The physics of the ladder is urgent! Physics problems, analysis of the force on the ladder

The force of the ladder pressing the wall is F1, the friction force of the ground to the ladder is F2, the length of the ladder is L, the lower end of the ladder is the fulcrum, and the balance condition is according to the lever.

f1*(lsin)=mg*((2L5)COS) is solved, f1=

The ladder has zero net force in the vertical direction.

f2=f1=choose d

The force in the horizontal direction at the bottom of the ladder is mainly from the reaction force of the wall. The higher the center of gravity and the smaller the angle, the greater the reaction force.

Start by equiving the center of gravity to two points at each end of the ladder, and the points below will not generate horizontal forces. The force on the upper point perpendicular to the wall (horizontal force) is the ladder pressing against the wall, which is also the friction force at the bottom of the ladder. The combined force of the pressure on the wall and the supporting force of the ladder counteracts the gravitational force at the upper point.

Hence the answer is.

Assuming the wall is on the left, the force of the wall on the stick is horizontal to the right, and the friction of the ground against the stick is canceled out.

The upward support force of the ground on the stick is equal to the gravitational force of the stick, so that it can be balanced.

With the point of contact between the stick and the ground as a lever, then gravity and the support force of the wall on the stick form a leverage, which must be balanced in the vertical direction of the stick so that the stick does not fall.

The force of the wall to the right horizontally on the stick is decomposed, the force perpendicular to the stick upwards is the FSINA gravity force to decompose, and the force perpendicular to the stick downwards is the mgcosa lever principle: FSINA * 1 = mgconsa * gives f =

Then the friction of the ground against the stick is also a choice D

Take the ladder and the human research object, the static equilibrium equation:

fx=0 -mg+nb=0 (1)

fy=0 f-na=0 (2)

mb=0 (3)

The upper three forms are solved together.

na=f=na=

NB=MG was taken as the research object.

The ladder supports people n=mg

As can be seen from above, b is correct.

Think of the ladder as a lever, a lever balance.

A simple ladder is of course a straight line, but the ladder is on the car and the car is moving backwards, so the people on the ground see that your trajectory is a curve.

Think of it as a kind of flat throwing motion, two directions perpendicular to each other, one with a constant velocity and one with a uniform acceleration, so it is a curvilinear motion.

If I'm not mistaken, you and I are making the same set of papers, and by the way, the "force of the ground facing the ladder" refers to the combined force of "the support force of the ground facing the ladder" and the "friction force of the ground facing the ladder".

It seems to be similar to the principle of a lever, with the contact point of the ground as the fulcrum, the force of the person on the ladder and the force of the wall on the ladder form two force arms, and the ladder is turned sideways, like this, the two force arms are right (the direction of action of the force is left alone for the time being)-

The gravity of the person remains the same, as the person goes up the stairs, the force of the wall to the ladder is also increasing, and the force of the ladder is analyzed again, because the gravity is constant, so the support force of the ground to the ladder is also unchanged, then only the wall to the ladder support force and the friction force of the ground to the ladder are changing, and these two forces are just opposite, and the support force of the wall increases, the friction force is also increasing, so the force of the ground facing the ladder is also increasing.

As for the answer, I don't know why the force between the ground and the ladder and the angle between the vertical direction increases.

The force arm explains, when calculating the pressure on the wall, the ground contact point is used as the fulcrum, the person counts the resistance point, and the pressure on the wall is the force point, and the resistance size remains the same, but the force arm becomes larger, and the force arm does not change, so the force becomes larger.

The pressure on the ground is the same, with the wall contact point as the fulcrum, the resistance arm becomes smaller, so the force becomes smaller.

Even though the person is always in a state of equilibrium, the point of action of the person on the ladder is constantly changing.

The ladder mass is negligible. People should be treated as particles

The supporting force of the object is determined by the gravitational force equal to that of the ladder, so it does not change a wrong.

The pressure of the b wall on the ladder is equal to the frictional force experienced by the ladder because c is right so b is wrong.

dReasonable and unchanged.

It is enough to analyze the force of the ladder, gravity is constant and friction is determined by pressure.

If you start sliding from A, the corresponding friction angle is A=Arctan(Fa Na)=<30 degrees.

If you slide from b, the corresponding friction angle is b=>30 degrees.

Therefore, people must start climbing from the A end, otherwise they will slip and fall, and the B end will be affected by the force of the ground along the BC direction.

The force at the A end is at an angle to the vertical direction, and according to the principle of the intersection of the three forces balance, the gravity must pass through the intersection of the two forces.

Let the distance between the person and the end A be x, and the length of the ladder is l, which is known by the geometric relationship.

xcota=(l--x)cot30。I can only do this here, I can't find the specific number, and I don't know if it's right!

<> question 1: What should be done?

Because the friction angle is less than 30° and the friction angle is greater than 30°, so if you think of the two ladders as a lever, when the friction is at its maximum, the friction angle on one side will be less than 30°, and the lift will turn up.

Question 2: H*CoT60+(H*CoT60) (Tan60*Tan 1)=L can be obtained by analysis

tanφ1=

Why this equation?

We can set the height to h and move the arrow of force along one side of the grape above the highest point, so that the friction on one side will also extend above the highest point, and then we can solve it according to the knowledge of geometry.

Note: To treat two ladders as a whole (this does not mean that the ladders are connected together), you can think of either point as a pivot when the lift is not really turning.

Because it's really hard to say, you might not be able to understand it, but if you can draw it on the spot, it's easy to understand.

It's tiring to draw and type.

It seems to be a very troublesome question.

The following corner mark 1 represents the section where the person climbs the ladder, and the corner mark 2 represents another section, and the length of the ladder is L, and the angle between each section of the ladder and the vertical direction is A, then the angle between the two ladders is 2A. Set the person on the ladder to climb the distance x (x is the length along the ladder).

The ladder 2 is balanced by the ground support force n2, the ground friction force f2, and the force n of the ladder 1 at the junction point of the two ladders

By the rotational equilibrium of ladder 2, the direction of force of n must be obliquely downward along ladder 2, and the equilibrium equation is columned:

Horizontal: f2=nsina

Vertical: n2=ncosa

For ladder 1, the ground support force n1, the ground friction force f1, the pressure of people on the ladder, the magnitude is equal to mg, and the force of ladder 2 to 1 is the reaction force of n, and the magnitude is also n, and the balance equation for ladder 1:

Horizontal: F1=nsina

Vertical direction: n1+ncosa=mg

Then by the rotational equilibrium of the ladder 1, the equation of the moment equilibrium is columned.

mg x sina=nl sin2a

From the above 5 equations, it can be obtained.

n1=mg(1-x/2l)

n2=mgx/2l

f1=f2=f=mgx tana/2l

Obviously n1 > n2, the two frictional forces are equal.

The relationship between these three quantities is discussed below.

a) If 1=

then 1n1> 2n2, to not fall, it must meet: 2n2>f

i.e. 0,2mgx 2L>mgx tana 2L

Solution: tana <

That is, as long as the angle between the two ladders is not large enough (tana<, it is safe enough to climb the ladder from = side and reach the top of the ladder, but as long as the tana >, if you climb the ladder from this side, the ladder will fall down as soon as you step on it.

b) If 1=

It is necessary to compare the maximum static friction first.

1n1-μ2n2=

1) When x 2n2

2) When, the ladder will fall down as soon as you step on it.

When x>l 2, it needs to be satisfied.

f<μ1n1

i.e. MGX Tana 2L <

Solution: x<2l (5tana+1).

When tana=, 2l (5tana+1)=l 2

When tana=, 2l (5tana+1)=l

So, at that time, it was not possible to use a ladder.

To sum up, to use a ladder, first of all, the angle a between the ladder and the vertical direction should at least meet the tana <

If tana <, then both sides are the same, and they can go up to the top, so in terms of safety, it is better to = that side is safer.

ps: Don't pay attention to the goods on the first floor large777.,That goods are just copying and pasting other people's answers.,The sticky answer is not complete.,That's the answer to another question I answered.,Don't believe it.。 Despise.

<> is balanced according to leverage:

Set: Up to h'Slide down, with the fulcrum at the top of the ladder.

mg*l1=μmg*h

Thereinto. l1=(h-h'*tan (is the angle between the ladder and the horizontal plane) tan = h (l 2-h 2).

Take the equation back and find h'

Let the velocity of point b be v, because it is a uniform acceleration and known conditions, let the acceleration be a, and the time of bc is t, then the time of ab is 2t, and there is v-at=3;v+at/2=6；The two-formula solution yields v=5

Well, the results are not the same in these two situations

First of all, analyze the situation of the jump:

People are in a free-falling elevator People and elevators are free-falling so that people are always in a state of being in a constant state At the moment before touching the ground, people jump upwards At this time, the speed of people relative to the ground is the speed of free fall minus the speed of their own upward jumping, but because the speed of people's own jumps is very small, so the speed of people relative to the ground is still very large, when falling back to the elevator, the elevator has landed and is stationary, then people will fall to the elevator at a great speed and then fall to their deaths Because the acceleration of the elevator at the moment of landing is much greater than the acceleration caused by the jump, so it ......）

Think about it another way: momentum is conserved, and the time when a person lands is very short, and the velocity is very large, then f must be very large in the case of mv=ft, so it is inevitable that people will die, unless the time of t is increased, that is, the buffer is increased.

Then analyze the situation under the horizontal bar:

When the person hangs on the horizontal bar and falls freely, the person's arms can be said to be not forceful, when the elevator + horizontal bar is a whole, it will instantly turn the speed into 0, and the person is still in the inertia of high speed, at this time, the arms must generate enough force to change the state of motion, in fact, it should be the person's arms that cannot withstand the huge tension and are torn (too bloody -, but because the arms give people more time to buffer, the possibility of people surviving is greater than the first one (of course, the possibility of hanging up with heavy bleeding is not considered).

In fact, the most practical way is to put enough thick clothes and other cushioning things for the elevator to lie on the elevator floor and resign yourself to fate.

Neither of these assumptions is valid, first of all, as far as the bouncing force of human zhi is concerned, dao is very special for gravitational acceleration. Even if there is a horizontal bar, no one can grasp it under that kind of inertia.

You ask questions like this, do you think the elevator is unsafe?

The current elevator would not have been in this situation if it were not for human vandalism. First of all, when the elevator goes down and overspeeds, the speed limiting device makes the safety gear move, and the safety gear will stop the elevator system on the guide rail in a very short time. Even if you fall down, there's a cushion device under the hoistway that absorbs the lethal velocity of the fall, and it's okay to have a human life.

I have read a lot of elevator accident cases, and it is rare to see people die or be injured in the car. Please feel free to ride.

1. Jump up, the downward velocity is relatively reduced (relatively negligible), ft=mv

f is almost unchanged.

2.The arm was broken first, and his life fell and he couldn't die.

I'll do the elevator, hehe.

The landlord's buddy is very thoughtful, once, the elevator fell to the ground from the 9th floor, and now the elevator protection measures are better, and there is a right to safety gear, clamp this guide rail. There is a spring-like cushion underneath, and the people in the elevator are professionals, the highest level of cripples!!

If you don't have these safety measures, you fall from dozens of floors, fall when you are overweight, and the speed and force you have been under will not make you jump!!

All elevators are protected! As the brother upstairs said, such as safety gear, buffer! I'll give you another answer for the upstairs brothers!

There is also a device for elevator protection in this world, called a speed limiter! When the elevator wire rope is disconnected (that is, the rope is pulled up and down the elevator), when the elevator falls down, the speed controller will work if the speed limit exceeds a certain speed, and the safety gear clamps the elevator in the middle of the guide rails on both sides, which means that the elevator is stuck in mid-air! At this time, the passengers inside must not pick the elevator door at will, and do not try to climb out of the elevator or anything!

Because being trapped in the elevator is a state of protection! It was originally very safe, but because I was smart and wanted to escape by myself, it was not worth causing an accident!