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<> take point C as the research object. Subjected to three forces. The Cb direction is the tensile force Mg, and the CE direction is the resultant force of the Cb tensile force and the AC tensile force FCEfce=m again'g.Thereafter, the FCE remains unchanged. The above three forces make up the blue vector triangle cef

From the meaning of the title to know fce=m'g does not change, mg decreases, and at the same time, the angle between the two ropes ac bc and the horizontal direction increases, and the vector triangle of the above three forces becomes a triangle with a red dot as an angle inside the blue triangle, and the tensile force corresponding to the ac side decreases first and then increases, when mg=0, fac=m'g

First decrease, then increase.

The pull force of the rope AC is f1, BC is f2, and the hook weight is f0, so that the force (f1) in the opposite direction of the same magnitude as f1 is the resultant force of f2 and f0.

In the 1 question condition, f1 is horizontally to the right, and as m decreases, the weight will move downward along the circumference around point a, and the simplest way is to consider two points.

1. Under the condition of 1 question, f0=f1=f1

2. When m=0, the weight is only subject to AC tension, downward, at this time f0=f1=f1 During this period, any point f0>f1, because there has always been a pull force of f2, you can feel free to draw a little bit during this period to try.

Therefore, at the beginning and end of the block, the AC pull force is equal to the heavy block, and the AC pull force is less than the heavy block during the period.

We can also continue to go deeper, the position of the minimum pull force of the rope AC is the angle of 90 degrees between the rope AC and BC, and the pull force of both AC and BC is (M root 2).

The second question I have a question about the topic, that is, the stem says "on the basis of this balance", if it refers to maintaining this state of balance at the same time, then it is relatively simple, after the sand leaks, the force of the BC rope decreases, the direction of the AC rope remains unchanged, and the corresponding triangle law can be, so the corresponding AC is constantly decreasing, but if you think that you do not maintain this equilibrium position, that is to say, when the AC rope is going to rotate down, then this problem is complicated, one is to deduce mathematically, But the process will be very troublesome, the other is to use the process method to see, the final force of AC must be mg, and then correspond to the value of AC obtained by the previous question, of course, it also depends on whether there is a special process in the middle, if I remember correctly, the original question of this question is actually only the previous question, and the latter question is estimated to be added by others. So if it corresponds to the latter situation, it will be more biased.

When the sand in the sand bucket gradually decreases, the sand bucket is accelerating upwards, and the remaining weight of MA+ sand is obviously less than the weight of the original sand bucket, otherwise it will not accelerate upwards.

In the process of sand reduction, the component received by the rope in the vertical direction is equal to the weight of the sand bucket + mA. When the sand gradually decreases, A will gradually increase, but the force and the mass of the sand bucket generated by the acceleration behind will always be less than that of the front, otherwise A will not increase.

The answer is even more obvious when acceleration is not taken into account.

Answer: Keep decreasing.

2 2) mg, always unchanged.

The force of ac is equal to the gravitational force of the weight when m disappears completely; In addition, it can be assumed that the AC angle is 45° downward, and the force analysis shows that the magnitude is still the weight gravity. The size of the two states is the same, and it is judged that it will not change.

Instantaneous power refers to the power of an object at a certain moment, which is found by the formula p=mv, where v represents the instantaneous velocity, and this problem is 2t. moment v=at=. m , when 3t.

Moment: v=. /m+。

m=。m, so 3t. The instantaneous power at the moment is 3F.

m, i.e. option b is correct.

The average power refers to the speed of the work done by the object in a certain period of time, which is found by the formula p=w t, and the total work done in this problem must be calculated by w=f*s, and s represents the displacement, then 2t. Moment: s1=at 2 2=.

2/2m,w1=f。*s1=2f。^2t。

2 m, while s2=vt+at 2 2=. ^2/m+。^2/2m=。

2 2m, then w2=3f. *s2=21f。^2t。

2 2m, so the total work w=w1+w2=25f. ^2t。2 2m, so the average power is:

p=w/3t。=25f。^2t。

6m, option d is correct

This symbol is so hard to play, I wish you progress in your studies!

Instantaneous power refers to the power of an object at a certain moment (moment), which is calculated by the formula p=mv, where v represents the instantaneous velocity;

The average power refers to the speed at which an object does work in a certain period of time, and is calculated by the formula p=w t.

You don't have a diagram, there is no way to help you explain the answer, there is a need hi me.

First of all, the horizontal force is balanced, so there is pressure f between any two books.

Then all the books as a whole weigh ng, so the friction of the left and right hands against the book is ng 2

Then the right half of the n 2 books do the whole. It can be seen that the friction between the two books in the middle is 0.