Problems with Newton s laws of physics, problems with Newton s laws of motion

Answer: 3 4umg

Let the friction between the two wooden blocks on the right be f1, the friction between the two back blocks is f2, and the tensile force is t for the whole: f=6mA (1).

For the small block on the upper right: f1-t=ma (2) and for the large block on the bottom: f-f1=2ma (3).

For the small block on the left: t-f2=ma (4).

2)+(4) obtains: f1-f2=2ma, we can know that f1 is greater than f2, and f1 reaches the maximum friction first, that is, f1=umg

t=umg-ma (5) is obtained from (2).

From the joint solution of (1) and (3), we can obtain: a=ug 4

Substituting (5) yields t=3 4umg

For your question, I was wrong at the beginning, because for the analysis of large wooden blocks that are not subject to force, because the friction force is not necessarily equal to UMG, from my analysis above, we can see that this friction force F2 will never reach UMG.

Let's start by looking at the four blocks as a whole. f=6mA, a=f 6m (because the whole is counted, the rope that pulls m is counted as the internal force).

The tensile force of the rope pulling m = 3mA = f 2 But f must be less than 2 mg, because the maximum friction force is mg, so the maximum rope pulling m is mg

Newton's laws are imperfect!

Regard the left 2m and two m as a whole, then the force of the horizontal acceleration of the whole is the static friction force of the right 2m on the upper m, the maximum force is mg, the maximum acceleration a= mg (2m+m+m)= g 4, when the maximum acceleration is reached, the left 2m and the m on it are regarded as a whole, the horizontal acceleration force is the tensile force given to it by the rope, the tensile force = ma = 3 mg 4, then the tensile force is less than the maximum static friction force, It can be seen that the m on the left does not slide on the 2m.

At this time, the maximum tensile force of the rope, when the external force f increases again, the right m slides relative to 2m and becomes sliding friction, and the sliding friction is less than the static friction, and the tensile force of the rope becomes smaller.

A1:

First look at a whole to find a, acceleration a=(f-m0g-mg) (m0+m), and then analyze the force of the object below, and get the reading f1=mg+m(f-m0g-mg) (m0+m), the main thing to do this kind of problem is to distinguish the internal relationship between the whole and the individual, and it is no longer difficult to clarify this kind of problem.

Answer 2: The reading of the spring dynamometer is actually to show the tension of the spring hook, and if the upward acceleration of the spring dynamometer is A, then the reading.

f1 = mg + ma, and then it is to find a, don't think about the spring problem in the middle, consider the whole, force f pulls the shell of m0 and the weight of m, and the acceleration a = (f-m0g-mg) (m0+m), so that the reading f1 = mg + m (f-m0g-mg) (m0 + m) = mf (m0 + m).

The key is to understand these two steps, the reading is very simple to be related to the weight below, and the overall consideration when finding the acceleration.

Hehe, I hope to help you, this kind of question is actually not difficult.

For the whole: a=f (m+mo) This phenomenon is overweight. Indication = tensile force of rope = mg + ma substitution: = mg + fm (m + mo).

First, take the whole as the research object, set the acceleration as a, multiply f=a (m0+m) to get a, and then take the spring as the research object, and set the spring reading to be x,..x should be the net force experienced by the spring. There is x=mg-ma, and then the known value a can be brought in to find x

1. It will do a uniform linear motion.

2. When object A hits object B with a force of 10n, how will object A and B move? It depends on the original state of A and B, and the situation of other forces, maybe the movement state of A and B is unchanged, or only one of them changes, if A is stopped by the reverse spike of 10N, B is moving or stopping? Not necessarily,

Question 1: When a rotating spinning top is suddenly not constrained by any force, the spinning top will keep rotating around another fixed axis while constantly rotating, which is the rotation of the gyroscope, also known as the rotation effect. Gyroscopic spirals are a common phenomenon in everyday life, as exemplified by the spinning tops that many people played with as children.

Question 2: The topic itself is not written in a standardized manner, A and B collide, whether it is inelastic or elastic collision, now I will analyze it according to your question.

1. When object A hits object B with a force of 10N, how will object A and object B move?

Answer: Objects A and B have at least one clear movement.

2 If A stops by a reaction force of 10N, does B move or stop? If it moves, where does the force come from to make it move?

Answer: If A stops by the reaction force of 10N, B must move, because according to the principle of conservation of momentum, the system composed of A and B, the dynamic resistance of the positive fiber is conserved, A is stationary, and B must move. The movement of b does not require force to maintain.

The motion of an object is not maintained by force. Force is not the cause of the motion of the object.

abFirst, the tangent slope of the V-T diagram represents the magnitude of the instantaneous acceleration at that point.

When t=0, f=kv=0

mgsin - mgcos -kv=ma a = ab slope in the figure) v=5 m s

The final velocity tends to be 10m s, and the force is balanced at this time.

mgsin - mgcos =kv v=10m s is solved by , .

k=20 kg/s

Hope it helps

b a can be found from the straight line of figure b, cd can be seen from the curve of figure b.

The essence of this problem examines the momentum theorem, and the quantitative proof wants you to do the math yourself, because typing is too cumbersome, and this problem only needs to use the conclusion...

Here's a way to think about it: consider that there is no gravity, so that you can flatten the two people who fall on both sides of the pulley upwards, imagine that they are lying on a horizontal smooth plane, and then A pulls the rope and B grabs, who gets to the midpoint first? Is the result of this question determined by the quality of A and B?

I don't know if it's understandable, but that's the first point.

The second model is to cancel the ground and restore gravity, just drop A and B on the side of the pulley. At this time, A and B, and the rope are connected together as a system, and in the end, it will slowly descend in the direction of A, and B will slowly rise; Or is it the other way around; Or A and B are of equal mass and hang still. Whether these are also determined by the quality of A and B.

This confusion is the second point.

Superimposing the results of the two models is the idea of the final answer. a: I don't know yet, please ask.

A According to the force analysis, if the mass of A is large, A is subjected to a downward force, and B is subjected to an upward force, so B arrives first.

I think it's also A, but I'm thinking about the centroid frame of reference, and the result is the same as the great god.

Acceleration downward, lift down or up is possible, only know the direction of acceleration downward, should be acceleration downward, the object is in a state of weightlessness, the combined external force is downward, but a part of gravity produces downward acceleration, reflected in the person, he shares a part of gravity, doesn't it seem that he can lift heavier objects.

Landlord, you're a freshman in high school, right? This is the simplest case for this kind of problem. In fact, the key lies in force analysis.

Think about it, why do they slide relative to each other? Imagine what happens after exercising, why does B move? It's because of the force that A exerts on it, right?

This force can only be the elastic force of the vertical contact surface. In this way, a is naturally subjected to a reaction force (which is important in force analysis), and this force "tops" it upwards, right? When this force is strong enough in the direction of gravity to counteract gravity, doesn't A "climb" upwards, and they don't slide relatively?

From this, the kinetic equations can be listed:

f=（ma+mb）a;

nsinα=mb*a;

The critical conditions are:

ncosα=ma*g

From this, the value of f can be solved.

To learn high school mechanics, it is good to carefully analyze the mechanical situation and figure out how the mechanical situation is generated, how the relevant forces act, and what the critical conditions are.

f-f=ma

5-f=ma1=m*1=m

8-f=ma2=2m

Lianli solution. The mass of the object m = 3kg

Friction f=2n