A typical physical triad converging equilibrium problem, i.e., ladder problem. This is a ladder wi

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.

Let's help you solve the second problem first.

If the ladder is divided into 8 units, the weight of the person is 24 unit weights, and there is a ladder of unit length at the left end of the center of gravity, and the person standing at the far left end has for the left end.

For the right end, <>

Left end = right end, proof.

Solution to the original question: <>

Let the support force at the right end be fn1 and the friction force is f1; The left end supports the force fn2=f1, and the friction force is f2,.

<> then, tan=fn1 f1=2

When only the ladder rests on an inclined plane, its center of gravity is located 1 2 away from the vertical wall.

And when a person stands at the very top of the ladder, there is nothing wrong with it, and the center of gravity should be shifted upward.

The center of gravity can be calculated according to the leverage balance. That is, the center of gravity is regarded as the fulcrum of leverage. Equations:

3m x = m x (1 2-x) Let the length of the ladder be in units of 1

Get x=1 8

Bar analysis: subject to gravity g (straight down), rope pull t (along the rope up, angular to the horizontal), horizontal force f.

Since the rod is stationary, there must be a resultant force of 0 and a resultant moment of 0.

It is obtained by the condition that the resultant force is 0.

tanα＝g / f

The length of the rod is L, and the A end of the rod is the shaft, and the resultant moment is 0.

g*(l 2)* cos f*l* sin i.e. tan sin cos g ( 2* f ) so tan 2* tan

Note: When using the condition that the net force is 0, the forces should be moved in parallel to the same point, as shown in the figure below.

When using the condition that the resultant moment is 0, it is necessary to follow the actual position of each force action point, as shown in the following figure.

Since the three forces are balanced, either the three forces are parallel, or the three forces are at the same point, in this problem, let f and mg intersect at the point o (the center of gravity is at the midpoint of the rod), then the tensile force t is along the rope direction and pass the point o, and the center of gravity is at the midpoint of the rod, let the center of gravity be m, and mg crosses the above line at b, so mo=mb, which can be proved.

Solution: Let the upward force of the wall to the wheel be x, and the force of the ground to the wheel upward is y, and the maximum weight of a is z x=0 x y z 210 y=0 x

m=0 100z 200 x solve the system of equations.

x＝30y＝480

Z 300300N is what you want.

Let the ground support force for the wheel be n1 (direction upward), the wall support force for the wheel is n2 (direction to the right), the friction force of the ground on the wheel is f1 (direction to the left), and the friction force of the wall on the wheel is f2 (direction upward), which is obtained by x=0: n2-f1=0

is obtained by y=0: n1+f2-g-f=0

is obtained by m=0: f1*r2+f2*r2-g*r1=0f1=f*n1

f2=f*n2

The above five formulas can be substituted into the number to obtain: g = 300n

When an object is balanced by the action of three forces that are not parallel to each other in the same plane, the lines of action of the three forces must converge at one point. That is, when an object is in equilibrium under the action of three forces that are not parallel to each other, the three forces must be coplanar and co-pointed, and the net force is zero.

1. Weak theorem: When a rigid body is balanced by three forces that are not parallel to each other but are coplanar, the lines of action of these three forces must converge at one point.

2. Strong theorem: if two of the three mutually balanced but not parallel forces acting on an object intersect at one point, then the three forces must be in the same plane, and the line of action of the third force passes through the intersection of the first two forces.

The content of the principle of the convergence of three forces is that if the object is in equilibrium by three forces, the three forces must be in common.

If the ball is suspended with a rope, the long line of the rope must pass through the center of the ball.

Mainly, if the three forces are balanced, the vector sum of the three forces is 0

If you figure out what a vector is, it's easy to understand.

bThere is no detailed process for this.

That is, suppose that three forces are divided into two groups, the smallest one is divided into one group, the largest is in a single group, and then the two groups of forces are homogeneous, and when they are in the same direction, can the two smaller groups be larger than the largest group, for example.

A 3+2<8 is not possible.

b 8+8>10 does.

c 5+2<8 is not possible.

It's about the same as the sum of the two sides of a triangle is greater than the third side.

Although this metaphor is inappropriate......）

Is the force surface smooth? If it is smooth, b: 10n=g, f1=8n, f2=8n, one to the left and one to the right, d is the same.

bPhysics and mathematics are interconnected.

Therefore, it is the principle of constructing a triangle based on 3 straight lines.

Only Yes.

Pure speculation.

The connection hinge can be used to resolve forces in both horizontal and vertical directions during analysis. Because two orthogonal directions in the plane can be combined into a force in either direction. The reason why it is orthogonal is because the question does not say that the whole is balanced, and the next moment the state of the whole may change.

Therefore, the force at this time is not necessarily vertically downward, but it must be in the plane.

You first go back and take a good look at the conditions for the intersection of the three forces, and then say, the force at point A is the internal force, and there are more than 3 forces on a single rod after decomposition (force couple), because the action point is not one, so you can only decompose the problem with force, and orthogonal at a is just a method.

The balance of an object is divided into two aspects: translational equilibrium and rotational equilibrium, for translational equilibrium, you can see the object as a particle, and all forces as a co-point force, but for rotational equilibrium, it makes no sense to treat the object as a particle, and there is no way to study rotation by moving the action point of all forces to a point, just like you can't say that a point is rotating, so don't avoid the problem of moment, otherwise you will never be able to solve the problem. Can you use the Three Forces Convergence to study leverage balance?

I can't anyway.

This question is selected from the CD

Seeing the hinge as the fulcrum, the moment of gravity increases, so the supporting force of the wall must increase, so B is wrong. This must be judged by the moment! The other three options can be judged by the common point force on the basis of b.

The ladder is balanced by force, so the gravity of the person (the pressure on the ladder), the support force of the wall, and the force of the hinge are balanced, and the support force of the wall increases, and the force of the hinge on the ladder must increase.

The overall analysis of the car, ladder and person is balanced by the ground support force, ground friction force, gravity force and wall support force. The support force of the wall increases, and it is horizontal, so the friction of the ground must increase. C pairs.

The force in the vertical direction is unchanged, so the elastic force on the ground against the car remains unchanged, and the D term is paired.

I don't know if I can convince you.

Think of people, ladders, and cars as a system, which is in equilibrium and has zero net force.

Vertically: G person + G ladder + G car = N ground (the gravity of the system is balanced with the support force given by the ground) D. √

In terms of torque: the person is upward, the torque increases, the moment of the wall to the ladder increases, and the elastic force of the wall to the ladder increases. b 。x

For the horizontal direction of the human and ladder system: n wall = n hinge (the elastic force n of the wall surface of the system is perpendicular to the wall surface, that is, horizontal, and balanced with the force of the hinge it is subjected to), n wall increases, n hinge increases. c。√

The n-hinge force is derived from the frictional force f on the ground against the car, and n increases, and f increases.

The answer is indeed A, D

A, the ladder as a lever, the point of contact between the ladder and the wall as the origin, the lever force is balanced, the force of the person on the ladder is gravity, gravity is unchanged, but in the process of people up, the force arm is getting smaller and smaller, so the moment is reduced, the moment of the hinge to the ladder should also be reduced to balance, the force arm of the hinge on the ladder has not changed, the force is of course reduced, a right.

B. The ladder is also regarded as a lever, but with the hinge as the origin, the force of the person on the ladder is not table, but the force arm becomes larger, so the moment becomes larger, and the force arm of the wall to the ladder remains the same, so the force becomes larger, B is wrong.

C. Considering the car, ladder, and people as a whole, gravity has not changed, and friction is only related to gravity and friction coefficient, and the friction coefficient has obviously not changed, so the friction force is unchanged, C is wrong. (It may be difficult to understand here, because when analyzing B, we know that the level of the wall on the ladder has increased, but looking back at the situation of analyzing A, the force of the hinge on the ladder is reduced, and according to the relative principle of force, the force of the hinged ladder on the hinge is reduced, and this part of the change just offsets the increase of the force of the wall on the ladder.) ）

d. The same gravity has not changed, and the elastic force of the ground to the car has not changed, D right.