Physics is concerned with the decomposition of motion.

There is no hypotenuse or right-angled edge, because the decomposition is whatever you want. As long as it is a closed graph, you can break it down into many sides. Just for the convenience of calculation, it is generally necessary to decompose a vector into two vectors perpendicular to each other, just like 10 can be equal to 1+9, 2+8, can also be equaled, and so on, there are infinite kinds of decomposition results.

The same goes for the decomposition of vectors.

Your force decomposition diagram is wrong, it's not drawn like this, if you analyze the block, the direction of the combined velocity is horizontal, and it can be broken down into the rope velocity and the velocity perpendicular to the rope.

Landlord, I'll teach you a trick: the nature of the force attached to a rope is exactly the same. The pulling force on the shore must be the same as the force pulling the object below, but the direction of the pulling object is tilted, so it should be calculated according to the hypotenuse.

This is a process of changing angles, so we can only do instantaneous force analysis, if you have any questions, please continue to ask, thank you.

For example, the object on your diagram will eventually move horizontally to the left, and the direction of the resultant force will be horizontal to the left, and because there is an oblique upward force, it needs to be balanced, so there is a force in the direction of the vertical oblique upward force.

The first thing to grasp is what the research object you take, and secondly, the absolute velocity is the speed of the research object you take relative to the absolute coordinate system, that is to say, the velocity of the research object relative to the ground is the resultant velocity, and the rest is the partial velocity.

In response to this problem, the speed of the boat relative to the ground horizontally to the left is the combined velocity, the speed of the rope is a sub-velocity, and the other sub-velocity is perpendicular to the rope diagonally to the left.

The direction in which the object actually moves is the direction in which the velocity is combined. This is also confused? In this problem, the object does not fly up with the rope, it just moves horizontally. No matter how you break it down, the direction of the synthesis is the direction in which it actually moves.

It's all as hypotenuse.

Generally you better look at the speed of the point where the boat and the rope are connected, obviously, the speed at that point is the same as the speed of the boat, assuming that it is horizontal, break it down, a sub-velocity along the rope, a sub-velocity perpendicular rope, along the rope is the speed of the rope along the direction of the rope, and the perpendicular is the speed of the rope rotation (understand it well here), that is, the speed of the rope is not only the speed of contraction in the direction of the rope, but also the rotation. So, it's wrong to use the speed of the rope as a hypotenuse, because that's not the actual speed of the rope at all, and the actual speed of the rope is complicated because each of its infinitesimal segments has a different speed (if the question is a rope pulling a boat across a pulley). This problem is better to look at the connection point between the rope and the boat, because that point is the common point between the rope and the boat, it can be said that it is the rope, it can also be said that it is the boat, and many problems in physics are to find common ground.

When the ship moves horizontally, the upper one is right and the lower one is wrong, because the horizontal velocity is the actual velocity, and the orthogonal decomposition of the velocity is to decompose the actual velocity into two perpendicular ones, so the actual velocity must be sandwiched between the decomposed velocity.

If the boat is not moving horizontally, but going up diagonally along the rope (if there is an inclined plane underneath), then the lower part is right.

Think about it.

Motion decomposition is decomposed according to the effect of action, and people generally orthogonal decomposition is for the convenience of calculation. Rope pull problem, rope pull has two effects: making the boat forward and making the rope contract. Let's analyze the specific problem specifically, if you want to calculate the general orthogonal decomposition.

Because the speed of the ship is the actual speed, the actual speed is always the speed, and it is necessary to do the oblique.

The minute velocity is 90 degrees, so the combined velocity is relatively large, and the hypotenuse is made, so it is good to remember.

If the steamed dumplings are decomposed, they will be right-angled triangles.

The hypotenuse is larger than the right-angled edge.

The combined velocity is the speed that you really perceive, and the point of light moves on the underside of the cloud, so the speed you perceive should be the speed at which the point of light moves.

This combined motion should be v, where v1 and v2 are just the result of the decomposition of v. How to understand these two points of speed is very important!

V1 is in the direction of the extension line of the ray, which means that there is velocity along the direction of the ray, and with the increase of the angle a, you see whether the ray is growing and purifying, and the growth is the displacement of V1 in the accumulation of time.

v2 is perpendicular to the light. It is because the angular velocity of the ray around the point is w rotation, should there be a linear velocity in the direction of the vertical ray to do circular motion, this v2 is the linear velocity of the bending slip.

Affected by the combined effect of V1 and V2, the speed it shows is V.

Therefore, the sub-motion can be synthesized into a combined motion, and at the same time, the combined motion can also be decomposed into a sub-movement of the buried wax!

Thank you, don't know how to ask again.

Hop motion is real motion. The point of light moves on the clouds, so it is the real speed of the square wheel and the direction of the balance is the envy in the direction of the clouds.

You can break down this real motion into two parts, one orbiting the axis and the other moving away from the axis (i.e., vertical and along the axis).

You're right, combined motion is the synthesis of sub-motion.

1. The boat has no sub-velocity in the direction of the current, and as the water moves downstream, the velocity of the water is v=180 10*60=

2. Set the speed of the boat in still water v1, set the river width d, and the two times are t1 = 10min = 600s, t2, there is.

The first vertical opposite shore reaches the opposite shore after t1=600s: v1t1=d. The second time it reaches the opposite bank, there is v1 along the river with a component of v1cosa=v=, and the partial velocity of the second vertical bank satisfies v1sina*t2=d....

Divide by. t1 t2sina = 1 substituting data to get sina = 4 5, so a = 53 °, cosa = cos53 ° =, substitution obtained, solution v1 = , v1 = substitution to obtain d =

Solution: The speed of the river water, the usable displacement is 180m, the time is 10min, and it is done, 180 600=

Let the width of the river be d, and the speed of the boat in still water is v, then by the title, the bow of the boat crosses the river vertically, d v=600

The bow of the ship crosses the river diagonally upstream, d (vsina) = 750, and a = 53 degrees.

v and d cannot be found. Only its ratio is known. v may be larger or smaller than large.

The key here is the understanding of gravity and the Lorentz force. Gravity has nothing to do with velocity and is a constant cherry force. Whereas, the Lorentz force is related to velocity.

When the velocity of point A is 0, it can be regarded as the synthesis of the horizontal rightward velocity v and the horizontal left velocity v, so that the motion of the charged particle can be decomposed into two motions, when gravity is regarded as a sub-motion acting on the horizontal rightward velocity v, then there is no gravity on the horizontal left velocity (because the velocity is a constant force, the magnitude and direction are constant), and the Lorentz force is different, it is related to the velocity, and the sub-motion of the horizontal rightward velocity has the Lorentz force, The sub-motion of the horizontal velocity to the left also has the Lorentz force. In this way, take the appropriate velocity magnitude such that the Lorentz force qvb = mg (qvb direction is vertically upward), then the horizontal direction of the partial motion is a uniform linear motion. And the other horizontal to the left velocity of the sub-motion is a uniform circular motion because it is only subjected to the Lorentz force.

When the ribbon particle reaches point b, the velocity is 0 again, and since the horizontal sub-motion does a uniform linear motion, the horizontal partial velocity at point b is still v, and the direction is horizontally to the right. The combined velocity of point b is 0, so the speed of the sub-motion of the charged particles doing uniform circular motion will be horizontal to the left, for the dung motion of uniform circular motion, points A and B are compared, and the sub-velocity direction of the charged particles doing uniform ridge width and disadvantage velocity circular motion is the same again, indicating that the sub-motion of uniform circular motion is just one week.

[The combined motion is a linear motion with a uniform variable speed, and at least one sub-motion is a linear motion with a uniform variable speed, is this sentence correct? 】

That's right. Because the combined motion is a linear motion with uniform velocity, it means that the object is subjected to a constant resultant external force. This resultant external force exists regardless of the decomposition (which may be decomposed into one or more sub-motions), so that at least one of the sub-motions is a linear motion with uniform velocity.

Whether the acceleration of the combined motion and the partial motion also satisfies the parallelogram rule], as long as it is a vector, it satisfies the parallelogram rule.

The key to this kind of problem is the understanding of the concept, and giving counter-examples is just a way to take advantage of it.

You can take a good look at the above explanations and understand them from the conceptual point of view.

1. The acceleration of the combined motion and the partial motion satisfies the parallelogram rule, because the acceleration is still a vector, in fact, as long as it is a vector, it satisfies the parallelogram rule.

2. The combined motion is a linear motion with uniform variable speed, then its velocity can be expressed as v=at, and a is the magnitude of acceleration. This can be illustrated with the following diagram:

In the diagram, the horizontal axis is time, the vertical axis is velocity, and v is the linear motion of the synthesized uniform velocity, and its expression is v=at, and a is the slope of the vector ov, that is, the acceleration. Suppose that this uniform linear motion is synthesized by 5 sub-motions, as shown in Figure v1 v5, it is obvious that v2 and v4 are uniform linear motions. The rest are uniformly variable speed motions, and these 5 motions can have an infinite number of forms by the superposition of vectors, but in any case, it is impossible to completely superimpose a diagonal line ov from a uniform linear motion, that is, a vector parallel to the time axis, so there must be at least one sub-motion that is uniformly accelerated.

In this diagram, assuming that the initial velocity of motion is 0, the analysis of non-0 is the same, and the number of sub-motions can be arbitrarily added or subtracted, but the principle remains the same.

PS: This is a kind of technical explanation to help you find out if there are counterexamples, and I advocate the baby034 method of considering the problem, which is more general and simple.

In mathematics, δf represents only the increase of the value, not the direction, so the added δf is actually in the same direction as f, and he also said that f1 is increasing. δf=f2-f1

The original object does a uniform acceleration linear motion, after adding f, the original velocity direction (also the original resultant force direction) and the current resultant force direction are different, so do a uniform variable speed curve motion.

45 degrees northwest direction, wind speed 4 2m s

In the case of no wind, people are subjected to a westward 4m s wind, and the 4m s due north wind forms a coordinate system with a direction of due northwest

45 degrees north-east, with people as the reference, the wind has two movements, one is from north to south, and the other is from east to west, and the speed of the two sub-movements is 4, so in the end people feel that the wind is blowing from 45 degrees northeast with a size of 4 2m s

It seems that you already know how to solve the problem, but you don't know why you do it, so I won't write the specific steps, but explain them briefly.

When the block moves to the left, the rope also moves with the block, so the rope generates velocity, which can be broken down into the direction of the rope and the direction perpendicular to the rope, and the speed along the direction of the rope causes the block on the right side of the pulley to rise; The speed perpendicular to the rope constantly changes the direction of the rope, so that the clamp angle between the rope and the vertical direction continues to increase.

At the junction of the rope and the block, the velocity of the two is the same, which can be broken down as shown below.

Comments: Finding the combined velocity is the key to this problem, you should understand that the actual velocity is the combined velocity Then decompose the velocity, make a parallelogram, and solve it according to the triangle The difficulty of this problem lies in the fact that the combined velocity is difficult to determine, and it belongs to the mid-range problem

Analysis: It should be clear that the actual motion of the contact point between the rod and the platform is that the direction of motion is perpendicular to the rod pointing to the upper left, and the vertical upward is a sub-velocity of it, and the velocity is decomposed and solved according to the knowledge of triangles

Choosing a is clear to the picture you drew yourself.

v orthogonal decomposition, v1=v*sina is the linear velocity.

Angular velocity w = v1 l

vsina/l

Answer: choose B to calculate the line speed first and divide by L to get it!