Physics and mechanics questions in senior high school, physics and mechanics questions in high schoo

Balance of Forces, Limit Thinking, Mathematical Method.

At the start (ob perpendicular to ab), fb=g, while fa=0;After moving A, the force is as shown in Figure 1, at this time FBG is obvious, then from Figure 1 to Figure 2, Fb is increasing, you can also use the limit idea, when pulling AOB to almost a straight line, Fb is infinite, which can also explain the process of increase. Taken together, FB decreases first and then increases. However, when the minimum value is not necessarily vertical, in our problem, except for G, everything else is changing, if the OA position remains the same (OA and the horizontal angle are unchanged), Fb is the minimum when vertical, as for when the minimum value is reached, we should use the cosine theorem in mathematics, it is recommended to use limit thinking.

Re-analyzing the fa, the beginning is 0, and then, looking at the plot, you can determine that the fa-has been increasing.

The idea of solving this problem is to draw a triangle at point p balanced by 3 forces.

One side (the tension of P to Point O) is fixed, and the other two sides change direction, and the pattern can be found by plotting.

The conclusion is: Fa has been increasing.

FB decreases first, reaches a minimum when Bo AO, and then starts to increase.

So you can only choose A

I don't understand hi me.

Three forces are balanced, and three forces can form a forward triangle, which is a problem of dynamic equilibrium analysis, and you can see it when you draw the diagram.

The force on the OA keeps getting bigger, and the force on the OB gets smaller first and then bigger.

I'm a physics teacher, and in this problem, the triangle of force, the balance of three forces, can be translated into a triangle, and the vertical force is gravity, and the left-side force is fa, and the right-side is fb

You'll see that I draw three** fa, which starts at 0 and gets bigger and bigger, and fb is originally g, and gradually gets smaller and bigger.

The answer to this question is A

b,d is definitely not right.

Initially, the fa is zero. Because in the horizontal direction the object is balanced. With the left shift of the A electric, the Fa gradually increases.

Assuming that the final state is that A o B is on the same level, then Fb is infinite, so option D is incorrect.

Brother is also just through the third year of high school, teach you some good methods, you can use it when you are unsure, Fa was 0 at the beginning, so it is impossible to reduce all the time, B is wrong, when A is infinite, the angle AOB is close to 180 degrees, it can be seen that at this time, FA, FB are infinite, so FB cannot be reduced all the time, D is wrong.

In fact, many multiple-choice questions can be tried with the elimination method when you don't know how to do it.

B is not true, there is no force at the beginning of FA (there is no force at the level) FB=MG

The FA FB horizontal component is the same after exercise.

Do vector triangles fa to become larger, fb first small and then large.

Choose A, the original force of Fa is 0, and then gradually increase, the original force of FB is equal to MG, when A moves, FB instantly becomes smaller, when AOB is greater than 90°, FB is equal to MG, and then it is larger and larger.

The parallelogram rule of force comes out easily.

The answer is A, how can your teacher say that the answer is D.

1) Since it didn't slide out at the end, the block was equivalent to a round trip in the rough part, i.e. 2l

By the momentum theorem: (m+m)v=mv0,v=mv0 (m+m).

Kinetic energy theorem: work done by friction = mg*2l=kinetic energy loss=mv0 2 2-(m+m)v 2 2

mv0^2/2-(mv0)^2/(2(m+m))

(mv0^2/2-(mv0)^2/(2(m+m)))/(mg*2l)

2) By the momentum theorem: when the spring has the maximum elastic potential energy, the speed of the trolley and the block are equal.

Work done by friction = mgl=mv0 2 4-(mv0) 2 (4(m+m)).

By the momentum theorem: (m+m)v=mv0,v=mv0 (m+m).

Then the elastic potential energy = initial kinetic energy - frictional work done - final kinetic energy = mv0 2 4-(mv0) 2 (4(m+m)).

1) The momentum of the whole process system is conserved, from which the final velocity of the system mv0=(m+m)v1 can be obtained, and then the work done by friction w,w=mgu*2l,(u kinetic friction factor) can be obtained from the conservation of energy

2) When the elastic potential energy is the largest, the mass point is the same as the trolley velocity, or the momentum is conserved, and the trolley velocity is obtained. Then the initial kinetic energy of the particle is subtracted from the initial kinetic energy of the particle at this time, and the sum of the kinetic energy of the particle and the trolley and mgul is the maximum elastic potential energy.

All rivers and lakes are masters!

Elastic collision, all kinetic energy friction heat, f (friction) l kinetic energy. Desolable friction factor. As for the second question, the same speed of the two cars is equal to the maximum potential energy, if you are a student in the first year of high school, I suggest you ask the math teacher for this question.

By establishing the Cartesian coordinate system of the parallel inclined plane and the perpendicular inclined plane, and decomposing the gravity and thrust force into these two directions, it can be seen that the component F1 along the upward slope of F and the component Ga1 of gravity along the downward inclined plane of A are balanced, and the sum of the components F2 and GA2 of F and the perpendicular inclined plane of GA is equal to the elastic force of the inclined plane A. After putting B up, and treating A and B as a whole, the downward component of gravity is greater than F1, so A and B accelerate their downward decline, and the component G(Ab)2 of the vertical inclined side of gravity becomes larger, so the elastic force of the inclined force on A becomes larger. In addition, B accelerates downward, the acceleration is downward along the inclined plane, there is a vertical downward sub-acceleration, and the resultant force in the vertical direction is downward, so the support force of A to B is less than the gravitational force of B, and the pressure of B on A is less than the gravitational force of B.

B, the whole system is in a static state, indicating that the net force in all directions is zero, the inclined plane is smooth and frictionless, the A object is stationary because of the horizontal to the left force, the magnitude is equal to the component of the A object along the horizontal to the right, if B is added, the equilibrium system is destroyed, and the AB two objects will accelerate the decline.

Hello: I didn't encounter this kind of problem when I was in high school, this is a simple drawing I want to draw, I hope it can help you.

Suppose the clothes are placed in the O1 position, the clothes are subjected to 3 forces at this time, gravity f1, the tension of the left section of the rope to the clothes, the tension of the right section of the rope to the clothes, the rope to the clothes through the tension to the clothes, the tension direction of the left section of the rope to the clothes is perpendicular to FA1, the tension direction of the right section of the rope to the clothes is perpendicular to FB1, the force of the right section of tension decomposition to the horizontal direction is definitely greater than the force of the left section of the tension decomposition to the horizontal direction (or there is no force to decompose the force of the clothes how can the clothes move), The reason is that the lower the position state of the object, the more stable it is. So the clothes moved in the direction of the left.

When the garment reaches point O, (point O is the lowest point in the range of movement), the left tension and the right tension are not the same (when point A and point B are the same horizontally, the tension on both sides is the same), but the left tension and the right tension are the same when they are broken down to the horizontal direction, and the resultant force of the left and right tension in the vertical direction is equal to the gravitational force. Of course, at this time, gravity f, pull fa, and pull fb are balanced. The tension fb and the two-stage tension are mutually embodied.

are balanced with gravity.

Since the rope is smooth, the hanger is also smooth, and the light rope is evenly held in the hand, that is, the pulling force exerted by point A and point B on the hanger is the same, draw a schematic diagram of force decomposition, the smaller the top angle of the isosceles triangle, the larger the waist (the gravity of the clothes and the hanger remains unchanged), it can be seen that when the angle of the rope is larger, the greater the tension.

Unchanged, because the forces on a rope are equal everywhere.

I feel that this topic is a bit of a problem, if you want to move the clothes is to move the clothes to a stationary place again, then the external force must act on the clothes for a period of time, during this time the force change of the rope must be related to the external force, the external force is different, the amount of change of the rope is different, personal understanding, for reference only, I don't know if the expression is right.

The rope is the whole of a system, and the tension is the same.

No change. The tension of the rope is equal everywhere. The resultant force of tension is equal to the constant gravitational force of the garment.

Two cases:1After jumping, the direction of movement of car A is in the same direction as the original, and the speed of car A must not be greater than the speed of car B. Just use the conservation of momentum to do it.

2.After jumping, car A moves in the opposite direction, but due to the existence of the slope, the direction will come again after a period of time, and the speed of car A should also be required to be no greater than the speed of car B.

The result of both cases is the range of speed at which the person jumps.

I won't tell you the specifics.

I probably calculated it, it's annoying, anyway, to ensure that it doesn't hit, car B and people must move in the opposite direction.

Two extremes, one is the other is.

1, Car A is still in the original direction, it should be.

2, Car A reverses and then chases Car B, it should be.

How it comes: The speed at which car A takes people to slide to the plane is 3, and car B is. If car A can't catch up with car B, car B's speed must be more than 3 (the direction is the opposite of that), so the weight of people and car B is the same). Momentum is conserved, not the addition of individual velocities.

Pick B. The hard rod is affected by gravity g, support force n, suspension line tension t, the resultant force of gravity and tension force and the support force and other general directions are opposite, you can draw the force diagram, in the process of moving right, the angle between the line and the vertical direction gradually increases, t times the cosine value = mg. From the angle, it can be seen that the cosine value gradually decreases, so the tensile force gradually increases.

Select: D only remove F2, and the slider B is still moving downward, and the inclined body is subjected to the force of B (pressure, friction The downward motion is mentioned in the question. Assuming that it is moving at a uniform velocity, then the net force of f1f2 is downward to the right, and the net force of other forces is and.

Answer: Zhongliang solution: to make the selling branch wide light rope not be broken, the maximum tensile force of the rope ft = 100N, first take b as the research object, according to Newton's second law, there is ft mbg = mba

Then take the disadvantages A and B as the whole.

In the same way, the equation f (ma+mb) g = (mA+mb) a is solved by f = (mA+mb)(g+a)=12

The guarantee is right, I don't know if it will help you.

1. (10 points) As shown in the figure, the two objects A and B are connected with a light rope with a maximum force of 100N, MA= 4KG, MB=8KG, and under the action of the tensile force F, it is added upwards and moves at a high speed, in order to prevent the light rope from being broken, what is the maximum value of F? (g takes 10m s2).

ft mbg = mba 2 points).

From the solution, f=(ma+mb)(g+a)=12 points) and force analysis diagrams are 2 points each.

The force analysis of the O junction of the rope is affected by the tensile force Ta of OA, the tensile force Tb of OB and the tensile force G of OC, respectively

If the three forces are balanced, then the three forces form a triangle that joins each other.

Analyzing this triangle, we can know that TB=G, TA=SQRT(2)G, since the maximum TB is 5N, the maximum gravity of the object can be known to be 5N, at this time, OA tensile force TA=5sqRT(2) 10N【Handsome Wolf Hunting】The team will answer for you.

fa=mg/cos45° (1)

fb=facos45°=mg (2)

According to (2), the maximum mass mg = FB = 5N

Bring in (1) to verify that FA=<10N

So the maximum gravitational force of the object suspended at the lower end of the OC is 5N

The ratio of oc ob oa pull force is 1:1:sqrt(2), so the maximum gravitational force is 5n