High school force analysis,High school force analysis, Help me put that...

You don't have all the conditions for this question, so you have to discuss it in a variety of situations.

1. The inclined plane is facing upwards to the left.

2. The inclined plane is facing upwards to the right.

3. The inclined plane is neither facing the upper left nor the upper right.

1. This is the simplest situation, at this time, the object must maintain the same acceleration a as the train in order to be stationary relative to the train, that is: a=10 and the direction is pointed to the left and parallel to the ground. Then to get a in this direction, it means that the net force experienced by this object is directed to the left and parallel to the ground.

Then the next step is a vector calculation.

Please pick up the pen. First of all, the bevel gives the object a fixed support force, and it can only point to the left 60 degrees from the ground. Then gravity, the direction and magnitude of gravity are fixed, is a vertical downward g=10.

Subtract the gravity of the resultant force a=10 parallel to the ground to the left and g=10 straight down (i.e., draw a line segment from the end of gravity to the end of the resultant force) to get a 45 degrees above the ground to the left.

That is to say, the vector combination of the acceleration of the support force of the inclined plane to the object and the frictional acceleration of the inclined plane to the object must be the 45 degrees of a =, and because the supporting force must be 60 degrees from the ground, the rest is clear from the diagram - the end of the supporting force is always to the right of the end of this resultant force, so it can be added with a force to the left to get such a resultant force, so the frictional force must be to the left. Then on this inclined plane, to the left also means down.

2. The same method obtains that the coefficient of friction must be very large, and the direction of friction is still to the left, and left on this inclined plane means upward.

3. The same method. However, this time, the vector calculation was performed using the 3D space axis, and the result was up and down. Depends on exactly which way the slope is facing.

The object and the car have the same acceleration to ensure that it is relatively stationary with the car, and the friction of the object provides this force, but in practice, the object on the train must maintain a linear relationship with the direction of the train's motion, and there will be no friction to act as an acceleration to the left.

Because there is such a condition, a is 10 when moving horizontally to the left.

That is, there is acceleration, so the net force is no longer 0 and should be equal to the mass multiplied by the acceleration in the direction of the left. You can draw a picture like this.

In a simple way, draw a diagram, when the train suddenly accelerates to the left, the object will move to the right relative to the inclined plane, that is, the upward movement trend along the inclined plane, then the friction force will naturally be downward along the inclined plane.

The specific calculation is that after the force analysis, you decompose the force other than the friction force pointing to the left of the longitudinal axis, that is, the two or three quadrants, and calculate whether the horizontal to the left resultant force can make the acceleration of the object reach 10, if not, then only the component of the friction force in the horizontal direction can be supplemented to the left, so it is along the inclined downward side, if it is just 10, the friction force does not exist, if it exceeds 10, then the friction force should be along the inclined surface up to offset part of the force to the left.

When the upper end of the lightweight spring is sheared, the instantaneous acceleration of both balls A and B is equal to the acceleration due to gravity.

You should have cut the string short, right? If so, the force is as follows:

Ball B is only subject to gravity, the acceleration is acceleration due to gravity, ball A is pulled by gravity and spring, and the resultant force is upward, the magnitude is equal to the gravitational force of ball B mb g

Therefore, the key to the problem of the acceleration of ball A = mb g ma is to pay attention to: the rope and rod are subjected to external force is sexually transformed, and after the external force is removed, the recovery deformation does not take time;

And springs, rubber bands and other objects with large deformation, after the external force is removed, it takes a period of time to return to the original state, therefore, the moment the string is cut, the rope force disappears, and the spring force remains unchanged.

That's how this question should be thought.

First of all, let's analyze the force before the spring is not sheared.

It can be seen that the spring gives a force to A, and this force pulls up A, B, which is M1G+M2G (M1 is the mass of A, and M2 is the mass of B).

The rope gives b a force of one m2g.

After shearing, the spring elastic force does not change, and the elastic force of the rope becomes zero.

So by f=ma

A1 can be found as m2g m1

a2 is g

Because the spring is "light", when the upper end of the lightweight spring is cut, the spring "immediately" returns to its original shape, i.e. the spring has no elastic force. Therefore, both balls are free falling, the rope has no elastic force, and the instantaneous acceleration of both balls A and B is g.

Three years ago, physics was very good, and I forgot about it, so don't be surprised if it's wrong. Personally, I feel that the lightweight spring here does not mean that the rope is broken, and there is no tension, but that the spring itself has no mass and is not affected by gravity. Let's do the following force analysis:

Before the rope is cut short, the A ball is subjected to its own gravity plus the pull force of the B ball, that is, (MA+MB)g is set as: F1 direction downward, and the spring pulls the A ball downward: F2 direction is up, because it is stationary, so the size of F1 is equal to the ball by its own gravity Set the direction of F3 downward and the pull force of A ball set F4, because it is stationary, the size of the pull force is equal to its own gravity.

When the rope is cut, the upward tension of ball A by the spring disappears, and the other forces remain unchanged, so the AA=(mA+MB) mA is obtained from F=MA, and the direction is downward, and the force on ball B remains unchanged, so the acceleration is 0

The upper ball is A, the lower ball is B, and the acceleration of B is G, and the direction is downward, only by gravity. A ball is subjected to the spring tension f=(ma+mb)g and its own gravity g=mag, so the acceleration is ma mb*g, and the direction is upward.

First of all, the diagonal beam has a mass m, which is subject to its own gravity mg;

Secondly, the diagonal beam is suspended by two ropes, and it is subjected to two tensile forces f1 and f2;

Again, because the object is placed on the inclined beam, the object is supported by the inclined beam, and the inclined beam is subjected to the pressure of the object at the same time, and these two forces are the action force and the reaction force, and the magnitude is the same and the direction is opposite;

Finally, if the object can remain stationary, the resultant force parallel to the inclined beam and perpendicular to the direction of the inclined beam is 0, and the force on the object perpendicular to the direction of the inclined beam is: the supporting force of the inclined beam to the object and the gravity of the object in this direction, the force parallel to the direction of the inclined beam is the component of the gravity of the object, and the direction is downward along the inclined beam, so that the object has a downward trend, but the object is stationary, therefore, it must be subjected to a static friction force along the inclined surface upward, so the inclined beam is subjected to the static friction force along the inclined surface downward at the same time, These two frictional forces are action and reaction forces in turn.

Your statement that the pressure of the object m on the diagonal beam is the component of gravity is wrong, this pressure is for the diagonal beam, and the component of gravity is for the object, and the object of analysis is different.

I hope the above description is helpful to you!

Because the system is in a parallel state, there are two states of the object, one is stationary, and the other is sliding at a uniform speed.

In this problem, it is easy to see that when the object moves in a circular motion around its axis, the centripetal force is provided by the resultant force of the tension of the rope and the gravitational force of the object. Using trigonometry knowledge, we can know that the radius r of the object is 1 2L, so we can use the centripetal force formula f=mv r and substitute the value to know.

1) When v = 1 6gl under the root number, f = 1 3mg, using trigonometric knowledge, it can be known that f pull=f cos30°=(2 3) 9mg

2) When v=3 2gl under the root number, f=3mg, using trigonometric knowledge, we can know that f-pull=f cos30°=2 3mg

Your values may be a little ambiguous, but here's how to solve a problem, using different values and the same formula.

I drew it myself, it's a bit rudimentary, don't mind.

Method: Integral force method.

Let's consider the acceleration of the object to the lower left, the acceleration is a, the angle of the bevel is m, the mass of the block is m, and the block and the inclined plane are a system.

Then according to the analysis of the inertial force, it is equivalent to the inertial force on the object obliquely to the right.

Therefore, the system needs a horizontal force to the left to balance, that is, the frictional force provides.

So the friction between the ground and the inclined plane is to the left.

The size is macOS

The acceleration of the block on the surface multiplied by the angle cosine between the inclined plane and the ground.

If the block moves in a straight line along the inclined plane with acceleration a, it accelerates downward. This acceleration can be broken down into horizontal and vertical components.

There is a horizontal component: a = acosa

Newton's second law is used for the whole: only friction is applied in the horizontal direction, and only the wooden block accelerates along the horizontal, so there is:

f=ma`=macosa

Since it is at rest, the net force is 0 (by external equilibrium force).

For A, the vertical direction is subjected to the vertical downward gravity g = mg = 10N, the hand to A and B to A vertical upward friction force Ta = 5n and TBA = 5N, gravity, friction is the equilibrium force GA = ta TBA, the horizontal direction is subject to the horizontal thrust f = 100N, the A reaction force of B pair FBA = 100N, each other is the balance force f = FBA.

For B, the vertical direction is subject to the vertical downward gravity GB=MG=10N, the vertical upward A to B and the wall to B vertical upward friction Tab=5N and TB=5N, gravity and friction are the equilibrium force, GB=Tab TB, the horizontal direction is subject to the reaction force of A to B Fab=100N (Fab is the FBA force and reaction force to each other) and the wall to B The reaction force FB=100N, each other is the equilibrium force.

Generally, the dynamic friction factor is used in moving objects, and the stationary objects are solved by balancing force, and both can be used for the friction force of objects moving at a constant speed).

In this question, the friction force should understand the process, b is affected by tb and tab, the same dynamic friction factor is the same roughness, then tb=tab= and tab are mutual force and reaction force, then tba=tab=5n, and tba ta=ga=10n, then ta=5n (this question uses a little knowledge of dynamic friction factors).

Well, ab is stationary, so the kinetic friction factor given is redundant and cannot be used.

AB is each subjected to the vertical downward gravity and the vertical upward and gravity equal to the static friction force, and there are two static friction forces, A is subjected to the horizontal F and the reverse B is equal to F to the reaction force of A.

b is subjected to the pressure of a on b f and the reaction force of an equal-sized wall on b.