A few questions on the overall method of force analysis in the third year of high school, ask for 10

Holistic method, one dynamic and one static analysis.

The wooden block and the inclined plane are regarded as a whole, the mass of the wooden block ** is m, the inclined plane is m, and the inclination angle of the inclined plane is a, 1It is recommended to use isolation, which is faster (the interaction is only the pressure, and the pressure is divided into levels, and the opposite is the f-direction).

Then there is only one friction force f in the horizontal direction (overall view), the horizontal acceleration of the wooden block is a1=gsina*sina, then in general, there is f=ma1+ma2 (a2 is the horizontal acceleration of the inclined plane), because the inclined plane is stationary, a2=0, so the f direction is the same as the a1 direction.

2.In the same way, the horizontal forces are f and fcosa

Then fcosa + f = ma1 + ma2 (acceleration is horizontal acceleration), because the wooden block is at a constant velocity, the inclined plane is stationary, so, a1 = a2 = 0

So f=-fcosa, the direction is opposite to the fsina direction.

3.In the same way, the horizontal acceleration of the wooden block is a1=(gsina + ugcosa) cosa

Then f=ma1+ma2, the inclined plane is stationary, and a2=0f is opposite to a1.

If you still have any questions or ** improper answers, welcome to hi me.

If you feel good, ** extra points. Severe lack of TT

Holistic approach? This kind of problem is very simple with non-inertial frames.

<>t*cosθ =gf=2g

So fast. t=f type 晌3= Li rents this 3f 3

<>t*cosθ=g

f=2g suspicious.

where cos = Qin cavity chaos 3 2 Yuanchang ( 30)t=f 3=f* 3 3

D solution: a, b total gravity g = 24n. Therefore, the force fx = GSIN 30° = 12n along the inclined plane

Thrust f = 16 N. Since it is now in equilibrium, the frictional force f on the inclined plane of a is the net force of thrust f and fx. and f is perpendicular to fx.

So: f = root number (12 + 16) = 20n

Then we analyze: the friction between a and b.

Because object B is at rest, the magnitude of the frictional force f is equal to the magnitude of the component force of the gravity of object B downward along the inclined plane.

So: f=6sin30°=3n

I didn't understand the phrase horizontal thrust acting on a (perpendicular to the paper face).

I guess you're mistaken.

The whole method is no problem! Compare the two small balls and the semi-cylinder as a whole, at rest. Therefore, the net force in the vertical direction is equal to 0, and the force is not affected in the horizontal direction, and the friction force = 0

Segregation: Treat the two balls as a whole because they are at rest. Therefore, the combined external force of the ball in both the vertical and horizontal directions is equal to 0

Isolate a semi-cylinder. In the horizontal direction, the reaction force is equal to the horizontal force of the two balls=0. So the frictional force is equal to 0

If you look at two balls as one element individually, then each ball is subjected to 3 forces: gravity, the support force of the semicircle, and the pull force of the line. Since the balls can remain motionless, then the net force at the level of each ball is 0, and the overall is also 0

If the combination of two small balls and a line is considered as a whole, then there are also three forces: total gravity, the support force on the left, and the support force on the right. Since the combination can remain motionless, then the resultant force of the combination is also 0 horizontally

You think that if you don't think 0, the ball 1 is affected by t1, g1, n1; Ball 2 is affected by t2, g2, n2. Because t1 = t2, and the angles of t1 and t2 are different, so that their horizontal components are not the same, so the horizontal components of n1 and n2 are also different, so the total horizontal resultant force should not be 0. Thinking about T1 and T2 is actually treating the two small balls as two individuals, and comparing N1 and N2 with them as a combination, the consideration of the object has changed twice, and the result is naturally strange.

In this question, the tensile force t1 and t2 should not be equal, don't be confused by the diagram, the rope and the ball should have a tangent point. If you think that the friction is not zero, of course, if you think that the friction is equal and then calculated, in fact, I have long been fed up with the questions for the exam, and many of the questions have loopholes and are unrealistic, child, remember to take the test when you do the questions, this question is to test your overall method.

This volleyball is subject to two forces of bai, one is.

Gravity, one is the net force of other volleyballs on it, and according to Newton's second law, f = ma

mg)²+ma)²=f²

So the answer is d, the direction of attention, and the schematic diagram of the force is easy to get.

Kid, this is a bit of an inappropriate topic for middle school students. The college entrance examination will not produce such unqualified questions. The rod is bent and deformed, and the force generated by the deformation is called "stress", which is what you understand as "restoring force".

This is not the scope of secondary school studies. This question is very bad.

Key points of force analysis:1Who is being analyzed? 2.Who is the force object?

Who is this question for? Rod or both ends of the rod? The topic says that the two ends of the rod, how to define the two ends? An endpoint is just a point. If it is a point, there is no mass, then there is no gravity.

Stress at the lower end: 1. Stress (The force object is the point on the rod adjacent to this point, and the direction is not to be said.) It's not something you learn) 2. Ground friction.

3. Ground support force. If this state is balanced, i.e. the state can continue forever, then the resultant force is 0In general, in this case, it is impossible to balance, and the resultant force will not be 0 basically

2. Upper endpoint, 1, stress, 2, pressure of the ball.

The gravity of the ball and the support force of the rod.

When the trolley is stationary, it is only subject to the pull and gravity of the rope. When the trolley accelerates to the right evenly, the acceleration of the ball is the same as the acceleration of the trolley, so the ball is subjected to the force generated by the acceleration of the trolley at this time m The trolley is driven horizontally to the right, as well as gravity, rope pull. There is an angle between the rope tension and the vertical direction.

The two ropes attached to the trolley remain unchanged, but the vertical rope has a smaller elastic force and the horizontal rope has a larger elastic force. I hope you understand me. Thank you.