg0 is brought in to obtain f=g(m+m)(u+tg0).

The three equations mentioned here are:

zhif * cos f m a dao fang special trip 1n genus f * sin mg 0 equation 2f n equation 3

The process by which they synthetically find f is as follows:

From Equation 2 we get n mg f * sin, and substituting it into Equation 3 gives f * mg f * sin ) mg f * sin

Substituting it into Equation 1.

f * cos (mg f * sin ) m a i.e. f * cos mg f * sin m a gets f*(cos sin ) mg m af* (cos sin) m *(g a), so f m *(g a) (cos sin).

Here the source speaks of the three equations.

The bai are:

f * cos f m a du equation zhi1n daof * sin mg 0 equation 2f n equation 3

The process by which they synthetically find f is as follows:

From Equation 2 we get n mg f * sin, and substituting it into Equation 3 gives f * mg f * sin ) mg f * sin

Substituting it into Equation 1.

f * cos (mg f * sin ) m a i.e. f * cos mg f * sin m a get f*(cos sin ) mg m af*(cos sin ) m *(g a) so f m *(g a) (cos sin ) no thanks! Haha