Physics A question about tension on a rope

I can't see the diagram and can only assume that the A rope is the rope between the trailer and car number one.

The b rope is the rope between car 1 and car 2.

The C rope is the rope between car No. 2 and car No. 3.

Two ways: 1) Holistic approach. First of all, the one, two, three and three cars are regarded as a whole, because their motion states are the same, and the amount of process is obviously not considered here (that is, the 10kg object leaves the first car and does not arrive at the second car), so the tension of the A rope remains unchanged.

If the tension of the A rope changes, then the tension of this whole (one, two, three, three cars) will also change, because m does not change, then according to A=F M, we know that A will also change, and here A remains the same, so the tension of the A rope does not change.

B rope is also used in a similar way, because a is unchanged, f=ma=(m(original)+10) a=m(original)a+10a, and the amount of change between this f and the original f=ma is 10a=

C rope, like section A, M does not change, A does not change, so the variable of the tension of C rope is also 0

3) Step-by-step approach. The last car No. 3 was judged first, then the No. 2 car, and then the No. 1 car, and the force analysis was performed on each car separately.

Car No. 3 is subjected to the pull of the C rope, which causes the acceleration to move. There are no other forces. According to f=ma, we know that there is no change in a and no change in m, so f does not change, so the tension variable of c rope is 0

Car No. 2 is pulled by rope B, and rope is pulled by rope C. Then there is.

f(b)-f(c)=m (car No. 2) a, and obtain f(b) = f(c) + m (car No. 2) a = f(c) + (m (original car No. 2) + 10) a

Because f(c) does not change, so f(original b) = f(c) + m (original car No. 2) a, so f(b)-f(original b) = 10a=

Finally, car No. 1 is analyzed. f(a)-f(b)=m(car one)a, get.

f(a) = f(b) + m (car No. 1) a = f(b) + (m (original car No. 1) - 10) a

And F (original A) = F (original B) + M (original car No. 1) A, according to F (original B) = F (C) + M (original No. 2 car) A

Then there is f (original a) = f (c) + m (original car No. 2) a + m (original car No. 1) a

f(b) = f(c) + (m (original car No. 2) + 10) a, so.

f(a)=f(b)+(m(original No. 1 car)-10)a=f(c)+(m(original No. 2 car)+10)a+(m(original No. 1 car)-10)a

F (C) + M (the original No. 2 car) A + M (the original No. 1 car) A

i.e. f(a) = f(formerly a).

So the tensile force of section A remains unchanged.

What about the figure? If nothing is clear, let's assume that the trailer is connected with a rope 1, 1-2 is connected with B, and 2-3 is connected with C. If the trolley *3 is considered as a whole, the force of the A rope remains the same no matter how the inside of the trolley changes.

In the same way, car No. 3 does not change its mass m, and the acceleration A does not change, so F does not change, so the C rope does not change. Car No. 1 becomes lighter by 10kg, the acceleration has not changed, so the resultant force has become smaller by 10*, then car No. 1 has received a rope pull and a B rope pull, and A has been launched unchanged, then only B has become larger, so far the inference is completed, there is no need to think about car No. 2 anymore, the result is the same.

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Problem description: If you pull a rope with both hands, where will it break from?

Analysis: In physics, a rope should be a light rope. A light rope is a physical model that does not elongate under tension regardless of weight.

In physics, the forces on a light rope are equal in all halls. If you pull a light rope with both hands, then it will break from everywhere at the same time and the light rope will disintegrate.

This may not be the case with the actual rope.

Upstairs is talking about the actual rope situation. She's not talking about a party, either. It should be that the stress is equal everywhere and it will break from the weakest point.

Is it so troublesome, just use the centroid motion theorem.

The coordinates of the centroid of the rope yc=y 2l

The acceleration of the centroid ac= d yc dt =[(dy dt) +y(d y dt )] l=[v +yg] l

By the inscription v = 2g(l-y).

So ac=g(2l-y) l

The centroid motion theorem is solved by n-mg=mac=mg(2l-y) l to obtain n= 3mg(l-y) l

mg(l-y) l is the weight of the chain that has fallen on the table, so the support force of the table on the chain is equal to 3 times the gravity of the chain that has fallen on the table.

When the force at this point is in the direction of the angular bisector of the angle with this point as the vertex and the rope at both ends and the wax as the side, the tension of the two sections of rope is the same.

The method of confirming the positive fixed force on the angular bisector is to make both sections of the rope in an elongated state, because as long as one of the sections is not in the elongated state, then the elastic force of this section of the rope is 0, which obviously does not meet the condition set by the question. In fact, when both sections of the rope are in an elongated state, the force you make must be in the direction of the bisector of the extension angle, and the elastic force in the rope at both ends is equal everywhere.

This conclusion is only true for the "same" rope (and one conclusion applies to two ropes), which is a common conclusion in the first year of high school physics.

For the answer to your doubts: The idealized light rope cannot be regarded as a spring with an infinite stiffness coefficient, and the idealized light rope should be regarded as a line with zero mass, zero elastic deformation, and zero thickness. If you still don't understand, then I'll tell you the difference between an idealized rope and an idealized spring

First of all, the balls connected by the two are conserved in momentum during the whole process of motion, and the idealized rope has no elastic potential energy because there is no elastic deformation, so the total momentum of the two balls does not change, but the total mechanical energy changes, because the disappeared kinetic energy does not become elastic potential energy like a spring, but becomes internal energy (specifically the internal energy of the rope) like the collision of two balls.

Namely: Idealized rope Inelastic collision Conservation of momentum Mechanical energy is not conserved.

Idealized Spring, Elastic Collision, Conservation of Momentum, Conservation of Mechanical Energy.

You can't understand it that way, and if you understand it that way, you're going to go into a misunderstanding, because you've already explained the difference between a rope and a spring in a physics test, and you have to understand that for two balls connected, idealized rope and idealized spring are two opposite extremes, and the problem we encounter in real life is somewhere in between (the reality is.

For the two balls tied to the rope, most of the total kinetic energy that disappears after action becomes the internal energy of the rope, and a small amount becomes the elastic potential energy.

For the two balls that are spring-tied, most of the total kinetic energy that disappears after action becomes elastic potential energy, and a small amount becomes the internal energy of the rope.

If you still don't understand why you can't analyze the rope as a hard spring in the exam-taking physics questions, then I will more vividly describe "the quantitative and qualitative changes from a rope to a spring:."

Idealized rope - Rope in reality - Elastic rope = Spring with stiff elasticity (such as a shock absorber in a car) - Spring with more elastic force such as a spring of a ballpoint pen - Amazing spring with high elasticity (1n force can be stretched several times longer) - Idealized spring.

I'm sorry to tell you that in all the physics questions in high school and junior high school (competition physics is not counted), the situation we encountered is only two extremes, idealized springs, idealized ropes, which of the situations in the middle will never be solved, because they have exceeded the scope of what we have learned, and your spirit of building a physical model to solve physical problems is very good.

You say that people who understand physics answer I don't know if I understand physics or not I can take the physics exam in high school every time I can score 100 points, and I didn't get a full score in the exam once or twice I didn't get a full score in the physics of the Shandong college entrance examination, but unfortunately I didn't pass the preliminary round of physics in the competition. . .

The light rope can be seen as a spring that omits the deformation process and goes directly to the final state.

A light rope is more like a "metal rod" that connects two objects, but it cannot be subjected to pressure. Its stiffness coefficient is infinity and its deformation becomes 0. It's just that I don't quite understand, even if you equate it with a spring, what different analysis can be done?

How to find the limit value of an infinite stiffness coefficient multiplied by an infinitesimal deformation related to their approaching velocity?

Then "the idea after that is not right. If it is a spring, when the two balls have the same velocity, the distance between the two balls is "teleported" to the original strength of the spring, and the spring does not produce elastic force. And this is where the light rope comes in.

If there is any conclusion related to the deformation of the spring, then it cannot be equated with a light rope.

As for what was said upstairs, " idealized springs elastic collisions conservation of momentum conservation of mechanical energy " I disagree. Idealizing a spring, it is clear that momentum is not necessarily conserved. Conservation of momentum and conservation of mechanical energy can exist only one.

But he was right when he said, "Idealized ropes, inelastic collisions, conservation of momentum, non-conservation of mechanical energy."

An idealized light rope is not elongated when a force is applied, but it can generate an infinite force. The stiffness coefficient is infinite, that is, an extremely weak type variable can produce an infinite force.

But in this problem we idealize the force of the rope as constant, and there will be no "linear change of force with spring elongation" of the spring. Because if we analyze the force at the angle of the spring, when the velocity of the two balls is equal, the distance between the two balls is the farthest, the deformation of the spring is the largest, and the elasticity is the largest. After that, the two balls are affected by the spring force to accelerate and decelerate respectively, and the movement speed is different.

The conclusion that it moves at the same speed as in the original question is not convincing, so it is not valid.

First, yes, because there is no tension.

Second, when the elastic force is the largest, when there is no external force acting on the first ball, that is, when the rope is tightened, the elastic force is the largest, and when the first ball has an external change of tension, look at the size of the external tension to calculate.

The answer is in the examples. It is calculated based on the conservation of the overall energy of the frictionless system, and the potential energy lost by the circumferential bridge and mountain ring increases the potential energy of the weight.

In the process of sliding down the ring, three forces are received, gravity, rope tension, and horizontal thrust of the rod, taking the ring as the research object, we can know that the gravity is unchanged, and the horizontal force is balanced, but the elimination is that the longitudinal component of the rope tension force changes with time.

Therefore, in the vertical direction, the force of the ring is unbalanced, and it cannot be calculated by forceful balance.

Even if the differential is used, the inertia effect must be considered, and the maximum decline is made, but the equilibrium position is not the middle of the fiber, and the equilibrium position we do not know. If you knew, you could just calculate m:m.

Because the force on the ring is not constant and varies with the angle of the rope, it is difficult to calculate the ratio. Therefore, it is necessary to use the mechanical energy to keep the law of the old wild and the constant, that is, the work of the two ridges before and after.

I'll talk about the method, do it yourself, ok? Trust you to do it.

Calculate the centripetal force of A Fa and the centripetal force of B Fb, no problem!

The tensile force T1 of the ob rope provides the centripetal force of the two objects of AB, that is, T1=FA+FB, and the tensile force T2 of the AB rope only provides the centripetal force of A, that is, T2=FA, the tensile force of the rope is the force of the rope to the object, the force of the rope, and the force of the object is the rope, and the two are the action force and the reaction force.

You add questions.

The spring dynamometer must be in equilibrium to measure the force. To hang an object under the spring, there must also be a force to balance the spring so that the spring is in a balanced state, so that the force can be measured. The spring dynamometer shows the elastic force of the spring, that is, the force of the spring to the object, and when the object is in equilibrium, it is equal to the pulling force of the object to the spring, which can only be one.

【 The force of the rope is exerted by the ball on the rope, and the tension of the rope is exerted by the rope on the ball] Let ob=ba=r, the mass of ab is m, and the angular velocity is w. Then the rope tension at a f1 = f (a direction) = m (2r) w 2 = 2mrw 2

The tensile force f2 = f(a) + f(b) = m(2r)w 2 + mrw 2 = 3mrw 2 at b

Therefore, the ratio of the tensile force of (ab,ob) f1:f2=2:3 [can also be brought into the calculation of specific numbers].

Supplementary question: Because the force is mutual, both people use a force of 6n to pull, which is equivalent to fixing the spring dynamometer on the wall with a nail, and hanging a 6n object below, the pulling force of the object on the spring is 6n, and the pulling force of the nail on the spring is also 6n (not counting the gravity of the spring dynamometer), the spring dynamometer measures the gravity of the object, so the reading of the spring dynamometer is 6n instead of 12n, and the analogy can be done

Now A is analyzed separately, A is only subjected to a force Fab, doing circular motion, Fab MW 2R, .,

Analyze b again, b is subjected to two forces f'ab and fob, the same, fob-f'ab=mw^2r', and then the calculation, that's it.