Newton s Second Law Applications! 100 channels

Pay attention to quality, not quantity, do more classic questions as shown in Figure (1), a mass of M system on two thin lines of length L1 and L2 respectively, one end of L1 is suspended on the ceiling, and the angle between L1 and the vertical direction is , L2 is straightened horizontally, and the object is in a state of equilibrium. Now the L2 line is cut to find the acceleration of the instantaneous object. (1) The following is a certain solution to the problem by a certain student:

Let the tensile force on the L1 line be T1, the tensile force on the L2 line be T2, and the gravity force be mg, and the object is in equilibrium under the action of the three forces. t1cos = mg, t1sin = t2, the solution is t2=mgtan, the moment the line is sheared, t2 suddenly disappears, but the object gains acceleration in the opposite direction of t2, because mgtan = ma, the acceleration a=gtan and the direction is in the opposite direction of t2. Do you think this result is correct?

Explain why. (2) If the thin line L1 in Figure (1) is changed to a light spring with the same length and no weight, as shown in Figure (2), other conditions remain unchanged, and the steps and results of the solution are exactly the same as those in (1), that is, A=GTAN, do you think this result is correct? Please explain why.

Analysis: (1) This result is wrong. When L2 is sheared, the tension on L1 changes abruptly due to the sudden disappearance of T2, which changes the force of the object, and the instantaneous acceleration is oblique downward along the vertical L1, which is A=GSIN

2) This result is correct. When L2 is sheared, T2 suddenly dissipates the disadvantages, and the spring has not had time to deform (there must be a process of change, not mutation), so the elastic force T1 of the spring remains unchanged, and its resultant force with gravity and T2 is a pair of balanced forces, which are equivalent to the opposite, so the instantaneous acceleration of L2 when shear is A=GTAN, and the direction is in the opposite direction of T2. Comments:

Newton's second law f and ma reflect the instantaneous correspondence between the acceleration of an object a and the external force it is subjected to The object is acted on by an external force and at the same time produces a corresponding acceleration, the external force is constant, and the acceleration of the object is also constant; When the external force changes with time, the acceleration also changes with time; At some point, the external force stops acting and its acceleration disappears at the same time.

The tension of the light rope of the 2m wooden block, the pressure and friction of the M wooden block, the support force and friction of the ground, these are 5, A is incorrect.

t m+2m+3m)*(m+2m)= so the light rope will not break, b is incorrect.

Therefore the light rope will not break, c correct.

t m+2m)*m=1 3t is not 2 3t d is incorrect.

A If the inclined plane is smooth, the force of A on the baffle is.

Then the acceleration of all objects in the inclined direction is equal and equal to GSIN, so the force of each object in the direction of the inclined plane is zero, and the spring returns to its natural length.

B If the inclined plane is rough, the pressure of A on the baffle is 0, if the inclined plane is rough, the pressure on the baffle plate of A must not be zero, and the size is: greater than zero, less than 2mg (determined by the acceleration of the object along the direction of the inclined plane).

C If the plank moves at a uniform speed along the inclined plane, the force of A on the baffle is if the plank moves at a uniform speed along the inclined plane, and the force of A on the baffle is: 2mgd If the plank moves uniformly along the inclined plane, the spring must be in a natural elongation state If the plank accelerates along the inclined plane, the spring is not necessarily in a natural elongation state, and the elongated state is determined by the acceleration of the object along the inclined plane. When the acceleration is:

GSIN, the spring is at its natural length, and when the acceleration is less than GSIN: it is in compression.

As a physics teacher, I tell you that if you can't read the questions, no one can help you, you can only rely on your reading comprehension skills. Only when you understand the problem, you will find that physics is very simple, but how to read a problem, except for you to read it over and over again, no one can really help you.

The key is that when your rope breaks, the force in the other rope will change abruptly (the spring will not). Because if it does not change, the acceleration level, then the rope must be loose, there will be no force, then it will be a contradiction. So the change in rope force will cause the object to move perpendicular to the end of the rope to make the rope taut.

After the OB line is burned out, the direction of the tensile force of the OA line changes all the time, and it is a centrifugal movement, and the tensile force provided by the OA line is the centripetal force.

The force on the rope can change instantaneously, when the OB is cut, that is, when the force of the OB is 0, the force of the OA segment will change instantaneously, so to do a single pendulum motion, the acceleration is of course not horizontal to the right.

The resultant force is indeed to the right, but the resultant force should be the resultant force of the centripetal force and another force (let's say it is called f1). What is required here is the acceleration produced by f1. should be the acceleration of the force of the perpendicular centripetal force (i.e., the perpendicular OA)...

It should be understood as such...

Maybe...

When the OB is sheared, i.e., the force of the OB is 0, the force of the OA segment changes instantaneously.

It should be understood that the moment it is burned off, the force acting on the ball O changes. It is only affected by a and gravity, and cannot be analyzed with the original equilibrium state.

At the moment when the rope is burned, the ball should be subjected to the combined force of the pull force and gravity given by the rope OA.

In the vertical direction, there is no motion of the object, let alone acceleration, so the force in the vertical direction is balanced. f*sin(θ)g (1)

In the horizontal direction, the object moves at a uniform speed and the acceleration is zero, so the force in the horizontal direction is also balanced. f*cos(θ)g*μ 2)

2) Divide by (1), cotangent( ) so there is =arc cotangent (

With orthogonal decomposition method: because the object is moving at a uniform speed, the net force is zero.

Horizontal FCOS FN

Vertically fsin +fn g

Get f g (cos + sin ) when the denominator is the largest, f is the smallest. Denominator (1 2) square root *sin( +a).

Where, the angle a satisfies the square root of sina 1 (1 2) and cosa 1 2) square root.

Obviously, when sin(+a) 1, f is minimal, then tana=1 and arctan1 = arctan at 90 degrees